Q3: Justify that the following reactions are redox reactions: (a) CuO(s)+H_2(~g) → Cu(s)+H_2 O(~g) (b) Fe_2 O_3(s)+3 CO(~g) → 2 Fe(s)+3 CO_2(~g) (c) 4…

Justification: A reaction is a redox reaction if the oxidation numbers of one or more elements change during the course of the reaction. (a) CuO(s)+H_2(~g) → Cu(s)+H_2 O(~g) Assigning oxidation numbers: stackrel+2Custackrel-2O + stackrel0H_2 → stackrel0Cu + stackrel+1H_2stackrel-2O Changes: The…

Q5: Calculate the oxidation number of sulphur, chromium and nitrogen in H_2 SO_5, Cr_2 O_7^2- and NO_3^-. Suggest structure of these compounds. Count for…

1. H_2 SO_5 (Peroxymonosulfuric acid or Caro's acid) Calculation using standard rules (leading to fallacy): Let the oxidation number of S be x. H is +1, O is -2. 2(+1) + x + 5(-2) = 0 2 + x - 10 = 0 implies x = +8. Fallacy: The maximum oxidation state for sulfur (Group 16) is +6. An oxidation state…

Q14: Consider the reactions: 2 S_2 O_3^2-(aq)+I_2(~s) → S_4 O_6^2-(aq)+2 I^-(aq) S_2 O_3^2-(aq)+2 Br_2(l)+5 H_2 O(l) → 2 SO_4^2-(aq)+4 Br^-(aq)+10 H^+(aq)…

The reason for the different reaction products is the difference in the oxidizing strength of iodine and bromine. Oxidizing Strength: Halogens are oxidizing agents, and their strength decreases down the group. Therefore, bromine ( Br_2 ) is a significantly stronger oxidizing agent than iodine ( I_2…

Q16: Why does the following reaction occur? XeO_6 ^4-(aq)+2 F^-(aq)+6 H^+(aq) → XeO_3(~g)+F_2(~g)+3 H_2 O(l) What conclusion about the compound Na_4 XeO_6…

Analysis of the Reaction: Let's first identify the oxidation and reduction processes by assigning oxidation numbers: stackrel+8XeO_6 ^4- + 2 stackrel-1F^- + 6 H^+ → stackrel+6XeO_3 + stackrel0F_2 + 3 H_2 O Oxidation: The fluoride ion ( F^- ) is oxidized from an oxidation state of -1 to 0 in…

Q17: Consider the reactions: (a) H_3 PO_2(aq)+4 AgNO_3(aq)+2 H_2 O(l) → H_3 PO_4(aq)+4 Ag(s)+4 HNO_3(aq) (b) H_3 PO_2(aq)+2 CuSO_4(aq)+2 H_2 O(l) → H_3…

Analysis of the Reactions: Let's analyze the oxidizing and reducing agents in each reaction. Reactions (a) and (b): Hypophosphorous acid ( H_3 PO_2 ) is acting as a reducing agent, getting oxidized to phosphoric acid ( H_3 PO_4 ). Both Ag^+ (from AgNO_3 ) and Cu^2+ (from CuSO_4 ) are strong enough…

Q18: Balance the following redox reactions by ion - electron method: (a) MnO_4^-(aq)+I^-(aq) → MnO_2(~s)+I_2(~s) (in basic medium) (b) MnO_4^-(aq)…

(a) MnO_4^-(aq)+I^-(aq) → MnO_2(~s)+I_2(~s) (in basic medium) Step 1 & 2: Half-reactions Oxidation: 2I^- → I_2 Reduction: MnO_4^- → MnO_2 Step 3 & 4: Balance atoms and O/H (basic medium) Oxidation: (Already balanced for atoms) Reduction: MnO_4^- + 2H_2O → MnO_2 + 4OH^- Step 5: Balance charge with…

Q20: What sorts of informations can you draw from the following reaction? (CN)_2(~g)+2 OH^-(aq) → CN^-(aq)+CNO^-(aq)+H_2 O(l)

From the given reaction, we can draw the following information: 1. It is a Redox Reaction: We can confirm this by assigning oxidation numbers. In cyanogen, (CN)_2, nitrogen is more electronegative than carbon, so N is -3. For the molecule to be neutral, C must be +3. In the cyanide ion, CN^-, N is…

Redox Reactions Class 11 Chemistry NCERT Solutions

Q1EXERCISES

Assign oxidation number to the underlined elements in each of the following species:

(a)

\mathrm{NaH}_{2} \underline{\mathrm{P}} \mathrm{O}_{4}

(b)

\mathrm{NaH} \underline{\mathrm{S}} \mathrm{O}_{4}

(c)

\mathrm{H}_{4} \underline{\mathrm{P}}_{2} \mathrm{O}_{7}

(d)

\mathrm{K}_{2} \underline{\mathrm{Mn}} \mathrm{O}_{4}

(e)

\mathrm{Ca} \underline{\mathrm{O}}_{2}

(f)

\mathrm{Na} \underline{\mathrm{B}} \mathrm{H}_{4}

(g)

\mathrm{H}_{2} \underline{\mathrm{S}}_{2} \mathrm{O}_{7}

(h)

\mathrm{KAl}(\underline{\mathrm{S}} \mathrm{O}_{4})_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O}

Solution

Rules for Assigning Oxidation Numbers:

O is usually -2 (except in peroxides, superoxides).
H is usually +1 (except in metal hydrides).
Alkali metals (like Na, K) are +1.
Alkaline earth metals (like Ca) are +2.
The sum of oxidation numbers in a neutral compound is 0.
The sum of oxidation numbers in a polyatomic ion equals the ion's charge.

Let the oxidation number of the underlined element be

x

.

(a) $\mathrm{NaH}_{2} \underline{\mathrm{P}} \mathrm{O}_{4}$

(+1) + 2(+1) + x + 4(-2) = 0

1 + 2 + x - 8 = 0

x - 5 = 0 \implies x = +5

Oxidation number of P is +5.

(b) $\mathrm{NaH} \underline{\mathrm{S}} \mathrm{O}_{4}$

(+1) + (+1) + x + 4(-2) = 0

2 + x - 8 = 0

x - 6 = 0 \implies x = +6

Oxidation number of S is +6.

(c) $\mathrm{H}_{4} \underline{\mathrm{P}}_{2} \mathrm{O}_{7}$

4(+1) + 2x + 7(-2) = 0

4 + 2x - 14 = 0

2x - 10 = 0 \implies 2x = 10 \implies x = +5

Oxidation number of P is +5.

(d) $\mathrm{K}_{2} \underline{\mathrm{Mn}} \mathrm{O}_{4}$

2(+1) + x + 4(-2) = 0

2 + x - 8 = 0

x - 6 = 0 \implies x = +6

Oxidation number of Mn is +6.

(e) $\mathrm{Ca} \underline{\mathrm{O}}_{2}$ This is calcium peroxide. Ca is an alkaline earth metal, so its oxidation number is +2. The compound contains the peroxide ion (

\mathrm{O}_{2}^{2-}

).

(+2) + 2x = 0

2x = -2 \implies x = -1

Oxidation number of O is -1.

(f) $\mathrm{Na} \underline{\mathrm{B}} \mathrm{H}_{4}$ This is sodium borohydride. H is bonded to a less electronegative element (B), so H has an oxidation number of -1. Na is +1.

(+1) + x + 4(-1) = 0

1 + x - 4 = 0

x - 3 = 0 \implies x = +3

Oxidation number of B is +3.

(g) $\mathrm{H}_{2} \underline{\mathrm{S}}_{2} \mathrm{O}_{7}$

2(+1) + 2x + 7(-2) = 0

2 + 2x - 14 = 0

2x - 12 = 0 \implies 2x = 12 \implies x = +6

Oxidation number of S is +6.

(h) $\mathrm{KAl}(\underline{\mathrm{S}} \mathrm{O}_{4})_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O}$ Water of hydration is neutral and can be ignored for calculation. We consider the KAl(SO₄)₂ part. K is +1, Al is +3. The sulphate ion is

\mathrm{SO}_{4}^{2-}

. Let's calculate for S within the sulphate ion:

x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6

. Alternatively, for the whole compound:

(+1) + (+3) + 2[x + 4(-2)] = 0

4 + 2(x - 8) = 0

4 + 2x - 16 = 0

2x - 12 = 0 \implies x = +6

Oxidation number of S is +6.

Q2EXERCISES

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?

(a)

\mathrm{K}\underline{\mathrm{I}}_{3}

(b)

\mathrm{H}_{2}\underline{\mathrm{S}}_{4}\mathrm{O}_{6}

(c)

\underline{\mathrm{Fe}}_{3}\mathrm{O}_{4}

(d)

\underline{\mathrm{C}}\mathrm{H}_{3}\underline{\mathrm{C}}\mathrm{H}_{2}\mathrm{OH}

(e)

\underline{\mathrm{C}}\mathrm{H}_{3}\underline{\mathrm{C}}\mathrm{OOH}

Solution

(a) $\mathrm{K}\underline{\mathrm{I}}_{3}$

Calculation of Average Oxidation Number: K is an alkali metal, so its oxidation number is +1. Let the oxidation number of I be $x$ . $(+1) + 3x = 0 \implies 3x = -1 \implies x = -1/3$ .
Rationalisation: The fractional oxidation number is an average. $\mathrm{KI}_{3}$ is composed of a $\mathrm{K}^{+}$ ion and an $\mathrm{I}_{3}^{-}$ ion. In the triiodide ion, $\mathrm{I}_{3}^{-}$ , one iodine atom has an oxidation state of -1, and the other two iodine atoms (forming a diatomic molecule $\mathrm{I}_{2}$ ) are bonded to it. The $\mathrm{I}_{2}$ molecule has an oxidation state of 0 for its atoms. So, the actual oxidation states of the three iodine atoms are -1, 0, and 0. The average is $\frac{-1+0+0}{3} = -1/3$ .

(b) $\mathrm{H}_{2}\underline{\mathrm{S}}_{4}\mathrm{O}_{6}$

Calculation of Average Oxidation Number: H is +1, O is -2. Let the oxidation number of S be $x$ . $2(+1) + 4x + 6(-2) = 0 \implies 2 + 4x - 12 = 0 \implies 4x = 10 \implies x = +2.5$ .
Rationalisation: This compound contains the tetrathionate ion, $\mathrm{S}_{4}\mathrm{O}_{6}^{2-}$ . The structure is $^{-}O_{3}S-S-S-SO_{3}^{-}$ . The two terminal sulfur atoms are bonded to three oxygen atoms and one sulfur atom, giving them an oxidation state of +5 each. The two middle sulfur atoms are bonded only to other sulfur atoms, so their oxidation state is 0. The actual oxidation states are +5, 0, 0, +5. The average is $\frac{5+0+0+5}{4} = \frac{10}{4} = +2.5$ .

(c) $\underline{\mathrm{Fe}}_{3}\mathrm{O}_{4}$

Calculation of Average Oxidation Number: O is -2. Let the oxidation number of Fe be $x$ . $3x + 4(-2) = 0 \implies 3x = 8 \implies x = +8/3$ .
Rationalisation: $\mathrm{Fe}_{3}\mathrm{O}_{4}$ (magnetite) is a mixed oxide, a combination of iron(II) oxide (FeO) and iron(III) oxide ( $\mathrm{Fe}_{2}\mathrm{O}_{3}$ ). It can be written as $\mathrm{FeO} \cdot \mathrm{Fe}_{2}\mathrm{O}_{3}$ . It contains one $\mathrm{Fe}^{2+}$ ion (oxidation state +2) and two $\mathrm{Fe}^{3+}$ ions (oxidation state +3). The average oxidation state is $\frac{(+2) + 2(+3)}{3} = \frac{8}{3}$ .

(d) $\underline{\mathrm{C}}\mathrm{H}_{3}\underline{\mathrm{C}}\mathrm{H}_{2}\mathrm{OH}$ (Ethanol)

Rationalisation: Oxidation numbers must be determined for each carbon atom individually based on the atoms bonded to it. Let the oxidation number of the first carbon (in $\mathrm{CH}_{3}$ $CH_{3}$ ) be $x_1$ $x_{1}$ and the second carbon (in $\mathrm{CH}_{2}OH$ $CH_{2} O H$ ) be $x_2$ $x_{2}$ .
- For the $\mathrm{CH}_{3}$ group: This carbon is bonded to 3 H atoms and 1 C atom. The C-C bond does not contribute to the oxidation number. So, $x_1 + 3(+1) = 0 \implies x_1 = -3$ .
- For the $\mathrm{CH}_{2}OH$ group: This carbon is bonded to 2 H atoms, 1 O atom, and 1 C atom. So, $x_2 + 2(+1) + (-2) + (+1) = 0 \implies x_2 - 1 = 0 \implies x_2 = -1$ . (The final +1 is from the H in OH). The oxidation numbers of the two carbons are -3 and -1.

(e) $\underline{\mathrm{C}}\mathrm{H}_{3}\underline{\mathrm{C}}\mathrm{OOH}$ (Acetic Acid)

Rationalisation: We determine the oxidation number for each carbon separately. Let the oxidation number of the methyl carbon ( $\mathrm{CH}_{3}$ $CH_{3}$ ) be $x_1$ $x_{1}$ and the carboxyl carbon (COOH) be $x_2$ $x_{2}$ .
- For the $\mathrm{CH}_{3}$ group: This carbon is bonded to 3 H atoms and 1 C atom. So, $x_1 + 3(+1) = 0 \implies x_1 = -3$ .
- For the COOH group: This carbon is bonded to two O atoms (one double bond, one single) and one C atom. So, $x_2 + (-2) + (-2) + (+1) = 0 \implies x_2 - 3 = 0 \implies x_2 = +3$ . (The two O atoms contribute -2 each, and the H contributes +1). The oxidation numbers of the two carbons are -3 and +3.

Q3EXERCISES

Justify that the following reactions are redox reactions:

(a)

\mathrm{CuO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g})

(b)

\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g})

(c)

4 \mathrm{BCl}_{3}(\mathrm{~g})+3 \mathrm{LiAlH}_{4}(\mathrm{~s}) \rightarrow 2 \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{LiCl}(\mathrm{s})+3 \mathrm{AlCl}_{3}(\mathrm{~s})

(d)

2 \mathrm{~K}(\mathrm{~s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(\mathrm{s})

(e)

4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})

Solution

Justification: A reaction is a redox reaction if the oxidation numbers of one or more elements change during the course of the reaction.

(a) $\mathrm{CuO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g})$

Assigning oxidation numbers: $\stackrel{+2}{\mathrm{Cu}}\stackrel{-2}{\mathrm{O}} + \stackrel{0}{\mathrm{H}}_{2} \rightarrow \stackrel{0}{\mathrm{Cu}} + \stackrel{+1}{\mathrm{H}}_{2}\stackrel{-2}{\mathrm{O}}$
Changes: The oxidation number of Cu decreases from +2 to 0 (reduction). The oxidation number of H increases from 0 to +1 (oxidation).
Conclusion: It is a redox reaction.

(b) $\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g})$

Assigning oxidation numbers: $\stackrel{+3}{\mathrm{Fe}}_{2}\stackrel{-2}{\mathrm{O}}_{3} + 3\stackrel{+2}{\mathrm{C}}\stackrel{-2}{\mathrm{O}} \rightarrow 2\stackrel{0}{\mathrm{Fe}} + 3\stackrel{+4}{\mathrm{C}}\stackrel{-2}{\mathrm{O}}_{2}$
Changes: The oxidation number of Fe decreases from +3 to 0 (reduction). The oxidation number of C increases from +2 to +4 (oxidation).
Conclusion: It is a redox reaction.

(c) $4 \mathrm{BCl}_{3}(\mathrm{~g})+3 \mathrm{LiAlH}_{4}(\mathrm{~s}) \rightarrow 2 \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{LiCl}(\mathrm{s})+3 \mathrm{AlCl}_{3}(\mathrm{~s})$

Assigning oxidation numbers: $4\stackrel{+3}{\mathrm{B}}\stackrel{-1}{\mathrm{Cl}}_{3} + 3\stackrel{+1}{\mathrm{Li}}\stackrel{+3}{\mathrm{Al}}\stackrel{-1}{\mathrm{H}}_{4} \rightarrow 2\stackrel{-3}{\mathrm{B}}_{2}\stackrel{+1}{\mathrm{H}}_{6} + 3\stackrel{+1}{\mathrm{Li}}\stackrel{-1}{\mathrm{Cl}} + 3\stackrel{+3}{\mathrm{Al}}\stackrel{-1}{\mathrm{Cl}}_{3}$
Changes: The oxidation number of B decreases from +3 to -3 (reduction). The oxidation number of H increases from -1 to +1 (oxidation).
Conclusion: It is a redox reaction.

(d) $2 \mathrm{~K}(\mathrm{~s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(\mathrm{s})$

Assigning oxidation numbers: $2\stackrel{0}{\mathrm{K}} + \stackrel{0}{\mathrm{F}}_{2} \rightarrow 2\stackrel{+1}{\mathrm{K}}^{+}\stackrel{-1}{\mathrm{F}}^{-}$
Changes: The oxidation number of K increases from 0 to +1 (oxidation). The oxidation number of F decreases from 0 to -1 (reduction).
Conclusion: It is a redox reaction.

(e) $4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})$

Assigning oxidation numbers: $4\stackrel{-3}{\mathrm{N}}\stackrel{+1}{\mathrm{H}}_{3} + 5\stackrel{0}{\mathrm{O}}_{2} \rightarrow 4\stackrel{+2}{\mathrm{N}}\stackrel{-2}{\mathrm{O}} + 6\stackrel{+1}{\mathrm{H}}_{2}\stackrel{-2}{\mathrm{O}}$
Changes: The oxidation number of N increases from -3 to +2 (oxidation). The oxidation number of O decreases from 0 to -2 (reduction).
Conclusion: It is a redox reaction.

Q4EXERCISES

Fluorine reacts with ice and results in the change: $\mathrm{H}_{2} \mathrm{O}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{~g})+\mathrm{HOF}(\mathrm{~g})$ Justify that this reaction is a redox reaction.

Solution

Given Reaction:

\mathrm{H}_{2} \mathrm{O}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{~g})+\mathrm{HOF}(\mathrm{~g})

Justification: To determine if this is a redox reaction, we need to assign oxidation numbers to all elements in the reactants and products and check for any changes.

Rules:

Oxidation number of F is always -1 in its compounds.
Oxidation number of H is +1 when bonded to non-metals.
Oxidation number of an element in its free state (like $\mathrm{F}_{2}$ ) is 0.

Assigning Oxidation Numbers:

Reactants:
- In $\mathrm{H}_{2} \mathrm{O}$ : H is +1, O is -2.
- In $\mathrm{F}_{2}$ : F is 0.
Products:
- In $\mathrm{HF}$ : H is +1, F is -1.
- In $\mathrm{HOF}$ : H is +1, F is -1. Let the oxidation number of O be $x$ . Then, $(+1) + x + (-1) = 0 \implies x = 0$ .

The reaction with oxidation numbers is:

\stackrel{+1}{\mathrm{H}}_{2}\stackrel{-2}{\mathrm{O}} + \stackrel{0}{\mathrm{F}}_{2} \rightarrow \stackrel{+1}{\mathrm{H}}\stackrel{-1}{\mathrm{F}} + \stackrel{+1}{\mathrm{H}}\stackrel{0}{\mathrm{O}}\stackrel{-1}{\mathrm{F}}

Identifying Changes:

Fluorine (F): The oxidation number of F decreases from 0 in $\mathrm{F}_{2}$ to -1 in both HF and HOF. This is reduction.
Oxygen (O): The oxidation number of O increases from -2 in $\mathrm{H}_{2} \mathrm{O}$ to 0 in HOF. This is oxidation.

Conclusion: Since the oxidation number of oxygen increases and the oxidation number of fluorine decreases, both oxidation and reduction are occurring simultaneously. Therefore, the reaction is a redox reaction.

It is interesting to note that this reaction appears to be a disproportionation of fluorine (0 to -1 and +1), but this is not the case because fluorine, being the most electronegative element, cannot have a positive oxidation state. The element being oxidized is oxygen.

Q5EXERCISES

Calculate the oxidation number of sulphur, chromium and nitrogen in $\mathrm{H}_{2} \mathrm{SO}_{5}, \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ and $\mathrm{NO}_{3}^{-}$ . Suggest structure of these compounds. Count for the fallacy.

Solution

1. $\mathrm{H}_{2} \mathrm{SO}_{5}$ (Peroxymonosulfuric acid or Caro's acid)

Calculation using standard rules (leading to fallacy): Let the oxidation number of S be $x$ . H is +1, O is -2. $2(+1) + x + 5(-2) = 0$ $2 + x - 10 = 0 \implies x = +8$ .
Fallacy: The maximum oxidation state for sulfur (Group 16) is +6. An oxidation state of +8 is not possible.
Reason and Structure: The fallacy arises because $\mathrm{H}_{2} \mathrm{SO}_{5}$ contains a peroxide linkage (O-O). In a peroxide linkage, each oxygen atom has an oxidation number of -1. The structure is: H-O-S(=O)₂-O-O-H
Correct Calculation from Structure: There are three oxygen atoms with O.N. = -2, two oxygen atoms in the peroxide link with O.N. = -1, and two hydrogen atoms with O.N. = +1. $2(+1) + x + 3(-2) + 2(-1) = 0$ $2 + x - 6 - 2 = 0$ $x - 6 = 0 \implies x = +6$ . The correct oxidation number of S in $\mathrm{H}_{2} \mathrm{SO}_{5}$ is +6.

2. $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ (Dichromate ion)

Calculation: Let the oxidation number of Cr be $x$ . O is -2. The total charge is -2. $2x + 7(-2) = -2$ $2x - 14 = -2$ $2x = 12 \implies x = +6$ .
Fallacy: There is no fallacy here. The calculated oxidation state of +6 is the highest common oxidation state for chromium and is correct.
Structure: The structure consists of two tetrahedral $\mathrm{CrO}_{4}$ units sharing one oxygen atom. [O₃Cr-O-CrO₃]²⁻ The oxidation number of Cr in $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ is +6.

3. $\mathrm{NO}_{3}^{-}$ (Nitrate ion)

Calculation: Let the oxidation number of N be $x$ . O is -2. The total charge is -1. $x + 3(-2) = -1$ $x - 6 = -1 \implies x = +5$ .
Fallacy: There is no fallacy. The calculated oxidation state of +5 is the highest possible oxidation state for nitrogen (Group 15) and is correct.
Structure: The structure is a trigonal planar arrangement with resonance, where the nitrogen atom is at the center. [N(=O)(O⁻)₂] ↔ forms The oxidation number of N in $\mathrm{NO}_{3}^{-}$ is +5.

Q6EXERCISES

Write formulas for the following compounds:

(a)

Mercury(II) chloride

(b)

Nickel(II) sulphate

(c)

Tin(IV) oxide

(d)

Thallium(I) sulphate

(e) Iron(III) sulphate

(f) Chromium(III) oxide

Solution

Concept: The Roman numeral in the Stock notation indicates the charge (oxidation state) of the metal cation. The formula is written by balancing the positive charge of the cation(s) with the negative charge of the anion(s).

Anions and their charges:

Chloride: $\mathrm{Cl}^{-}$ (charge -1)
Sulphate: $\mathrm{SO}_{4}^{2-}$ (charge -2)
Oxide: $\mathrm{O}^{2-}$ (charge -2)

(a) Mercury(II) chloride

Cation: Mercury(II) $\implies \mathrm{Hg}^{2+}$
Anion: Chloride $\implies \mathrm{Cl}^{-}$
Balancing charges: One $\mathrm{Hg}^{2+}$ needs two $\mathrm{Cl}^{-}$ ions. $(+2) + 2(-1) = 0$ .
Formula: $\mathrm{HgCl}_{2}$

(b) Nickel(II) sulphate

Cation: Nickel(II) $\implies \mathrm{Ni}^{2+}$
Anion: Sulphate $\implies \mathrm{SO}_{4}^{2-}$
Balancing charges: The charges (+2 and -2) are already balanced.
Formula: $\mathrm{NiSO}_{4}$

(c) Tin(IV) oxide

Cation: Tin(IV) $\implies \mathrm{Sn}^{4+}$
Anion: Oxide $\implies \mathrm{O}^{2-}$
Balancing charges: One $\mathrm{Sn}^{4+}$ needs two $\mathrm{O}^{2-}$ ions. $(+4) + 2(-2) = 0$ .
Formula: $\mathrm{SnO}_{2}$

(d) Thallium(I) sulphate

Cation: Thallium(I) $\implies \mathrm{Tl}^{+}$
Anion: Sulphate $\implies \mathrm{SO}_{4}^{2-}$
Balancing charges: Two $\mathrm{Tl}^{+}$ ions are needed for one $\mathrm{SO}_{4}^{2-}$ ion. $2(+1) + (-2) = 0$ .
Formula: $\mathrm{Tl}_{2}SO_{4}$

(e) Iron(III) sulphate

Cation: Iron(III) $\implies \mathrm{Fe}^{3+}$
Anion: Sulphate $\implies \mathrm{SO}_{4}^{2-}$
Balancing charges: The least common multiple of 3 and 2 is 6. We need two $\mathrm{Fe}^{3+}$ ions ( $2 \times +3 = +6$ ) and three $\mathrm{SO}_{4}^{2-}$ ions ( $3 \times -2 = -6$ ).
Formula: $\mathrm{Fe}_{2}(SO_{4})_{3}$

(f) Chromium(III) oxide

Cation: Chromium(III) $\implies \mathrm{Cr}^{3+}$
Anion: Oxide $\implies \mathrm{O}^{2-}$
Balancing charges: The least common multiple of 3 and 2 is 6. We need two $\mathrm{Cr}^{3+}$ ions ( $2 \times +3 = +6$ ) and three $\mathrm{O}^{2-}$ ions ( $3 \times -2 = -6$ ).
Formula: $\mathrm{Cr}_{2}O_{3}$

Q7EXERCISES

Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

Solution

Carbon (Oxidation States from -4 to +4)

-4: Methane ( $\mathrm{CH}_{4}$ )
-3: Ethane ( $\mathrm{C}_{2}\mathrm{H}_{6}$ )
-2: Ethyl chloride ( $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{Cl}$ ), Ethene ( $\mathrm{C}_{2}\mathrm{H}_{4}$ )
-1: Ethyne ( $\mathrm{C}_{2}\mathrm{H}_{2}$ ), Acetaldehyde ( $\mathrm{CH}_{3}\mathrm{CHO}$ )
0: Dichloromethane ( $\mathrm{CH}_{2}\mathrm{Cl}_{2}$ ), Formaldehyde (HCHO)
+1: (Not common in simple compounds)
+2: Carbon monoxide (CO), Chloroform ( $\mathrm{CHCl}_{3}$ )
+3: Oxalic acid ( $\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}$ )
+4: Carbon dioxide ( $\mathrm{CO}_{2}$ ), Carbon tetrachloride ( $\mathrm{CCl}_{4}$ ), Carbonate ion ( $\mathrm{CO}_{3}^{2-}$ )

Nitrogen (Oxidation States from -3 to +5)

-3: Ammonia ( $\mathrm{NH}_{3}$ ), Ammonium ion ( $\mathrm{NH}_{4}^{+}$ )
-2: Hydrazine ( $\mathrm{N}_{2}\mathrm{H}_{4}$ )
-1: Hydroxylamine ( $\mathrm{NH}_{2}\mathrm{OH}$ )
0: Nitrogen gas ( $\mathrm{N}_{2}$ )
+1: Nitrous oxide ( $\mathrm{N}_{2}\mathrm{O}$ )
+2: Nitric oxide (NO)
+3: Nitrogen trioxide ( $\mathrm{N}_{2}\mathrm{O}_{3}$ ), Nitrous acid ( $\mathrm{HNO}_{2}$ ), Nitrite ion ( $\mathrm{NO}_{2}^{-}$ )
+4: Nitrogen dioxide ( $\mathrm{NO}_{2}$ ), Dinitrogen tetroxide ( $\mathrm{N}_{2}\mathrm{O}_{4}$ )
+5: Dinitrogen pentoxide ( $\mathrm{N}_{2}\mathrm{O}_{5}$ ), Nitric acid ( $\mathrm{HNO}_{3}$ ), Nitrate ion ( $\mathrm{NO}_{3}^{-}$ )

Q8EXERCISES

While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?

Solution

The ability of a substance to act as an oxidizing agent, a reducing agent, or both depends on the oxidation state of its central element relative to its possible range of oxidation states.

1. Sulphur Dioxide ( $\mathrm{SO}_{2}$ )

The oxidation state of sulfur in $\mathrm{SO}_{2}$ is +4.
The range of oxidation states for sulfur is from -2 (e.g., in $\mathrm{H}_{2}S$ ) to +6 (e.g., in $\mathrm{SO}_{3}, \mathrm{H}_{2}SO_{4}$ ).
Since +4 is an intermediate oxidation state, sulfur in $\mathrm{SO}_{2}$ $SO_{2}$ can be:
- Oxidized to +6 (acting as a reducing agent). Example: $2\mathrm{SO}_{2} + \mathrm{O}_{2} \rightarrow 2\mathrm{SO}_{3}$
- Reduced to 0 or -2 (acting as an oxidizing agent). Example: $\mathrm{SO}_{2} + 2\mathrm{H}_{2}S \rightarrow 3\mathrm{S} + 2\mathrm{H}_{2}O$

2. Hydrogen Peroxide ( $\mathrm{H}_{2} \mathrm{O}_{2}$ )

The oxidation state of oxygen in $\mathrm{H}_{2} \mathrm{O}_{2}$ is -1 (peroxide).
The range of oxidation states for oxygen is from -2 (in most oxides, water) to 0 (in $\mathrm{O}_{2}$ ) and positive values with fluorine.
Since -1 is an intermediate state, oxygen in $\mathrm{H}_{2} \mathrm{O}_{2}$ $H_{2} O_{2}$ can be:
- Oxidized to 0 (acting as a reducing agent). Example: $2\mathrm{KMnO}_{4} + 3\mathrm{H}_{2}SO_{4} + 5\mathrm{H}_{2}O_{2} \rightarrow \mathrm{K}_{2}SO_{4} + 2\mathrm{MnSO}_{4} + 8\mathrm{H}_{2}O + 5\mathrm{O}_{2}$
- Reduced to -2 (acting as an oxidizing agent). Example: $2\mathrm{Fe}^{2+} + \mathrm{H}_{2}O_{2} + 2\mathrm{H}^{+} \rightarrow 2\mathrm{Fe}^{3+} + 2\mathrm{H}_{2}O$

3. Ozone ( $\mathrm{O}_{3}$ )

The oxidation state of oxygen in ozone is 0.
Ozone is a very unstable allotrope of oxygen and readily decomposes to the more stable dioxygen ( $\mathrm{O}_{2}$ ), releasing a nascent oxygen atom. $\mathrm{O}_{3} \rightarrow \mathrm{O}_{2} + [\mathrm{O}]$
This process involves gaining electrons, causing the oxygen to be reduced to the -2 state in compounds. It has a very high tendency to accept electrons and get reduced. It cannot be easily oxidized further. Therefore, ozone acts only as a powerful oxidizing agent.

4. Nitric Acid ( $\mathrm{HNO}_{3}$ )

The oxidation state of nitrogen in $\mathrm{HNO}_{3}$ is +5.
The range of oxidation states for nitrogen is from -3 (in $\mathrm{NH}_{3}$ ) to +5.
Since +5 is the highest possible oxidation state for nitrogen, it cannot be oxidized further. It can only be reduced to lower oxidation states (e.g., +4, +2, 0, -3).
Therefore, nitric acid acts only as an oxidizing agent.

Q9EXERCISES

Consider the reactions:

(a)

6 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{~g})

(b)

\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{O}_{2}(\mathrm{~g})

Why it is more appropriate to write these reactions as :

(a)

6 \mathrm{CO}_{2}(\mathrm{~g})+12 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+6 \mathrm{O}_{2}(\mathrm{g})

(b)

\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Solution

Reason for More Appropriate Representation:

The more detailed representations of the reactions provide insight into the reaction mechanism and the origin of the product molecules, which the simpler, net equations obscure.

(a) Photosynthesis Reaction:

Simple equation: $6 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{~g})$
More appropriate equation: $6 \mathrm{CO}_{2}(\mathrm{~g})+12 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+6 \mathrm{O}_{2}(\mathrm{~g})$
Justification: The first equation incorrectly suggests that 6 oxygen molecules (12 atoms) are formed from 6 water molecules (6 oxygen atoms). The second, more accurate equation, reflects the known mechanism of photosynthesis. It shows that all 12 oxygen atoms in the evolved $6\mathrm{O}_{2}$ molecules come from the 12 reactant water molecules. The oxygen atoms from $\mathrm{CO}_{2}$ are incorporated into the glucose ( $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$ ) and the 6 water molecules produced.

(b) Ozone and Hydrogen Peroxide Reaction:

Simple equation: $\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{O}_{2}(\mathrm{~g})$
More appropriate equation: $\mathrm{O}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$
Justification: The mechanism of this reaction involves the transfer of an oxygen atom from ozone to hydrogen peroxide, which then decomposes. Writing the product as $\mathrm{O}_{2} + \mathrm{O}_{2}$ helps to trace the origin of the oxygen atoms. One molecule of $\mathrm{O}_{2}$ comes from the ozone, and the other molecule of $\mathrm{O}_{2}$ comes from the hydrogen peroxide. The simple equation $2\mathrm{O}_{2}$ masks this mechanistic detail.

Technique to Investigate the Reaction Path:

The most effective technique to investigate the path of these reactions and confirm the origin of atoms is isotopic labeling (or tracer technique).

For reaction (a): One can perform the experiment using water containing a heavy isotope of oxygen, $\mathrm{H}_{2}{}^{18}\mathrm{O}$ , and carbon dioxide with the normal isotope, $\mathrm{C}^{16}\mathrm{O}_{2}$ . $6 \mathrm{C}^{16}\mathrm{O}_{2} + 12 \mathrm{H}_{2}{}^{18}\mathrm{O} \rightarrow \mathrm{C}_{6} \mathrm{H}_{12}{}^{16}\mathrm{O}_{6} + 6 \mathrm{H}_{2}{}^{16}\mathrm{O} + 6 {}^{18}\mathrm{O}_{2}$ By analyzing the products using a mass spectrometer, it is found that the evolved oxygen gas is exclusively ${}^{18}\mathrm{O}_{2}$ , proving that it comes from water and not from carbon dioxide.
For reaction (b): Similarly, one could use hydrogen peroxide labeled with the heavy isotope, $\mathrm{H}_{2}{}^{18}\mathrm{O}_{2}$ , and normal ozone, ${}^{16}\mathrm{O}_{3}$ . ${}^{16}\mathrm{O}_{3} + \mathrm{H}_{2}{}^{18}\mathrm{O}_{2} \rightarrow \mathrm{H}_{2}{}^{16}\mathrm{O} + {}^{16}\mathrm{O}_{2} + {}^{18}\mathrm{O}_{2}$ Analysis of the products would show that one molecule of the evolved oxygen is ${}^{16}\mathrm{O}_{2}$ and the other is ${}^{18}\mathrm{O}_{2}$ , confirming that the oxygen molecules originate from different reactants.

Q10EXERCISES

The compound $\mathrm{AgF}_{2}$ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?

Solution

The stability of an ion or a compound is related to its electronic configuration and its tendency to achieve a more stable state.

Stable Oxidation State of Silver: The most common and stable oxidation state for silver (Ag) is +1. The electronic configuration of Ag is $[\text{Kr}] 4d^{10} 5s^{1}$ . When it forms the $\mathrm{Ag}^{+}$ ion, it loses the single 5s electron, resulting in a very stable $[\text{Kr}] 4d^{10}$ configuration with a completely filled d-orbital.
Unstable Oxidation State in $\mathrm{AgF}_{2}$ : In the compound $\mathrm{AgF}_{2}$ , fluorine has an oxidation state of -1. To balance the charge, silver must have an oxidation state of +2. This corresponds to the $\mathrm{Ag}^{2+}$ ion.
Electronic Configuration of $\mathrm{Ag}^{2+}$ : To form the $\mathrm{Ag}^{2+}$ ion, silver has to lose two electrons, one from the 5s orbital and one from the stable, completely filled 4d orbital. The resulting configuration is $[\text{Kr}] 4d^{9}$ . This configuration is much less stable than the $4d^{10}$ configuration of $\mathrm{Ag}^{+}$ .
Tendency to Act as an Oxidizing Agent: Because the +2 oxidation state is highly unstable for silver, the $\mathrm{Ag}^{2+}$ ion has a very strong tendency to gain an electron to revert to the much more stable $\mathrm{Ag}^{+}$ state. $\mathrm{Ag}^{2+} + \mathrm{e}^{-} \rightarrow \mathrm{Ag}^{+}$ A substance that readily accepts electrons from other species is, by definition, a strong oxidizing agent.

Conclusion:

\mathrm{AgF}_{2}

is a very strong oxidizing agent because it contains silver in the unstable +2 oxidation state. It readily accepts an electron to reduce itself to the stable +1 oxidation state, thereby oxidizing another substance in the process.

Q11EXERCISES

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Solution

Justification: The statement is true because the extent of oxidation of the reducing agent depends on the availability of the oxidizing agent. If the oxidizing agent is abundant (in excess), it can oxidize the reducing agent to its highest possible oxidation state. If the oxidizing agent is limited (i.e., the reducing agent is in excess), the reducing agent will only be partially oxidized to a lower, intermediate oxidation state.

Here are three illustrations:

Illustration 1: Reaction of Carbon with Oxygen

Reducing Agent: Carbon (C)
Oxidizing Agent: Oxygen ( $\mathrm{O}_{2}$ )
Case 1: Excess Reducing Agent (Limited Oxygen): Carbon is only partially oxidized from 0 to +2. $2\stackrel{0}{\mathrm{C}}(\mathrm{s}) + \stackrel{0}{\mathrm{O}}_{2}(\mathrm{g}) \xrightarrow{\Delta} 2\stackrel{+2}{\mathrm{C}}\mathrm{O}(\mathrm{g})$ A compound of a lower oxidation state (Carbon Monoxide, C = +2) is formed.
Case 2: Excess Oxidizing Agent (Excess Oxygen): Carbon is completely oxidized from 0 to its highest oxidation state of +4. $\stackrel{0}{\mathrm{C}}(\mathrm{s}) + \stackrel{0}{\mathrm{O}}_{2}(\mathrm{g}) \xrightarrow{\Delta} \stackrel{+4}{\mathrm{C}}\mathrm{O}_{2}(\mathrm{g})$ A compound of a higher oxidation state (Carbon Dioxide, C = +4) is formed.

Illustration 2: Reaction of Phosphorus with Chlorine

Reducing Agent: Phosphorus (P)
Oxidizing Agent: Chlorine ( $\mathrm{Cl}_{2}$ )
Case 1: Excess Reducing Agent (Limited Chlorine): Phosphorus is oxidized from 0 to +3. $\stackrel{0}{\mathrm{P}}_{4}(\mathrm{s}) + 6\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4\stackrel{+3}{\mathrm{P}}\mathrm{Cl}_{3}(\mathrm{l})$ A compound of a lower oxidation state (Phosphorus Trichloride, P = +3) is formed.
Case 2: Excess Oxidizing Agent (Excess Chlorine): Phosphorus is oxidized from 0 to its higher oxidation state of +5. $\stackrel{0}{\mathrm{P}}_{4}(\mathrm{s}) + 10\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4\stackrel{+5}{\mathrm{P}}\mathrm{Cl}_{5}(\mathrm{s})$ A compound of a higher oxidation state (Phosphorus Pentachloride, P = +5) is formed.

Illustration 3: Reaction of Sodium with Oxygen

Reducing Agent: Sodium (Na)
Oxidizing Agent: Oxygen ( $\mathrm{O}_{2}$ )
Case 1: Excess Reducing Agent (Limited Oxygen): Normal sodium oxide is formed, where oxygen is in the -2 oxidation state. $4\stackrel{0}{\mathrm{Na}}(\mathrm{s}) + \stackrel{0}{\mathrm{O}}_{2}(\mathrm{g}) \rightarrow 2\mathrm{Na}_{2}\stackrel{-2}{\mathrm{O}}(\mathrm{s})$ The oxidizing agent is reduced to a lower oxidation state.
Case 2: Excess Oxidizing Agent (Excess Oxygen): Sodium peroxide is formed, where oxygen is in the -1 oxidation state. $2\stackrel{0}{\mathrm{Na}}(\mathrm{s}) + \stackrel{0}{\mathrm{O}}_{2}(\mathrm{g}) \rightarrow \mathrm{Na}_{2}\stackrel{-1}{\mathrm{O}}_{2}(\mathrm{s})$ The oxidizing agent is reduced to a higher (less negative) oxidation state. This example shows the effect on the oxidizing agent itself, which is also a valid interpretation of the principle.

Q12EXERCISES

How do you count for the following observations ?

(a)

Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.

(b)

When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl , but if the mixture contains bromide then we get red vapour of bromine. Why ?

Solution

(a) Use of Alcoholic Potassium Permanganate for Toluene Oxidation

Reason: The choice of solvent is crucial for a reaction to occur efficiently. Toluene ( $\mathrm{C}_{6}H_{5}CH_{3}$ ) is an organic compound and is non-polar. It is immiscible with water. Potassium permanganate ( $\mathrm{KMnO}_{4}$ ) is an ionic compound and is soluble in water but not in toluene. If we use an aqueous solution of $\mathrm{KMnO}_{4}$ (either acidic or alkaline), the reaction would be extremely slow because the reactants (toluene and $\mathrm{KMnO}_{4}$ ) would be in separate phases (an oily layer and an aqueous layer) with very limited contact at their interface. Alcohol (like ethanol) is a solvent that can dissolve both the non-polar toluene and, to some extent, the polar potassium permanganate. Using an alcoholic medium creates a homogeneous solution where the reactant molecules can mix and react effectively. Therefore, alcoholic $\mathrm{KMnO}_{4}$ is used to ensure both reactants are in the same phase.
Balanced Redox Equation (in alkaline medium): The reaction is typically carried out in a basic medium, and the initial product is potassium benzoate, which is then acidified to get benzoic acid. $\mathrm{C}_{6}H_{5}CH_{3} + 2\mathrm{KMnO}_{4} \rightarrow \mathrm{C}_{6}H_{5}COOK + 2\mathrm{MnO}_{2} + \mathrm{KOH} + \mathrm{H}_{2}O$ In this reaction, the methyl group (-3) of toluene is oxidized to the carboxylate group (+3), and Mn (+7) in $\mathrm{KMnO}_{4}$ is reduced to Mn (+4) in $\mathrm{MnO}_{2}$ .

(b) Reaction of Concentrated Sulphuric Acid with Chloride vs. Bromide

Reason: The difference in observation is due to the different reducing strengths of the halide ions ( $\mathrm{Cl}^{-}$ , $\mathrm{Br}^{-}$ ) and the oxidizing power of concentrated sulphuric acid ( $\mathrm{H}_{2}SO_{4}$ ). Concentrated $\mathrm{H}_{2}SO_{4}$ is a moderately strong oxidizing agent.
With Chloride ( $\mathrm{Cl}^{-}$ ): The chloride ion is a very weak reducing agent. Concentrated $\mathrm{H}_{2}SO_{4}$ is not strong enough to oxidize $\mathrm{Cl}^{-}$ to $\mathrm{Cl}_{2}$ . Therefore, only an acid-base reaction occurs, where the non-volatile acid ( $\mathrm{H}_{2}SO_{4}$ ) displaces the more volatile acid (HCl) from its salt. $\mathrm{NaCl}(\mathrm{s}) + \mathrm{H}_{2}SO_{4}(\text{conc.}) \rightarrow \mathrm{NaHSO}_{4}(\mathrm{s}) + \mathrm{HCl}(\mathrm{g})$ The product is hydrogen chloride gas, which is colourless and has a pungent smell.
With Bromide ( $\mathrm{Br}^{-}$ ): The bromide ion is a stronger reducing agent than the chloride ion. It is strong enough to be oxidized by concentrated $\mathrm{H}_{2}SO_{4}$ . The reaction occurs in two steps:
1. First, an acid-base reaction similar to the chloride case produces hydrogen bromide gas (HBr). $\mathrm{NaBr}(\mathrm{s}) + \mathrm{H}_{2}SO_{4}(\text{conc.}) \rightarrow \mathrm{NaHSO}_{4}(\mathrm{s}) + \mathrm{HBr}(\mathrm{g})$
2. Then, the HBr produced is oxidized by the concentrated $\mathrm{H}_{2}SO_{4}$ . In this redox reaction, $\mathrm{Br}^{-}$ is oxidized to $\mathrm{Br}_{2}$ , and sulfur in $\mathrm{H}_{2}SO_{4}$ (oxidation state +6) is reduced to $\mathrm{SO}_{2}$ (oxidation state +4). $2\mathrm{HBr}(\mathrm{g}) + \mathrm{H}_{2}SO_{4}(\text{conc.}) \rightarrow \mathrm{Br}_{2}(\mathrm{g}) + \mathrm{SO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}O(\mathrm{l})$ The product, bromine ( $\mathrm{Br}_{2}$ ), is a reddish-brown vapour. This is why red vapours are observed.

Q13EXERCISES

Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

(a)

2 \mathrm{AgBr}(\mathrm{s})+\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{HBr}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(\mathrm{aq})

(b)

\mathrm{HCHO}(\mathrm{l})+2[\mathrm{Ag}(\mathrm{NH}_{3})_{2}]^{+}(\mathrm{aq})+3\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{HCOO}^{-}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) +2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})

(c)

\mathrm{HCHO}(\mathrm{l})+2 \mathrm{Cu}^{2+}(\mathrm{aq})+5\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{2}O(\mathrm{s})+\mathrm{HCOO}^{-}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})

(d)

\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})

(e)

\mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2 \mathrm{PbSO}_{4}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})

Solution

(a) $2 \mathrm{AgBr}(\mathrm{s})+\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{HBr}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(\mathrm{aq})$

Changes: $\mathrm{Ag}^{+}$ in AgBr (O.N. +1) is converted to Ag (O.N. 0). This is reduction. Hydroquinone ( $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$ ) is converted to quinone ( $\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}$ ) by removal of hydrogen. This is oxidation.
Substance Oxidised: $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$ (hydroquinone)
Substance Reduced: AgBr
Oxidising Agent: AgBr
Reducing Agent: $\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}$

(b) $\mathrm{HCHO}(\mathrm{l})+2[\mathrm{Ag}(\mathrm{NH}_{3})_{2}]^{+}(\mathrm{aq})+3\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{HCOO}^{-}(\mathrm{aq})+4 \mathrm{NH}_{3}(\mathrm{aq}) +2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Changes: Carbon in HCHO (O.N. 0) is oxidized to Carbon in $\mathrm{HCOO}^{-}$ (O.N. +2). Silver in $[\mathrm{Ag}(\mathrm{NH}_{3})_{2}]^{+}$ (O.N. +1) is reduced to Ag (O.N. 0).
Substance Oxidised: HCHO (formaldehyde)
Substance Reduced: $[\mathrm{Ag}(\mathrm{NH}_{3})_{2}]^{+}$ (Tollens' reagent)
Oxidising Agent: $[\mathrm{Ag}(\mathrm{NH}_{3})_{2}]^{+}$
Reducing Agent: HCHO

(c) $\mathrm{HCHO}(\mathrm{l})+2 \mathrm{Cu}^{2+}(\mathrm{aq})+5\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{2}O(\mathrm{s})+\mathrm{HCOO}^{-}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Changes: Carbon in HCHO (O.N. 0) is oxidized to Carbon in $\mathrm{HCOO}^{-}$ (O.N. +2). Copper in $\mathrm{Cu}^{2+}$ (O.N. +2) is reduced to Copper in $\mathrm{Cu}_{2}O$ (O.N. +1).
Substance Oxidised: HCHO (formaldehyde)
Substance Reduced: $\mathrm{Cu}^{2+}$ (Fehling's solution)
Oxidising Agent: $\mathrm{Cu}^{2+}$
Reducing Agent: HCHO

(d) $\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Changes: Nitrogen in $\mathrm{N}_{2} \mathrm{H}_{4}$ (O.N. -2) is oxidized to $\mathrm{N}_{2}$ (O.N. 0). Oxygen in $\mathrm{H}_{2} \mathrm{O}_{2}$ (O.N. -1) is reduced to Oxygen in $\mathrm{H}_{2} \mathrm{O}$ (O.N. -2).
Substance Oxidised: $\mathrm{N}_{2} \mathrm{H}_{4}$ (hydrazine)
Substance Reduced: $\mathrm{H}_{2} \mathrm{O}_{2}$ (hydrogen peroxide)
Oxidising Agent: $\mathrm{H}_{2} \mathrm{O}_{2}$
Reducing Agent: $\mathrm{N}_{2} \mathrm{H}_{4}$

(e) $\mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow 2 \mathrm{PbSO}_{4}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Changes: Lead in Pb(s) (O.N. 0) is oxidized to Lead in $\mathrm{PbSO}_{4}$ (O.N. +2). Lead in $\mathrm{PbO}_{2}$ (O.N. +4) is reduced to Lead in $\mathrm{PbSO}_{4}$ (O.N. +2). This is a comproportionation reaction, a type of redox reaction.
Substance Oxidised: Pb(s)
Substance Reduced: $\mathrm{PbO}_{2}$
Oxidising Agent: $\mathrm{PbO}_{2}$
Reducing Agent: Pb(s)

Q14EXERCISES

Consider the reactions : $2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{~s}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})$ $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+2 \mathrm{Br}_{2}(\mathrm{l})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+4 \mathrm{Br}^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq})$ Why does the same reductant, thiosulphate react differently with iodine and bromine?

Solution

The reason for the different reaction products is the difference in the oxidizing strength of iodine and bromine.

Oxidizing Strength: Halogens are oxidizing agents, and their strength decreases down the group. Therefore, bromine ( $\mathrm{Br}_{2}$ ) is a significantly stronger oxidizing agent than iodine ( $\mathrm{I}_{2}$ ). ( $\mathrm{F}_{2} > \mathrm{Cl}_{2} > \mathrm{Br}_{2} > \mathrm{I}_{2}$ )
Reaction with Iodine ( $\mathrm{I}_{2}$ ): Iodine is a mild oxidizing agent. It is only strong enough to oxidize the thiosulphate ion ( $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ ) to the tetrathionate ion ( $\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$ ). Let's look at the change in the average oxidation state of sulfur: In $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ , the average O.S. of S is +2. In $\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$ , the average O.S. of S is +2.5. This is a mild oxidation. $2 \stackrel{+2}{\mathrm{S}}_{2} \mathrm{O}_{3}^{2-} + \mathrm{I}_{2} \rightarrow \stackrel{+2.5}{\mathrm{S}}_{4} \mathrm{O}_{6}^{2-} + 2\mathrm{I}^{-}$
Reaction with Bromine ( $\mathrm{Br}_{2}$ ): Bromine is a strong oxidizing agent. It is powerful enough to oxidize the sulfur in the thiosulphate ion all the way to the sulphate ion ( $\mathrm{SO}_{4}^{2-}$ ), breaking the S-S bond in the process. Let's look at the change in the average oxidation state of sulfur: In $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ , the average O.S. of S is +2. In $\mathrm{SO}_{4}^{2-}$ , the O.S. of S is +6. This is a much more extensive oxidation. $\stackrel{+2}{\mathrm{S}}_{2} \mathrm{O}_{3}^{2-} + 2\mathrm{Br}_{2} + 5\mathrm{H}_{2}O \rightarrow 2\stackrel{+6}{\mathrm{SO}}_{4}^{2-} + 4\mathrm{Br}^{-} + 10\mathrm{H}^{+}$

Conclusion: The same reductant, thiosulphate, reacts differently because the stronger oxidizing agent, bromine, causes a more complete oxidation of sulfur (from +2 to +6) compared to the weaker oxidizing agent, iodine, which only causes a partial oxidation (from +2 to +2.5).

Q15EXERCISES

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Solution

Justification based on Standard Electrode Potentials (E°): The tendency of a species to act as an oxidant or reductant is best quantified by its standard electrode potential. A higher positive E° value indicates a stronger oxidizing agent. A lower (more negative) E° value indicates a stronger reducing agent (for the reduced form).

$\mathrm{F}_{2}(\mathrm{g}) + 2e^{-} \rightarrow 2\mathrm{F}^{-}(\mathrm{aq}) \qquad E° = +2.87 \mathrm{V}$
$\mathrm{Cl}_{2}(\mathrm{g}) + 2e^{-} \rightarrow 2\mathrm{Cl}^{-}(\mathrm{aq}) \qquad E° = +1.36 \mathrm{V}$
$\mathrm{Br}_{2}(\mathrm{l}) + 2e^{-} \rightarrow 2\mathrm{Br}^{-}(\mathrm{aq}) \qquad E° = +1.09 \mathrm{V}$
$\mathrm{I}_{2}(\mathrm{s}) + 2e^{-} \rightarrow 2\mathrm{I}^{-}(\mathrm{aq}) \qquad E° = +0.54 \mathrm{V}$

Fluorine as the Best Oxidant:

Justification: Fluorine ( $\mathrm{F}_{2}$ ) has the highest standard reduction potential (+2.87 V) among all halogens. This indicates it has the strongest tendency to accept electrons and get reduced, making it the most powerful oxidizing agent.
Reaction Evidence: Due to its high oxidizing power, fluorine can oxidize all other halide ions to their respective halogens. It is so powerful it can even oxidize water.
1. Oxidation of other halides: $\mathrm{F}_{2}(\mathrm{g}) + 2\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow 2\mathrm{F}^{-}(\mathrm{aq}) + \mathrm{Cl}_{2}(\mathrm{g})$ $\mathrm{F}_{2}(\mathrm{g}) + 2\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow 2\mathrm{F}^{-}(\mathrm{aq}) + \mathrm{Br}_{2}(\mathrm{l})$ $\mathrm{F}_{2}(\mathrm{g}) + 2\mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2\mathrm{F}^{-}(\mathrm{aq}) + \mathrm{I}_{2}(\mathrm{s})$
2. Oxidation of water: $2\mathrm{F}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{l}) \rightarrow 4\mathrm{HF}(\mathrm{aq}) + \mathrm{O}_{2}(\mathrm{g})$

Hydroiodic Acid (HI) as the Best Reductant:

Justification: A hydrohalic acid (HX) acts as a reductant through its halide ion (X⁻). The reducing power is the tendency to lose electrons (get oxidized). This corresponds to the reverse of the reduction reactions listed above. The reaction $2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2} + 2e^{-}$ has the lowest oxidation potential (-0.54 V), meaning the iodide ion ( $\mathrm{I}^{-}$ ) is the most easily oxidized among the halides. Therefore, hydroiodic acid (HI) is the strongest reducing agent among the hydrohalic acids.
Reaction Evidence: Iodide ions can reduce stronger oxidizing agents like copper(II) ions, while bromide and chloride cannot.
1. Reduction of Copper(II): $2\mathrm{Cu}^{2+}(\mathrm{aq}) + 4\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cu}_{2}\mathrm{I}_{2}(\mathrm{s}) + \mathrm{I}_{2}(\mathrm{aq})$
2. Reduction of Sulphuric Acid: HI is strong enough to reduce concentrated $\mathrm{H}_{2}SO_{4}$ to $\mathrm{H}_{2}S$ . $8\mathrm{HI}(\mathrm{aq}) + \mathrm{H}_{2}SO_{4}(\text{conc.}) \rightarrow 4\mathrm{I}_{2}(\mathrm{s}) + \mathrm{H}_{2}S(\mathrm{g}) + 4\mathrm{H}_{2}O(\mathrm{l})$

Q16EXERCISES

Why does the following reaction occur ? $\mathrm{XeO}_{6}{ }^{4-}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{XeO}_{3}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ What conclusion about the compound $\mathrm{Na}_{4} \mathrm{XeO}_{6}$ (of which $\mathrm{XeO}_{6}^{4-}$ is a part) can be drawn from the reaction.

Solution

Analysis of the Reaction: Let's first identify the oxidation and reduction processes by assigning oxidation numbers:

\stackrel{+8}{\mathrm{Xe}}\mathrm{O}_{6}{ }^{4-} + 2 \stackrel{-1}{\mathrm{F}}^{-} + 6 \mathrm{H}^{+} \rightarrow \stackrel{+6}{\mathrm{Xe}}\mathrm{O}_{3} + \stackrel{0}{\mathrm{F}}_{2} + 3 \mathrm{H}_{2} \mathrm{O}

Oxidation: The fluoride ion ( $\mathrm{F}^{-}$ ) is oxidized from an oxidation state of -1 to 0 in fluorine gas ( $\mathrm{F}_{2}$ ). $2\mathrm{F}^{-} \rightarrow \mathrm{F}_{2} + 2e^{-}$
Reduction: The xenon atom in the perxenate ion ( $\mathrm{XeO}_{6}^{4-}$ ) is reduced from an oxidation state of +8 to +6 in xenon trioxide ( $\mathrm{XeO}_{3}$ ). $\mathrm{XeO}_{6}^{4-} \rightarrow \mathrm{XeO}_{3}$

Reason for the Reaction's Occurrence: The reaction occurs because the perxenate ion (

\mathrm{XeO}_{6}^{4-}

) is an exceptionally powerful oxidizing agent. We know that fluorine (

\mathrm{F}_{2}

) is the strongest oxidizing agent among all elements, with a very high standard reduction potential (

E°(\mathrm{F}_{2}/\mathrm{F}^{-}) = +2.87 \mathrm{V}

).

For the given reaction to be spontaneous, the oxidizing agent (

\mathrm{XeO}_{6}^{4-}

) must have a stronger tendency to be reduced than the product (

\mathrm{F}_{2}

) has to be reduced. This means the standard reduction potential of the perxenate ion must be greater than that of fluorine.

E°(\mathrm{XeO}_{6}^{4-}/\mathrm{XeO}_{3}) > E°(\mathrm{F}_{2}/\mathrm{F}^{-})

E°(\mathrm{XeO}_{6}^{4-}/\mathrm{XeO}_{3}) > +2.87 \mathrm{V}

The reaction proceeds because perxenate ion is a more powerful oxidizing agent than fluorine gas.

Conclusion about $\mathrm{Na}_{4} \mathrm{XeO}_{6}$ : Since the perxenate ion (

\mathrm{XeO}_{6}^{4-}

) is the active species from the compound sodium perxenate (

\mathrm{Na}_{4} \mathrm{XeO}_{6}

), we can draw a significant conclusion about the compound itself.

The conclusion is that $\mathrm{Na}_{4} \mathrm{XeO}_{6}$ is an extremely strong oxidizing agent, even stronger than fluorine gas. It is one of the most powerful oxidizing agents known in chemistry because it is capable of oxidizing fluoride ions to fluorine gas, a feat that is very difficult to achieve chemically.

Q17EXERCISES

Consider the reactions:

(a)

\mathrm{H}_{3} \mathrm{PO}_{2}(\mathrm{aq})+4 \mathrm{AgNO}_{3}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+4 \mathrm{Ag}(\mathrm{s})+4 \mathrm{HNO}_{3}(\mathrm{aq})

(b)

\mathrm{H}_{3} \mathrm{PO}_{2}(\mathrm{aq})+2 \mathrm{CuSO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})

(c)

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(\mathrm{l})+2[\mathrm{Ag}(\mathrm{NH}_{3})_{2}]^{+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+ 4 \mathrm{NH}_{3}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})

(d)

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(\mathrm{l})+2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow

No change observed.

What inference do you draw about the behaviour of

\mathrm{Ag}^{+}

and

\mathrm{Cu}^{2+}

from these reactions?

Solution

Analysis of the Reactions: Let's analyze the oxidizing and reducing agents in each reaction.

Reactions (a) and (b): Hypophosphorous acid ( $\mathrm{H}_{3} \mathrm{PO}_{2}$ ) is acting as a reducing agent, getting oxidized to phosphoric acid ( $\mathrm{H}_{3} \mathrm{PO}_{4}$ ). Both $\mathrm{Ag}^{+}$ (from $\mathrm{AgNO}_{3}$ ) and $\mathrm{Cu}^{2+}$ (from $\mathrm{CuSO}_{4}$ ) are strong enough to oxidize $\mathrm{H}_{3} \mathrm{PO}_{2}$ .
- In (a), $\mathrm{Ag}^{+}$ is reduced to Ag.
- In (b), $\mathrm{Cu}^{2+}$ is reduced to Cu.
Reactions (c) and (d): Benzaldehyde ( $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}$ ) is the reducing agent.
- In (c), $\mathrm{Ag}^{+}$ (in the form of Tollens' reagent, $[\mathrm{Ag}(\mathrm{NH}_{3})_{2}]^{+}$ ) successfully oxidizes benzaldehyde to the benzoate ion ( $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}$ ). This indicates $\mathrm{Ag}^{+}$ is a sufficiently strong oxidizing agent for this reaction.
- In (d), $\mathrm{Cu}^{2+}$ (in the form of Fehling's or Benedict's solution) does not oxidize benzaldehyde. This indicates that $\mathrm{Cu}^{2+}$ is not a strong enough oxidizing agent to carry out this specific reaction. (Note: Aromatic aldehydes do not give a positive Fehling's test).

Inference: The reactions show a difference in the oxidizing power of

\mathrm{Ag}^{+}

and

\mathrm{Cu}^{2+}

.

Both $\mathrm{Ag}^{+}$ and $\mathrm{Cu}^{2+}$ are capable of oxidizing a relatively strong reducing agent like hypophosphorous acid.
However, when faced with a weaker reducing agent like benzaldehyde, only $\mathrm{Ag}^{+}$ is strong enough to oxidize it, while $\mathrm{Cu}^{2+}$ is not.

Conclusion: The inference drawn is that $\mathrm{Ag}^{+}$ ion is a stronger oxidizing agent than the $\mathrm{Cu}^{2+}$ ion. This is consistent with their standard electrode potentials:

$E°(\mathrm{Ag}^{+}/\mathrm{Ag}) = +0.80 \mathrm{V}$
$E°(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34 \mathrm{V}$ The higher positive potential for silver confirms it is the stronger oxidant.

Q18EXERCISES

Balance the following redox reactions by ion - electron method :

(a)

\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{I}_{2}(\mathrm{~s})

(in basic medium)

(b)

\mathrm{MnO}_{4}^{-}(\mathrm{aq}) +\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) +\mathrm{HSO}_{4}^{-}(\mathrm{aq})

(in acidic solution)

(c)

\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

(in acidic solution)

(d)

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}{ }^{2-}(\mathrm{aq})

(in acidic solution)

Solution

(a) $\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s})+\mathrm{I}_{2}(\mathrm{~s})$ (in basic medium)

Step 1 & 2: Half-reactions
- Oxidation: $2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2}$
- Reduction: $\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{MnO}_{2}$
Step 3 & 4: Balance atoms and O/H (basic medium)
- Oxidation: (Already balanced for atoms)
- Reduction: $\mathrm{MnO}_{4}^{-} + 2\mathrm{H}_{2}O \rightarrow \mathrm{MnO}_{2} + 4\mathrm{OH}^{-}$
Step 5: Balance charge with electrons
- Oxidation: $2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2} + 2e^{-}$
- Reduction: $\mathrm{MnO}_{4}^{-} + 2\mathrm{H}_{2}O + 3e^{-} \rightarrow \mathrm{MnO}_{2} + 4\mathrm{OH}^{-}$
Step 6: Equalize electrons
- Multiply oxidation half by 3: $6\mathrm{I}^{-} \rightarrow 3\mathrm{I}_{2} + 6e^{-}$
- Multiply reduction half by 2: $2\mathrm{MnO}_{4}^{-} + 4\mathrm{H}_{2}O + 6e^{-} \rightarrow 2\mathrm{MnO}_{2} + 8\mathrm{OH}^{-}$
Step 7: Add half-reactions $2\mathrm{MnO}_{4}^{-} + 6\mathrm{I}^{-} + 4\mathrm{H}_{2}O \rightarrow 2\mathrm{MnO}_{2} + 3\mathrm{I}_{2} + 8\mathrm{OH}^{-}$

(b) $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) +\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) +\mathrm{HSO}_{4}^{-}(\mathrm{aq})$ (in acidic solution)

Step 1 & 2: Half-reactions
- Oxidation: $\mathrm{SO}_{2} \rightarrow \mathrm{HSO}_{4}^{-}$
- Reduction: $\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}$
Step 3 & 4: Balance atoms and O/H (acidic medium)
- Oxidation: $\mathrm{SO}_{2} + 2\mathrm{H}_{2}O \rightarrow \mathrm{HSO}_{4}^{-} + 3\mathrm{H}^{+}$
- Reduction: $\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}O$
Step 5: Balance charge with electrons
- Oxidation: $\mathrm{SO}_{2} + 2\mathrm{H}_{2}O \rightarrow \mathrm{HSO}_{4}^{-} + 3\mathrm{H}^{+} + 2e^{-}$
- Reduction: $\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}O$
Step 6: Equalize electrons
- Multiply oxidation half by 5: $5\mathrm{SO}_{2} + 10\mathrm{H}_{2}O \rightarrow 5\mathrm{HSO}_{4}^{-} + 15\mathrm{H}^{+} + 10e^{-}$
- Multiply reduction half by 2: $2\mathrm{MnO}_{4}^{-} + 16\mathrm{H}^{+} + 10e^{-} \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_{2}O$
Step 7: Add half-reactions and simplify $2\mathrm{MnO}_{4}^{-} + 5\mathrm{SO}_{2} + 2\mathrm{H}_{2}O + \mathrm{H}^{+} \rightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{HSO}_{4}^{-}$

(c) $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ (in acidic solution)

Step 1 & 2: Half-reactions
- Oxidation: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}$
- Reduction: $\mathrm{H}_{2}O_{2} \rightarrow \mathrm{H}_{2}O$
Step 3 & 4: Balance atoms and O/H (acidic medium)
- Oxidation: (Already balanced)
- Reduction: $\mathrm{H}_{2}O_{2} + 2\mathrm{H}^{+} \rightarrow 2\mathrm{H}_{2}O$
Step 5: Balance charge with electrons
- Oxidation: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + 1e^{-}$
- Reduction: $\mathrm{H}_{2}O_{2} + 2\mathrm{H}^{+} + 2e^{-} \rightarrow 2\mathrm{H}_{2}O$
Step 6: Equalize electrons
- Multiply oxidation half by 2: $2\mathrm{Fe}^{2+} \rightarrow 2\mathrm{Fe}^{3+} + 2e^{-}$
Step 7: Add half-reactions $\mathrm{H}_{2}O_{2} + 2\mathrm{Fe}^{2+} + 2\mathrm{H}^{+} \rightarrow 2\mathrm{Fe}^{3+} + 2\mathrm{H}_{2}O$

(d) $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}{ }^{2-}(\mathrm{aq})$ (in acidic solution)

Step 1 & 2: Half-reactions
- Oxidation: $\mathrm{SO}_{2} \rightarrow \mathrm{SO}_{4}^{2-}$
- Reduction: $\mathrm{Cr}_{2}O_{7}^{2-} \rightarrow 2\mathrm{Cr}^{3+}$
Step 3 & 4: Balance atoms and O/H (acidic medium)
- Oxidation: $\mathrm{SO}_{2} + 2\mathrm{H}_{2}O \rightarrow \mathrm{SO}_{4}^{2-} + 4\mathrm{H}^{+}$
- Reduction: $\mathrm{Cr}_{2}O_{7}^{2-} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}O$
Step 5: Balance charge with electrons
- Oxidation: $\mathrm{SO}_{2} + 2\mathrm{H}_{2}O \rightarrow \mathrm{SO}_{4}^{2-} + 4\mathrm{H}^{+} + 2e^{-}$
- Reduction: $\mathrm{Cr}_{2}O_{7}^{2-} + 14\mathrm{H}^{+} + 6e^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}O$
Step 6: Equalize electrons
- Multiply oxidation half by 3: $3\mathrm{SO}_{2} + 6\mathrm{H}_{2}O \rightarrow 3\mathrm{SO}_{4}^{2-} + 12\mathrm{H}^{+} + 6e^{-}$
Step 7: Add half-reactions and simplify $\mathrm{Cr}_{2}O_{7}^{2-} + 3\mathrm{SO}_{2} + 2\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 3\mathrm{SO}_{4}^{2-} + \mathrm{H}_{2}O$

Q19EXERCISES

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

(a)

\mathrm{P}_{4}(\mathrm{~s})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{PH}_{3}(\mathrm{~g})+\mathrm{HPO}_{2}^{-}(\mathrm{aq})

(b)

\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{~g})+\mathrm{Cl}^{-}(\mathrm{g})

(c)

\mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{ClO}_{2}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}^{+}

Solution

(a) $\mathrm{P}_{4}(\mathrm{~s})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{PH}_{3}(\mathrm{~g})+\mathrm{H}_{2}PO_{2}^{-}(\mathrm{aq})$ (Note: product is $\mathrm{H}_{2}PO_{2}^{-}$ ) This is a disproportionation reaction. P(0) is reduced to P(-3) and oxidized to P(+1).

Ion-Electron Method:
- Reduction: $\mathrm{P}_{4} + 12\mathrm{H}_{2}O + 12e^{-} \rightarrow 4\mathrm{PH}_{3} + 12\mathrm{OH}^{-}$
- Oxidation: $\mathrm{P}_{4} + 8\mathrm{OH}^{-} \rightarrow 4\mathrm{H}_{2}PO_{2}^{-} + 4e^{-}$
- Equalize e⁻ (multiply oxidation by 3): $\mathrm{3P}_{4} + 24\mathrm{OH}^{-} \rightarrow 12\mathrm{H}_{2}PO_{2}^{-} + 12e^{-}$
- Add and simplify: $\mathrm{4P}_{4} + 12\mathrm{OH}^{-} + 12\mathrm{H}_{2}O \rightarrow 4\mathrm{PH}_{3} + 12\mathrm{H}_{2}PO_{2}^{-}$
- Final: $\mathrm{P}_{4} + 3\mathrm{OH}^{-} + 3\mathrm{H}_{2}O \rightarrow \mathrm{PH}_{3} + 3\mathrm{H}_{2}PO_{2}^{-}$
Oxidation Number Method:
- O.N. changes: P(0) $\rightarrow$ P(-3) (decrease by 3); P(0) $\rightarrow$ P(+1) (increase by 1).
- To balance change, we need 3 atoms of P oxidized for every 1 atom reduced. Total 4 P atoms. So, $\mathrm{P}_{4} \rightarrow \mathrm{PH}_{3} + 3\mathrm{H}_{2}PO_{2}^{-}$ .
- Balance charge: LHS charge = 0. RHS charge = 0 + 3(-1) = -3. Add $3\mathrm{OH}^{-}$ to LHS.
- Balance atoms: $\mathrm{P}_{4} + 3\mathrm{OH}^{-} \rightarrow \mathrm{PH}_{3} + 3\mathrm{H}_{2}PO_{2}^{-}$ . Add $3\mathrm{H}_{2}O$ to LHS to balance H and O.
- Final: $\mathrm{P}_{4} + 3\mathrm{OH}^{-} + 3\mathrm{H}_{2}O \rightarrow \mathrm{PH}_{3} + 3\mathrm{H}_{2}PO_{2}^{-}$
Agents: $\mathrm{P}_{4}$ is both the oxidising agent and the reducing agent.

(b) $\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{~g})+\mathrm{Cl}^{-}(\mathrm{g})$

Ion-Electron Method:
- Oxidation: $\mathrm{N}_{2}H_{4} + 8\mathrm{OH}^{-} \rightarrow 2\mathrm{NO} + 6\mathrm{H}_{2}O + 8e^{-}$
- Reduction: $\mathrm{ClO}_{3}^{-} + 3\mathrm{H}_{2}O + 6e^{-} \rightarrow \mathrm{Cl}^{-} + 6\mathrm{OH}^{-}$
- Equalize e⁻ (LCM is 24): Multiply oxidation by 3, reduction by 4.
- $3\mathrm{N}_{2}H_{4} + 24\mathrm{OH}^{-} \rightarrow 6\mathrm{NO} + 18\mathrm{H}_{2}O + 24e^{-}$
- $4\mathrm{ClO}_{3}^{-} + 12\mathrm{H}_{2}O + 24e^{-} \rightarrow 4\mathrm{Cl}^{-} + 24\mathrm{OH}^{-}$
- Add and simplify: $3\mathrm{N}_{2}H_{4} + 4\mathrm{ClO}_{3}^{-} \rightarrow 6\mathrm{NO} + 4\mathrm{Cl}^{-} + 6\mathrm{H}_{2}O$
Oxidation Number Method:
- O.N. changes: N(-2) $\rightarrow$ N(+2) (increase by 4 per N, total 8 for $\mathrm{N}_{2}H_{4}$ ). Cl(+5) $\rightarrow$ Cl(-1) (decrease by 6).
- LCM of 8 and 6 is 24. Multiply N species by 3, Cl species by 4.
- $3\mathrm{N}_{2}H_{4} + 4\mathrm{ClO}_{3}^{-} \rightarrow 6\mathrm{NO} + 4\mathrm{Cl}^{-}$
- Balance O: LHS has 12 O. RHS has 6 O. Add $6\mathrm{H}_{2}O$ to RHS.
- Final: $3\mathrm{N}_{2}H_{4} + 4\mathrm{ClO}_{3}^{-} \rightarrow 6\mathrm{NO} + 4\mathrm{Cl}^{-} + 6\mathrm{H}_{2}O$
Agents: Oxidising agent: $\mathrm{ClO}_{3}^{-}$ , Reducing agent: $\mathrm{N}_{2}H_{4}$ .

(c) $\mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{ClO}_{2}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{~g})$ (in basic medium) (The question mentions

\mathrm{H}^{+}

as a product, which suggests acidic medium. We will balance in basic medium as requested.)

Ion-Electron Method:
- Oxidation: $\mathrm{H}_{2}O_{2} + 2\mathrm{OH}^{-} \rightarrow \mathrm{O}_{2} + 2\mathrm{H}_{2}O + 2e^{-}$
- Reduction: $\mathrm{Cl}_{2}O_{7} + 6\mathrm{H}_{2}O + 8e^{-} \rightarrow 2\mathrm{ClO}_{2}^{-} + 12\mathrm{OH}^{-}$
- Equalize e⁻ (multiply oxidation by 4): $4\mathrm{H}_{2}O_{2} + 8\mathrm{OH}^{-} \rightarrow 4\mathrm{O}_{2} + 8\mathrm{H}_{2}O + 8e^{-}$
- Add and simplify: $\mathrm{Cl}_{2}O_{7} + 4\mathrm{H}_{2}O_{2} \rightarrow 2\mathrm{ClO}_{2}^{-} + 4\mathrm{O}_{2} + 2\mathrm{H}_{2}O + 4\mathrm{OH}^{-}$
- Wait, the charges don't balance. Let's re-do reduction. $\mathrm{Cl}_{2}O_{7} + 3H_2O + 8e^{-} \rightarrow 2ClO_2^{-} + 6OH^{-}$ . No. $\mathrm{Cl}_{2}O_{7} \rightarrow 2ClO_2^{-}$ . Add $3H_2O$ to right. Add $6H^{+}$ to left. Add $6OH^{-}$ to both sides. $\mathrm{Cl}_{2}O_{7} + 3H_2O + 6OH^{-} \rightarrow 2ClO_2^{-} + 6H_2O$ . Simplify. $\mathrm{Cl}_{2}O_{7} + 6OH^{-} \rightarrow 2ClO_2^{-} + 3H_2O$ . Now charge. LHS -6. RHS -2. Add 4e- to left. This is incorrect. Let's use O.N. method.
Oxidation Number Method:
- O.N. changes: Cl(+7) $\rightarrow$ Cl(+3) (decrease by 4 per Cl, total 8 for $\mathrm{Cl}_{2}O_{7}$ ). O(-1) $\rightarrow$ O(0) (increase by 1 per O, total 2 for $\mathrm{H}_{2}O_{2}$ ).
- To balance change, multiply $\mathrm{H}_{2}O_{2}$ species by 4.
- $\mathrm{Cl}_{2}O_{7} + 4\mathrm{H}_{2}O_{2} \rightarrow 2\mathrm{ClO}_{2}^{-} + 4\mathrm{O}_{2}$
- Balance charge: LHS is 0. RHS is -2. Add $2\mathrm{OH}^{-}$ to LHS.
- $\mathrm{Cl}_{2}O_{7} + 4\mathrm{H}_{2}O_{2} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{ClO}_{2}^{-} + 4\mathrm{O}_{2}$
- Balance atoms: Add $3\mathrm{H}_{2}O$ to RHS to balance H and O.
- Final: $\mathrm{Cl}_{2}O_{7} + 4\mathrm{H}_{2}O_{2} + 2\mathrm{OH}^{-} \rightarrow 2\mathrm{ClO}_{2}^{-} + 4\mathrm{O}_{2} + 3\mathrm{H}_{2}O$
Agents: Oxidising agent: $\mathrm{Cl}_{2}O_{7}$ , Reducing agent: $\mathrm{H}_{2}O_{2}$ .

Q20EXERCISES

What sorts of informations can you draw from the following reaction ? $(\mathrm{CN})_{2}(\mathrm{~g})+2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{CN}^{-}(\mathrm{aq})+\mathrm{CNO}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Solution

From the given reaction, we can draw the following information:

It is a Redox Reaction: We can confirm this by assigning oxidation numbers.
- In cyanogen, $(\mathrm{CN})_{2}$ , nitrogen is more electronegative than carbon, so N is -3. For the molecule to be neutral, C must be +3.
- In the cyanide ion, $\mathrm{CN}^{-}$ , N is -3, so C must be +2 for the overall charge to be -1.
- In the cyanate ion, $\mathrm{CNO}^{-}$ , N is -3 and O is -2. For the overall charge to be -1, C must be +4.
The reaction with oxidation numbers is: $(\stackrel{+3}{C}\stackrel{-3}{N})_{2} + 2OH^{-} \rightarrow \stackrel{+2}{C}\stackrel{-3}{N}^{-} + \stackrel{+4}{C}\stackrel{-3}{N}\stackrel{-2}{O}^{-} + H_{2}O$
- The oxidation state of carbon changes from +3 to +2 (a reduction) and from +3 to +4 (an oxidation).
It is a Disproportionation Reaction: Since the same element, carbon, in the reactant $(\mathrm{CN})_{2}$ is simultaneously oxidized (to $\mathrm{CNO}^{-}$ ) and reduced (to $\mathrm{CN}^{-}$ ), this reaction is a classic example of a disproportionation reaction.
Basic Medium: The presence of $\mathrm{OH}^{-}$ as a reactant indicates that the reaction takes place in a basic or alkaline medium. Disproportionation reactions of many non-metals (like halogens, phosphorus, sulfur) are common in basic solutions.
Analogy to Halogens: This reaction is analogous to the disproportionation of halogens (like $\mathrm{Cl}_{2}$ ) in a basic medium, where the halogen forms a halide ion (lower oxidation state) and a hypohalite ion (higher oxidation state). $\mathrm{Cl}_{2} + 2\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-} + \mathrm{ClO}^{-} + \mathrm{H}_{2}O$ This suggests that cyanogen, $(\mathrm{CN})_{2}$ , behaves as a pseudohalogen.

Q21EXERCISES

The $\mathrm{Mn}^{3+}$ ion is unstable in solution and undergoes disproportionation to give $\mathrm{Mn}^{2+}$ , $\mathrm{MnO}_{2}$ , and $\mathrm{H}^{+}$ ion. Write a balanced ionic equation for the reaction.

Solution

Method: Half-Reaction (Ion-Electron) Method

Step 1: Write the skeletal equation. The reactant is

\mathrm{Mn}^{3+}

. The products are

\mathrm{Mn}^{2+}

,

\mathrm{MnO}_{2}

, and

\mathrm{H}^{+}

.

\mathrm{Mn}^{3+}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) + \mathrm{MnO}_{2}(\mathrm{~s}) + \mathrm{H}^{+}(\mathrm{aq})

Step 2: Separate into two half-reactions. In this disproportionation reaction,

\mathrm{Mn}^{3+}

is both oxidized and reduced.

Reduction Half-Reaction: The oxidation state of Mn decreases from +3 to +2. $\mathrm{Mn}^{3+}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})$
Oxidation Half-Reaction: The oxidation state of Mn increases from +3 to +4 (in $\mathrm{MnO}_{2}$ ). $\mathrm{Mn}^{3+}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s})$

Step 3: Balance atoms other than O and H. In both half-reactions, the Mn atoms are already balanced (one on each side).

Step 4: Balance O and H atoms (in acidic medium, as $\mathrm{H}^{+}$ is a product).

Reduction Half-Reaction: No O or H atoms to balance.
Oxidation Half-Reaction:
- Balance O atoms by adding $\mathrm{H}_{2} \mathrm{O}$ . There are 2 O atoms on the right, so add $2\mathrm{H}_{2} \mathrm{O}$ to the left. $\mathrm{Mn}^{3+}(\mathrm{aq}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s})$
- Balance H atoms by adding $\mathrm{H}^{+}$ . There are 4 H atoms on the left, so add $4\mathrm{H}^{+}$ to the right. $\mathrm{Mn}^{3+}(\mathrm{aq}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s}) + 4\mathrm{H}^{+}(\mathrm{aq})$

Step 5: Balance the charges by adding electrons (e⁻).

Reduction Half-Reaction: Charge on LHS is +3. Charge on RHS is +2. Add 1 e⁻ to the left. $\mathrm{Mn}^{3+}(\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})$
Oxidation Half-Reaction: Charge on LHS is +3. Charge on RHS is +4. Add 1 e⁻ to the right. $\mathrm{Mn}^{3+}(\mathrm{aq}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s}) + 4\mathrm{H}^{+}(\mathrm{aq}) + \mathrm{e}^{-}$

Step 6: Equalize electrons and add the half-reactions. The number of electrons (1 e⁻) is already equal in both half-reactions. We can add them directly.

(\mathrm{Mn}^{3+} + \mathrm{e}^{-}) + (\mathrm{Mn}^{3+} + 2\mathrm{H}_{2} \mathrm{O}) \rightarrow (\mathrm{Mn}^{2+}) + (\mathrm{MnO}_{2} + 4\mathrm{H}^{+} + \mathrm{e}^{-})

Combine like terms and cancel the electrons from both sides.

2\mathrm{Mn}^{3+}(\mathrm{aq}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) + \mathrm{MnO}_{2}(\mathrm{~s}) + 4\mathrm{H}^{+}(\mathrm{aq})

Final Answer: The balanced ionic equation for the disproportionation of

\mathrm{Mn}^{3+}

is:

2\mathrm{Mn}^{3+}(\mathrm{aq}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) + \mathrm{MnO}_{2}(\mathrm{~s}) + 4\mathrm{H}^{+}(\mathrm{aq})

Q22EXERCISES

Consider the elements: Cs, Ne, I and F

(a)

Identify the element that exhibits only negative oxidation state.

(b)

Identify the element that exhibits only postive oxidation state.

(c)

Identify the element that exhibits both positive and negative oxidation states.

(d)

Identify the element which exhibits neither the negative nor does the positive oxidation state.

Solution

(a) Identify the element that exhibits only negative oxidation state.

Answer: F (Fluorine)
Reason: Fluorine is the most electronegative element in the periodic table. In any compound it forms, it will always attract the bonding electrons more strongly than the other element, resulting in a fixed oxidation state of -1 (except in its elemental form $\mathrm{F}_{2}$ , where it is 0).

(b) Identify the element that exhibits only positive oxidation state.

Answer: Cs (Caesium)
Reason: Caesium is an alkali metal and is one of the most electropositive elements. It has a very strong tendency to lose its single valence electron to achieve a stable noble gas configuration. In its compounds, it always exhibits an oxidation state of +1 (except in its elemental form, where it is 0).

(c) Identify the element that exhibits both positive and negative oxidation states.

Answer: I (Iodine)
Reason: Iodine is a halogen with intermediate electronegativity.
- When it combines with more electropositive elements (like metals or hydrogen), it exhibits a negative oxidation state of -1 (e.g., in KI, HI).
- When it combines with more electronegative elements (like oxygen or fluorine), it exhibits positive oxidation states, such as +1, +3, +5, +7 (e.g., in $\mathrm{ICl}$ , $\mathrm{IF}_{5}$ , $\mathrm{HIO}_{3}$ ).

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Answer: Ne (Neon)
Reason: Neon is a noble gas. It has a completely filled valence electron shell, which makes it extremely stable and chemically inert. It does not readily form compounds with other elements, and therefore it does not typically exhibit positive or negative oxidation states. Its oxidation state is considered to be 0.

Q23EXERCISES

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Solution

Method: Half-Reaction (Ion-Electron) Method in Aqueous (Acidic) Medium

Step 1: Identify reactants and products and write the skeletal equation. Chlorine (

\mathrm{Cl}_{2}

) reacts with sulphur dioxide (

\mathrm{SO}_{2}

) in water. Chlorine, the oxidizing agent, is reduced to chloride ions (

\mathrm{Cl}^{-}

). Sulphur dioxide, the reducing agent, is oxidized to sulphate ions (

\mathrm{SO}_{4}^{2-}

).

\mathrm{Cl}_{2}(\mathrm{aq}) + \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq}) + \mathrm{SO}_{4}^{2-}(\mathrm{aq})

Step 2: Separate into two half-reactions.

Reduction Half-Reaction: Chlorine is reduced. $\mathrm{Cl}_{2}(\mathrm{aq}) \rightarrow \mathrm{Cl}^{-}(\mathrm{aq})$
Oxidation Half-Reaction: Sulphur dioxide is oxidized. $\mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{4}^{2-}(\mathrm{aq})$

Step 3: Balance atoms other than O and H.

Reduction: Balance Cl atoms. $\mathrm{Cl}_{2}(\mathrm{aq}) \rightarrow 2\mathrm{Cl}^{-}(\mathrm{aq})$
Oxidation: S atoms are already balanced.

Step 4: Balance O and H atoms.

Reduction: No O or H atoms.
Oxidation: Balance O atoms with $\mathrm{H}_{2} \mathrm{O}$ . Add $2\mathrm{H}_{2} \mathrm{O}$ to the left. $\mathrm{SO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{SO}_{4}^{2-}(\mathrm{aq})$ Balance H atoms with $\mathrm{H}^{+}$ . Add $4\mathrm{H}^{+}$ to the right. $\mathrm{SO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{SO}_{4}^{2-}(\mathrm{aq}) + 4\mathrm{H}^{+}(\mathrm{aq})$

Step 5: Balance the charges by adding electrons (e⁻).

Reduction: Charge on LHS is 0. Charge on RHS is -2. Add 2 e⁻ to the left. $\mathrm{Cl}_{2}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-}(\mathrm{aq})$
Oxidation: Charge on LHS is 0. Charge on RHS is $(-2) + (+4) = +2$ . Add 2 e⁻ to the right. $\mathrm{SO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{SO}_{4}^{2-}(\mathrm{aq}) + 4\mathrm{H}^{+}(\mathrm{aq}) + 2\mathrm{e}^{-}$

Step 6: Add the two half-reactions. The number of electrons (2 e⁻) is already equal in both half-reactions. We can add them directly and cancel the electrons.

(\mathrm{Cl}_{2} + 2\mathrm{e}^{-}) + (\mathrm{SO}_{2} + 2\mathrm{H}_{2} \mathrm{O}) \rightarrow (2\mathrm{Cl}^{-}) + (\mathrm{SO}_{4}^{2-} + 4\mathrm{H}^{+} + 2\mathrm{e}^{-})

Final Answer: The balanced equation for the redox change is:

\mathrm{Cl}_{2}(\mathrm{aq}) + \mathrm{SO}_{2}(\mathrm{~g}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2\mathrm{Cl}^{-}(\mathrm{aq}) + \mathrm{SO}_{4}^{2-}(\mathrm{aq}) + 4\mathrm{H}^{+}(\mathrm{aq})

(The products are effectively hydrochloric acid and sulphuric acid in solution.)

Q24EXERCISES

Refer to the periodic table given in your book and now answer the following questions:

(a)

Select the possible non metals that can show disproportionation reaction.

(b)

Select three metals that can show disproportionation reaction.

Solution

Concept: A disproportionation reaction requires an element to be in an intermediate oxidation state, from which it can be both oxidized to a higher state and reduced to a lower state. Therefore, elements that exhibit multiple oxidation states are candidates for disproportionation.

(a) Select the possible non-metals that can show disproportionation reaction.

Many non-metals, particularly from groups 15, 16, and 17 (excluding fluorine), can show disproportionation as they have a wide range of accessible oxidation states.

Phosphorus (P): Can exist in states from -3 to +5. White phosphorus (P₄, oxidation state 0) disproportionates in alkali to phosphine (-3) and hypophosphite (+1).
Sulphur (S): Can exist in states from -2 to +6. Sulphur (S₈, oxidation state 0) disproportionates in alkali to sulphide (-2) and thiosulphate (+2).
Chlorine (Cl): Can exist in states from -1 to +7. Chlorine (Cl₂, oxidation state 0) disproportionates in cold alkali to chloride (-1) and hypochlorite (+1), and in hot alkali to chloride (-1) and chlorate (+5).
Bromine (Br) and Iodine (I): Behave similarly to chlorine.

Possible non-metals: Phosphorus, Sulphur, Chlorine, Bromine, Iodine. (Note: Fluorine cannot as it only shows 0 and -1 states. Nitrogen disproportionation is less common but possible, e.g.,

\mathrm{HNO}_{2}

).

(b) Select three metals that can show disproportionation reaction.

Metals that can exhibit at least three different oxidation states (including the 0 state) are candidates. This is common for transition metals.

Copper (Cu): Copper can exist in 0, +1, and +2 oxidation states. The Copper(I) ion ( $\mathrm{Cu}^{+}$ ) is unstable in aqueous solution and disproportionates into Copper metal (0) and Copper(II) ion ( $\mathrm{Cu}^{2+}$ ). $2\mathrm{Cu}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}(\mathrm{s}) + \mathrm{Cu}^{2+}(\mathrm{aq})$
Manganese (Mn): Manganese has numerous oxidation states (+2, +3, +4, +6, +7). The Manganese(III) ion ( $\mathrm{Mn}^{3+}$ ) is unstable and disproportionates into Manganese(II) ion ( $\mathrm{Mn}^{2+}$ ) and Manganese dioxide ( $\mathrm{MnO}_{2}$ , with Mn in +4 state). $2\mathrm{Mn}^{3+}(\mathrm{aq}) + 2\mathrm{H}_{2}O(\mathrm{l}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) + \mathrm{MnO}_{2}(\mathrm{s}) + 4\mathrm{H}^{+}(\mathrm{aq})$
Gold (Au): Gold can exist in 0, +1, and +3 oxidation states. The Gold(I) ion ( $\mathrm{Au}^{+}$ ) can disproportionate into Gold metal (0) and Gold(III) ion ( $\mathrm{Au}^{3+}$ ). $3\mathrm{Au}^{+}(\mathrm{aq}) \rightarrow 2\mathrm{Au}(\mathrm{s}) + \mathrm{Au}^{3+}(\mathrm{aq})$

Three selected metals: Copper (Cu), Manganese (Mn), and Gold (Au).

Q25EXERCISES

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

Solution

Step 1: Write the balanced chemical equation. The reaction is the oxidation of ammonia (

\mathrm{NH}_{3}

) by oxygen (

\mathrm{O}_{2}

) to form nitric oxide (NO) and steam (

\mathrm{H}_{2}O

).

4\mathrm{NH}_{3}(\mathrm{g}) + 5\mathrm{O}_{2}(\mathrm{g}) \rightarrow 4\mathrm{NO}(\mathrm{g}) + 6\mathrm{H}_{2}O(\mathrm{g})

Step 2: Calculate the molar masses.

Molar mass of $\mathrm{NH}_{3}$ = 14.01 + 3(1.01) = 17.04 g/mol
Molar mass of $\mathrm{O}_{2}$ = 2(16.00) = 32.00 g/mol
Molar mass of NO = 14.01 + 16.00 = 30.01 g/mol

Step 3: Calculate the moles of each reactant.

Moles of $\mathrm{NH}_{3}$ = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{10.00 \mathrm{~g}}{17.04 \mathrm{~g/mol}} = 0.587 \mathrm{~mol}$
Moles of $\mathrm{O}_{2}$ = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{20.00 \mathrm{~g}}{32.00 \mathrm{~g/mol}} = 0.625 \mathrm{~mol}$

Step 4: Determine the limiting reactant. From the balanced equation, 4 moles of

\mathrm{NH}_{3}

react with 5 moles of

\mathrm{O}_{2}

. Let's find out how many moles of

\mathrm{O}_{2}

are required to react completely with the given moles of

\mathrm{NH}_{3}

.

Moles of $\mathrm{O}_{2}$ required = $0.587 \mathrm{~mol~NH}_{3} \times \frac{5 \mathrm{~mol~O}_{2}}{4 \mathrm{~mol~NH}_{3}} = 0.734 \mathrm{~mol~O}_{2}$

We have only 0.625 mol of

\mathrm{O}_{2}

, which is less than the required 0.734 mol. Therefore, oxygen ( $\mathrm{O}_{2}$ ) is the limiting reactant. The amount of product formed will be determined by the amount of

\mathrm{O}_{2}

.

Step 5: Calculate the moles of product (NO) formed. The calculation must be based on the limiting reactant,

\mathrm{O}_{2}

. From the balanced equation, 5 moles of

\mathrm{O}_{2}

produce 4 moles of NO.

Moles of NO produced = $0.625 \mathrm{~mol~O}_{2} \times \frac{4 \mathrm{~mol~NO}}{5 \mathrm{~mol~O}_{2}} = 0.500 \mathrm{~mol~NO}$

Step 6: Calculate the mass of NO produced.

Mass of NO = Moles of NO $\times$ Molar mass of NO
Mass of NO = $0.500 \mathrm{~mol} \times 30.01 \mathrm{~g/mol} = 15.005 \mathrm{~g}$

Final Answer: The maximum weight of nitric oxide that can be obtained is 15.01 g.

Q26EXERCISES

Using the standard electrode potentials given in the Table 7.1, predict if the reaction between the following is feasible:

(a)

\mathrm{Fe}^{3+}(\mathrm{aq})

and

\mathrm{I}^{-}(\mathrm{aq})

(b)

\mathrm{Ag}^{+}(\mathrm{aq})

and

\mathrm{Cu}(\mathrm{s})

(c)

\mathrm{Fe}^{3+}(\mathrm{aq})

and

\mathrm{Cu}(\mathrm{s})

(d)

\mathrm{Ag}(\mathrm{s})

and

\mathrm{Fe}^{3+}(\mathrm{aq})

(e)

\mathrm{Br}_{2}(\mathrm{aq})

and

\mathrm{Fe}^{2+}(\mathrm{aq})

.

Solution

Concept: A redox reaction is feasible (spontaneous) if the standard cell potential (

E°_{\text{cell}}

) is positive.

E°_{\text{cell}} = E°_{\text{cathode}} (\text{reduction}) - E°_{\text{anode}} (\text{oxidation})

Alternatively, a reaction is spontaneous if the species with the higher reduction potential gets reduced (acts as the oxidizing agent) and the species with the lower reduction potential gets oxidized (acts as the reducing agent).

Standard Electrode Potentials from Table 7.1:

$E°(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = +0.77 \mathrm{V}$
$E°(\mathrm{I}_{2}/\mathrm{I}^{-}) = +0.54 \mathrm{V}$
$E°(\mathrm{Ag}^{+}/\mathrm{Ag}) = +0.80 \mathrm{V}$
$E°(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34 \mathrm{V}$
$E°(\mathrm{Br}_{2}/\mathrm{Br}^{-}) = +1.09 \mathrm{V}$

(a) $\mathrm{Fe}^{3+}(\mathrm{aq})$ and $\mathrm{I}^{-}(\mathrm{aq})$

Possible reduction: $\mathrm{Fe}^{3+} + e^{-} \rightarrow \mathrm{Fe}^{2+}; E° = +0.77 \mathrm{V}$
Possible oxidation: $2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2} + 2e^{-}; E°_{\text{red}} = +0.54 \mathrm{V}$
Since $E°(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) > E°(\mathrm{I}_{2}/\mathrm{I}^{-})$ , $\mathrm{Fe}^{3+}$ will be reduced and $\mathrm{I}^{-}$ will be oxidized.
$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0.77 - 0.54 = +0.23 \mathrm{V}$ .
Since $E°_{\text{cell}}$ is positive, the reaction is feasible.

(b) $\mathrm{Ag}^{+}(\mathrm{aq})$ and $\mathrm{Cu}(\mathrm{s})$

Possible reduction: $\mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag}; E° = +0.80 \mathrm{V}$
Possible oxidation: $\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2e^{-}; E°_{\text{red}} = +0.34 \mathrm{V}$
Since $E°(\mathrm{Ag}^{+}/\mathrm{Ag}) > E°(\mathrm{Cu}^{2+}/\mathrm{Cu})$ , $\mathrm{Ag}^{+}$ will be reduced and Cu will be oxidized.
$E°_{\text{cell}} = 0.80 - 0.34 = +0.46 \mathrm{V}$ .
Since $E°_{\text{cell}}$ is positive, the reaction is feasible.

(c) $\mathrm{Fe}^{3+}(\mathrm{aq})$ and $\mathrm{Cu}(\mathrm{s})$

Possible reduction: $\mathrm{Fe}^{3+} + e^{-} \rightarrow \mathrm{Fe}^{2+}; E° = +0.77 \mathrm{V}$
Possible oxidation: $\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2e^{-}; E°_{\text{red}} = +0.34 \mathrm{V}$
Since $E°(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) > E°(\mathrm{Cu}^{2+}/\mathrm{Cu})$ , $\mathrm{Fe}^{3+}$ will be reduced and Cu will be oxidized.
$E°_{\text{cell}} = 0.77 - 0.34 = +0.43 \mathrm{V}$ .
Since $E°_{\text{cell}}$ is positive, the reaction is feasible.

(d) $\mathrm{Ag}(\mathrm{s})$ and $\mathrm{Fe}^{3+}(\mathrm{aq})$

Possible reduction: $\mathrm{Fe}^{3+} + e^{-} \rightarrow \mathrm{Fe}^{2+}; E° = +0.77 \mathrm{V}$
Possible oxidation: $\mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^{-}; E°_{\text{red}} = +0.80 \mathrm{V}$
Here, the potential oxidizing agent ( $\mathrm{Fe}^{3+}$ ) has a lower reduction potential than the potential reducing agent (Ag). The reaction will not proceed spontaneously in this direction.
$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0.77 - 0.80 = -0.03 \mathrm{V}$ .
Since $E°_{\text{cell}}$ is negative, the reaction is not feasible.

(e) $\mathrm{Br}_{2}(\mathrm{aq})$ and $\mathrm{Fe}^{2+}(\mathrm{aq})$

Possible reduction: $\mathrm{Br}_{2} + 2e^{-} \rightarrow 2\mathrm{Br}^{-}; E° = +1.09 \mathrm{V}$
Possible oxidation: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^{-}; E°_{\text{red}} = +0.77 \mathrm{V}$
Since $E°(\mathrm{Br}_{2}/\mathrm{Br}^{-}) > E°(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+})$ , $\mathrm{Br}_{2}$ will be reduced and $\mathrm{Fe}^{2+}$ will be oxidized.
$E°_{\text{cell}} = 1.09 - 0.77 = +0.32 \mathrm{V}$ .
Since $E°_{\text{cell}}$ is positive, the reaction is feasible.

Q27EXERCISES

Predict the products of electrolysis in each of the following:

(i)

An aqueous solution of $\mathrm{AgNO}_{3}$ with silver electrodes

(ii)

An aqueous solution $\mathrm{AgNO}_{3}$ with platinum electrodes

(iii)

A dilute solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$ with platinum electrodes

(iv)

An aqueous solution of $\mathrm{CuCl}_{2}$ with platinum electrodes.

Solution

Concept of Electrolysis:

Cathode (Negative Electrode): Reduction occurs. Cations migrate here. The species with the higher reduction potential is reduced.
Anode (Positive Electrode): Oxidation occurs. Anions migrate here. The species with the lower reduction potential (higher oxidation potential) is oxidized. Inert electrodes (like Pt, graphite) do not participate, while active electrodes (like Ag, Cu) can be oxidized.

(i) An aqueous solution of $\mathrm{AgNO}_{3}$ with silver electrodes

Species present: $\mathrm{Ag}^{+}(\mathrm{aq}), \mathrm{NO}_{3}^{-}(\mathrm{aq}), \mathrm{H}_{2}O(\mathrm{l})$ and Ag electrodes.
At Cathode: Possible reductions are of $\mathrm{Ag}^{+}$ $Ag^{+}$ and $\mathrm{H}_{2}O$ $H_{2} O$ .
- $\mathrm{Ag}^{+}(\mathrm{aq}) + e^{-} \rightarrow \mathrm{Ag}(\mathrm{s}); E° = +0.80 \mathrm{V}$
- $2\mathrm{H}_{2}O(\mathrm{l}) + 2e^{-} \rightarrow \mathrm{H}_{2}(\mathrm{g}) + 2\mathrm{OH}^{-}(\mathrm{aq}); E° = -0.83 \mathrm{V}$ Since the reduction potential of $\mathrm{Ag}^{+}$ is much higher, $\mathrm{Ag}^{+}$ ions are reduced. Product at Cathode: Silver (Ag) metal deposits.
At Anode (Active Ag electrode): Possible oxidations are of the Ag electrode, $\mathrm{NO}_{3}^{-}$ $NO_{3}^{-}$ , and $\mathrm{H}_{2}O$ $H_{2} O$ .
- $\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}^{+}(\mathrm{aq}) + e^{-}; E°_{ox} = -0.80 \mathrm{V}$
- $2\mathrm{H}_{2}O(\mathrm{l}) \rightarrow \mathrm{O}_{2}(\mathrm{g}) + 4\mathrm{H}^{+}(\mathrm{aq}) + 4e^{-}; E°_{ox} = -1.23 \mathrm{V}$ Oxidation of $\mathrm{NO}_{3}^{-}$ is very difficult. The oxidation of the silver electrode has a much higher oxidation potential (less negative) than water. Product at Anode: The silver anode dissolves to form $\mathrm{Ag}^{+}$ ions.

(ii) An aqueous solution of $\mathrm{AgNO}_{3}$ with platinum electrodes

Species present: $\mathrm{Ag}^{+}(\mathrm{aq}), \mathrm{NO}_{3}^{-}(\mathrm{aq}), \mathrm{H}_{2}O(\mathrm{l})$ and inert Pt electrodes.
At Cathode: Same as above. Product at Cathode: Silver (Ag) metal deposits.
At Anode (Inert Pt electrode): Possible oxidations are of $\mathrm{NO}_{3}^{-}$ $NO_{3}^{-}$ and $\mathrm{H}_{2}O$ $H_{2} O$ . The electrode does not react.
- $2\mathrm{H}_{2}O(\mathrm{l}) \rightarrow \mathrm{O}_{2}(\mathrm{g}) + 4\mathrm{H}^{+}(\mathrm{aq}) + 4e^{-}; E° = +1.23 \mathrm{V}$ Oxidation of $\mathrm{NO}_{3}^{-}$ has a much lower potential. Thus, water is oxidized. Product at Anode: Oxygen ( $\mathrm{O}_{2}$ ) gas evolves.

(iii) A dilute solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$ with platinum electrodes

Species present: $\mathrm{H}^{+}(\mathrm{aq}), \mathrm{SO}_{4}^{2-}(\mathrm{aq}), \mathrm{H}_{2}O(\mathrm{l})$ and inert Pt electrodes. This is essentially the electrolysis of water.
At Cathode: Possible reductions are of $\mathrm{H}^{+}$ $H^{+}$ and $\mathrm{H}_{2}O$ $H_{2} O$ . Reduction of $\mathrm{H}^{+}$ $H^{+}$ is preferred.
- $2\mathrm{H}^{+}(\mathrm{aq}) + 2e^{-} \rightarrow \mathrm{H}_{2}(\mathrm{g}); E° = 0.00 \mathrm{V}$ Product at Cathode: Hydrogen ( $\mathrm{H}_{2}$ ) gas evolves.
At Anode: Possible oxidations are of $\mathrm{SO}_{4}^{2-}$ $SO_{4}^{2 -}$ and $\mathrm{H}_{2}O$ $H_{2} O$ .
- $2\mathrm{H}_{2}O(\mathrm{l}) \rightarrow \mathrm{O}_{2}(\mathrm{g}) + 4\mathrm{H}^{+}(\mathrm{aq}) + 4e^{-}; E° = +1.23 \mathrm{V}$ Oxidation of sulphate ion ( $\mathrm{S}_{2}O_{8}^{2-}$ ) requires a much higher potential. Thus, water is oxidized. Product at Anode: Oxygen ( $\mathrm{O}_{2}$ ) gas evolves.

(iv) An aqueous solution of $\mathrm{CuCl}_{2}$ with platinum electrodes

Species present: $\mathrm{Cu}^{2+}(\mathrm{aq}), \mathrm{Cl}^{-}(\mathrm{aq}), \mathrm{H}_{2}O(\mathrm{l})$ and inert Pt electrodes.
At Cathode: Possible reductions are of $\mathrm{Cu}^{2+}$ $Cu^{2 +}$ and $\mathrm{H}_{2}O$ $H_{2} O$ .
- $\mathrm{Cu}^{2+}(\mathrm{aq}) + 2e^{-} \rightarrow \mathrm{Cu}(\mathrm{s}); E° = +0.34 \mathrm{V}$
- $2\mathrm{H}_{2}O(\mathrm{l}) + 2e^{-} \rightarrow \mathrm{H}_{2}(\mathrm{g}) + 2\mathrm{OH}^{-}(\mathrm{aq}); E° = -0.83 \mathrm{V}$ Since the reduction potential of $\mathrm{Cu}^{2+}$ is higher, it is reduced. Product at Cathode: Copper (Cu) metal deposits.
At Anode: Possible oxidations are of $\mathrm{Cl}^{-}$ $Cl^{-}$ and $\mathrm{H}_{2}O$ $H_{2} O$ .
- $2\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{g}) + 2e^{-}; E° = +1.36 \mathrm{V}$
- $2\mathrm{H}_{2}O(\mathrm{l}) \rightarrow \mathrm{O}_{2}(\mathrm{g}) + 4\mathrm{H}^{+}(\mathrm{aq}) + 4e^{-}; E° = +1.23 \mathrm{V}$ Although the standard potential for water oxidation is lower, due to the phenomenon of overpotential of oxygen on platinum, the oxidation of $\mathrm{Cl}^{-}$ occurs preferentially. Product at Anode: Chlorine ( $\mathrm{Cl}_{2}$ ) gas evolves.

Q28EXERCISES

Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn .

Solution

Concept: A metal can displace another metal from its salt solution if it is more reactive. The reactivity of metals is determined by their tendency to lose electrons (get oxidized). A more reactive metal has a more negative standard reduction potential (

E°

). The metal with the most negative

E°

will displace all others, and the metal with the least negative (or most positive)

E°

will be displaced by all others.

Step 1: Find the standard electrode potentials ( $E°$ ) for the given metals. From a standard electrochemical series table (like Table 7.1):

$E°(\mathrm{Mg}^{2+}/\mathrm{Mg}) = -2.36 \mathrm{V}$
$E°(\mathrm{Al}^{3+}/\mathrm{Al}) = -1.66 \mathrm{V}$
$E°(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76 \mathrm{V}$
$E°(\mathrm{Fe}^{2+}/\mathrm{Fe}) = -0.44 \mathrm{V}$
$E°(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34 \mathrm{V}$

Step 2: Arrange the metals in order of decreasing reactivity. The order of decreasing reactivity (or increasing standard reduction potential) is: Mg > Al > Zn > Fe > Cu

Step 3: State the displacement order. This order represents the ability of a metal to displace the ones that follow it from their salt solutions.

Mg can displace Al, Zn, Fe, and Cu.
Al can displace Zn, Fe, and Cu (but not Mg).
Zn can displace Fe and Cu (but not Mg and Al).
Fe can displace Cu (but not Mg, Al, and Zn).
Cu cannot displace any of the other metals from their salt solutions.

Final Answer: The order in which the metals displace each other from their salt solutions is: Mg > Al > Zn > Fe > Cu

Q29EXERCISES

Given the standard electrode potentials, $\mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}, \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}$ , $\mathrm{Hg}^{2+} / \mathrm{Hg}=0.79 \mathrm{~V}$ $\mathrm{Mg}^{2+} / \mathrm{Mg}=-2.37 \mathrm{~V} . \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.74 \mathrm{~V}$ arrange these metals in their increasing order of reducing power.

Solution

Concept: The reducing power of a substance is its ability to donate electrons and get oxidized. A stronger reducing agent is more easily oxidized. In terms of standard electrode potentials (

E°

), a more negative value indicates a greater tendency for the metal to be oxidized (and thus a stronger reducing agent).

Step 1: List the given metals and their standard reduction potentials ( $E°$ ).

$\mathrm{K}^{+} / \mathrm{K}: E° = -2.93 \mathrm{V}$
$\mathrm{Mg}^{2+} / \mathrm{Mg}: E° = -2.37 \mathrm{V}$
$\mathrm{Cr}^{3+} / \mathrm{Cr}: E° = -0.74 \mathrm{V}$
$\mathrm{Hg}^{2+} / \mathrm{Hg}: E° = +0.79 \mathrm{V}$
$\mathrm{Ag}^{+} / \mathrm{Ag}: E° = +0.80 \mathrm{V}$

Step 2: Arrange the metals based on their $E°$ values. To arrange the metals in increasing order of reducing power, we need to arrange them from the weakest reducing agent to the strongest. This corresponds to arranging them from the highest (most positive)

E°

value to the lowest (most negative)

E°

value.

Arranging the

E°

values from most positive to most negative: +0.80 V > +0.79 V > -0.74 V > -2.37 V > -2.93 V

Step 3: Write the corresponding metals in this order. This order corresponds to the following metals: Ag < Hg < Cr < Mg < K

Final Answer: The metals in their increasing order of reducing power are: Ag < Hg < Cr < Mg < K

Q30EXERCISES

Depict the galvanic cell in which the reaction $\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ takes place, Further show:

(i)

which of the electrode is negatively charged,

(ii)

the carriers of the current in the cell, and

(iii)

individual reaction at each electrode.

Solution

Depiction of the Galvanic Cell: The galvanic cell can be depicted using standard cell notation: $\mathrm{Zn}(\mathrm{s}) | \mathrm{Zn}^{2+}(\mathrm{aq}) || \mathrm{Ag}^{+}(\mathrm{aq}) | \mathrm{Ag}(\mathrm{s})$

This notation represents:

A zinc electrode immersed in a solution of zinc ions (the anode compartment).
A silver electrode immersed in a solution of silver ions (the cathode compartment).
A salt bridge (represented by ||) connecting the two compartments.

(iii) Individual reaction at each electrode. First, we identify the oxidation and reduction half-reactions from the overall equation.

Oxidation: Zinc loses electrons, its oxidation number increases from 0 to +2. Oxidation occurs at the anode. Anode Reaction: $\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-}$
Reduction: Silver ions gain electrons, their oxidation number decreases from +1 to 0. Reduction occurs at the cathode. Cathode Reaction: $\mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})$ (To balance electrons for the overall reaction, this is multiplied by 2: $2\mathrm{Ag}^{+} + 2e^{-} \rightarrow 2\mathrm{Ag}$ )

(i) Which of the electrode is negatively charged. In a galvanic (voltaic) cell, the anode is the site of oxidation, where electrons are produced. This build-up of electrons makes the anode the negative electrode.

Answer: The Zinc (Zn) electrode (the anode) is negatively charged.

(ii) The carriers of the current in the cell. The current is carried by different species in different parts of the cell.

External Circuit (the wire): The current is carried by the flow of electrons from the negative anode (Zn) to the positive cathode (Ag).
Internal Circuit (the solutions and salt bridge): The circuit is completed by the movement of ions. Cations (positive ions like $\mathrm{Zn}^{2+}$ and cations from the salt bridge, e.g., $\mathrm{K}^{+}$ ) move towards the cathode, and anions (negative ions like anions from the salt bridge, e.g., $\mathrm{Cl}^{-}$ or $\mathrm{NO}_{3}^{-}$ ) move towards the anode.
Answer: The carriers of current are electrons in the external wire and ions (cations and anions) in the electrolyte solutions and the salt bridge.

Q1In-text Problems

In the reactions given below, identify the species undergoing oxidation and reduction:

(i)

$\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{~g})+\mathrm{S}(\mathrm{s})$

(ii)

$3 \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+8 \mathrm{Al}(\mathrm{s}) \rightarrow 9 \mathrm{Fe}(\mathrm{s}) +4 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})$

(iii)

$2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaH}(\mathrm{s})$

Solution

Solution:

(i) $\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{~g})+\mathrm{S}(\mathrm{s})$

Oxidation: Hydrogen is removed from Hydrogen Sulphide ( $\mathrm{H}_{2} \mathrm{~S}$ ) to form Sulphur (S). Removal of hydrogen is oxidation. So, $\mathrm{H}_{2} \mathrm{~S}$ is oxidized.
Reduction: Hydrogen is added to Chlorine ( $\mathrm{Cl}_{2}$ ) to form Hydrogen Chloride (HCl). Addition of hydrogen is reduction. So, $\mathrm{Cl}_{2}$ is reduced.

(ii) $3 \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+8 \mathrm{Al}(\mathrm{s}) \rightarrow 9 \mathrm{Fe}(\mathrm{s}) +4 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})$

Oxidation: Oxygen is added to Aluminium (Al) to form Aluminium Oxide ( $\mathrm{Al}_{2} \mathrm{O}_{3}$ ). Addition of oxygen is oxidation. So, Al is oxidized.
Reduction: Oxygen is removed from Ferrous ferric oxide ( $\mathrm{Fe}_{3} \mathrm{O}_{4}$ ) to form Iron (Fe). Removal of oxygen is reduction. So, $\mathrm{Fe}_{3} \mathrm{O}_{4}$ is reduced.

(iii) $2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaH}(\mathrm{s})$

Oxidation: Sodium (Na) is an electropositive element. It combines with hydrogen to form Sodium Hydride (NaH). In this ionic compound ( $\mathrm{Na}^{+} \mathrm{H}^{-}$ ), sodium has lost an electron. Loss of electrons is oxidation. So, Na is oxidized.
Reduction: Hydrogen ( $\mathrm{H}_{2}$ ) is an electronegative element in this context. It combines with sodium, gaining an electron to form the hydride ion ( $\mathrm{H}^{-}$ ). Gain of electrons is reduction. So, $\mathrm{H}_{2}$ is reduced.

Q2In-text Problems

Justify that the reaction: $2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaH}(\mathrm{s})$ is a redox change.

Solution

Given Reaction:

2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaH}(\mathrm{s})

Justification: A redox reaction involves the transfer of electrons, where one species is oxidized (loses electrons) and another is reduced (gains electrons).

The product, Sodium Hydride (NaH), is an ionic compound composed of Sodium ions (

\mathrm{Na}^{+}

) and Hydride ions (

\mathrm{H}^{-}

). We can represent the formation of these ions from the neutral elements by splitting the overall reaction into two half-reactions:

Oxidation Half-Reaction: Each sodium atom loses one electron to become a sodium ion. This is oxidation. $2 \mathrm{Na}(\mathrm{s}) \rightarrow 2 \mathrm{Na}^{+}(\mathrm{g})+2 \mathrm{e}^{-}$
Reduction Half-Reaction: The hydrogen molecule gains two electrons, and each hydrogen atom becomes a hydride ion. This is reduction. $\mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}^{-}(\mathrm{g})$

Since the overall reaction involves both the loss of electrons by sodium (oxidation) and the gain of electrons by hydrogen (reduction) occurring simultaneously, it is a redox change.

Final Answer: The reaction is a redox change because sodium is oxidized and hydrogen is reduced.

Q3In-text Problems

Using Stock notation, represent the following compounds: $\mathrm{HAuCl}_{4}, \mathrm{Tl}_{2} \mathrm{O}, \mathrm{FeO}, \mathrm{Fe}_{2} \mathrm{O}_{3}, \mathrm{CuI}, \mathrm{CuO}, \mathrm{MnO}$ and $\mathrm{MnO}_{2}$ .

Solution

Concept: Stock notation represents the oxidation state of a metal in a compound using a Roman numeral in parentheses immediately following the symbol of the metal.

Calculations of Oxidation States: Let the oxidation state of the metal be

x

.

$\mathrm{HAuCl}_{4}$ : H is +1, Cl is -1. So, $(+1) + x + 4(-1) = 0 \implies x - 3 = 0 \implies x = +3$ . For Au.
$\mathrm{Tl}_{2} \mathrm{O}$ : O is -2. So, $2x + (-2) = 0 \implies 2x = 2 \implies x = +1$ . For Tl.
$\mathrm{FeO}$ : O is -2. So, $x + (-2) = 0 \implies x = +2$ . For Fe.
$\mathrm{Fe}_{2} \mathrm{O}_{3}$ : O is -2. So, $2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies x = +3$ . For Fe.
$\mathrm{CuI}$ : I is -1. So, $x + (-1) = 0 \implies x = +1$ . For Cu.
$\mathrm{CuO}$ : O is -2. So, $x + (-2) = 0 \implies x = +2$ . For Cu.
$\mathrm{MnO}$ : O is -2. So, $x + (-2) = 0 \implies x = +2$ . For Mn.
$\mathrm{MnO}_{2}$ : O is -2. So, $x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4$ . For Mn.

Representation using Stock Notation:

$\mathrm{HAuCl}_{4} \rightarrow \mathrm{HAu(III)Cl}_{4}$
$\mathrm{Tl}_{2} \mathrm{O} \rightarrow \mathrm{Tl}_{2}(\mathrm{I}) \mathrm{O}$
$\mathrm{FeO} \rightarrow \mathrm{Fe(II)O}$
$\mathrm{Fe}_{2} \mathrm{O}_{3} \rightarrow \mathrm{Fe}_{2}(\mathrm{III}) \mathrm{O}_{3}$
$\mathrm{CuI} \rightarrow \mathrm{Cu(I)I}$
$\mathrm{CuO} \rightarrow \mathrm{Cu(II)O}$
$\mathrm{MnO} \rightarrow \mathrm{Mn(II)O}$
$\mathrm{MnO}_{2} \rightarrow \mathrm{Mn(IV)O}_{2}$

Q4In-text Problems

Justify that the reaction: $2 \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{Cu}_{2} \mathrm{S}(\mathrm{s}) \rightarrow 6 \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{~g})$ is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.

Solution

Given Reaction:

2 \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{Cu}_{2} \mathrm{S}(\mathrm{s}) \rightarrow 6 \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{~g})

Step 1: Assign oxidation numbers to all elements. Let's assign oxidation numbers to each element in the reactants and products.

Reactants:
- In $\mathrm{Cu}_{2} \mathrm{O}$ : Oxygen (O) is -2. So, $2 \times (\text{O.N. of Cu}) + (-2) = 0 \implies$ O.N. of Cu is +1.
- In $\mathrm{Cu}_{2} \mathrm{S}$ : Sulphur (S) is -2. So, $2 \times (\text{O.N. of Cu}) + (-2) = 0 \implies$ O.N. of Cu is +1.
Products:
- In Cu(s): Copper is in its elemental state, so its O.N. is 0.
- In $\mathrm{SO}_{2}$ : Oxygen (O) is -2. So, $(\text{O.N. of S}) + 2 \times (-2) = 0 \implies$ O.N. of S is +4.

The reaction with oxidation numbers is:

2 \stackrel{+1}{\mathrm{Cu}}_{2} \stackrel{-2}{\mathrm{O}}(\mathrm{s}) + \stackrel{+1}{\mathrm{Cu}}_{2} \stackrel{-2}{\mathrm{S}}(\mathrm{s}) \rightarrow 6 \stackrel{0}{\mathrm{Cu}}(\mathrm{s}) + \stackrel{+4}{\mathrm{S}} \stackrel{-2}{\mathrm{O}}_{2}(\mathrm{~g})

Step 2: Identify changes in oxidation numbers.

Copper (Cu): The oxidation number of Cu decreases from +1 in both $\mathrm{Cu}_{2} \mathrm{O}$ and $\mathrm{Cu}_{2} \mathrm{S}$ to 0 in Cu(s). A decrease in oxidation number is reduction.
Sulphur (S): The oxidation number of S increases from -2 in $\mathrm{Cu}_{2} \mathrm{S}$ to +4 in $\mathrm{SO}_{2}$ . An increase in oxidation number is oxidation.

Step 3: Justify and identify agents.

Justification: Since the oxidation numbers of copper and sulphur change during the reaction, it is a redox reaction.
Species Oxidised: Sulphur in $\mathrm{Cu}_{2} \mathrm{S}$ is oxidised.
Species Reduced: Copper in both $\mathrm{Cu}_{2} \mathrm{O}$ and $\mathrm{Cu}_{2} \mathrm{S}$ is reduced.
Oxidant (Oxidising Agent): The oxidant is the species that gets reduced and causes oxidation. Here, $\mathrm{Cu}_{2} \mathrm{O}$ contains copper(I) which is reduced. It provides oxygen to oxidise sulphur. Therefore, $\mathrm{Cu}_{2} \mathrm{O}$ is the oxidant.
Reductant (Reducing Agent): The reductant is the species that gets oxidised and causes reduction. Here, $\mathrm{Cu}_{2} \mathrm{S}$ contains sulphur which is oxidised. It helps reduce copper from +1 to 0. Therefore, $\mathrm{Cu}_{2} \mathrm{S}$ is the reductant.

Q5In-text Problems

Which of the following species, do not show disproportionation reaction and why? $\mathrm{ClO}^{-}, \mathrm{ClO}_{2}^{-}, \mathrm{ClO}_{3}^{-}$ and $\mathrm{ClO}_{4}^{-}$ Also write reaction for each of the species that disproportionates.

Solution

Concept: A disproportionation reaction is a redox reaction in which a species is simultaneously oxidized and reduced. This is only possible if the element in the species is in an intermediate oxidation state, meaning it can both increase and decrease its oxidation state.

Step 1: Determine the oxidation state of Chlorine (Cl) in each oxoanion. Let the oxidation state of Cl be

x

. Oxygen is -2.

$\mathrm{ClO}^{-}$ : $x + (-2) = -1 \implies x = +1$
$\mathrm{ClO}_{2}^{-}$ : $x + 2(-2) = -1 \implies x - 4 = -1 \implies x = +3$
$\mathrm{ClO}_{3}^{-}$ : $x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5$
$\mathrm{ClO}_{4}^{-}$ : $x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7$

Step 2: Identify which species cannot disproportionate. The possible oxidation states of chlorine range from -1 (in

\mathrm{Cl}^{-}

) to +7. For disproportionation, an element must be able to be both oxidized (increase its oxidation state) and reduced (decrease its oxidation state).

In the perchlorate ion,

\mathrm{ClO}_{4}^{-}

, chlorine is in its highest possible oxidation state of +7. It can be reduced (e.g., to +5, +3, +1, -1), but it cannot be oxidized further. Therefore,

\mathrm{ClO}_{4}^{-}

cannot undergo a disproportionation reaction.

Step 3: Write disproportionation reactions for the other species.

Hypochlorite ion ( $\mathrm{ClO}^{-}$ ): Chlorine (+1) disproportionates to chloride (-1) and chlorate (+5). $3 \stackrel{+1}{\mathrm{ClO}}^{-}(\mathrm{aq}) \rightarrow 2 \stackrel{-1}{\mathrm{Cl}}^{-}(\mathrm{aq}) + \stackrel{+5}{\mathrm{ClO}}_{3}^{-}(\mathrm{aq})$
Chlorite ion ( $\mathrm{ClO}_{2}^{-}$ ): Chlorine (+3) disproportionates to chloride (-1) and chlorate (+5) or to chlorate (+5) and perchlorate (+7) depending on conditions. A common disproportionation is to chlorite (+3) and chlorate (+5). $2 \stackrel{+3}{\mathrm{ClO}}_{2}^{-}(\mathrm{aq}) \rightarrow \stackrel{+1}{\mathrm{ClO}}^{-}(\mathrm{aq}) + \stackrel{+5}{\mathrm{ClO}}_{3}^{-}(\mathrm{aq})$ Another example: $3 \stackrel{+3}{\mathrm{ClO}}_{2}^{-}(\mathrm{aq}) \rightarrow 2 \stackrel{+5}{\mathrm{ClO}}_{3}^{-}(\mathrm{aq}) + \stackrel{-1}{\mathrm{Cl}}^{-}(\mathrm{aq})$
Chlorate ion ( $\mathrm{ClO}_{3}^{-}$ ): Chlorine (+5) disproportionates upon heating to perchlorate (+7) and chloride (-1). $4 \stackrel{+5}{\mathrm{ClO}}_{3}^{-}(\mathrm{aq}) \xrightarrow{\Delta} \stackrel{-1}{\mathrm{Cl}}^{-}(\mathrm{aq}) + 3 \stackrel{+7}{\mathrm{ClO}}_{4}^{-}(\mathrm{aq})$

Final Answer:

\mathrm{ClO}_{4}^{-}

does not show a disproportionation reaction because chlorine is in its highest oxidation state (+7) and cannot be further oxidized.

Q6In-text Problems

Suggest a scheme of classification of the following redox reactions

(a)

\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{~g})

(b)

2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s}) \rightarrow 2 \mathrm{PbO}(\mathrm{s})+4 \mathrm{NO}_{2}(\mathrm{~g})+ \mathrm{O}_{2}(\mathrm{~g})

(c)

\mathrm{NaH}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})

(d)

2 \mathrm{NO}_{2}(\mathrm{~g})+2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}_{2}^{-}(\mathrm{aq})+ \mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

Solution

Scheme of Classification: Redox reactions can be classified into the following types: Combination, Decomposition, Displacement, and Disproportionation.

(a) $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{~g})$

Analysis: Two elements, Nitrogen and Oxygen, combine to form a single compound, Nitric oxide. Both reactants are in their elemental form (oxidation state 0) and form a compound where their oxidation states have changed (N is +2, O is -2).
Classification: This is a combination redox reaction.

(b) $2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{s}) \rightarrow 2 \mathrm{PbO}(\mathrm{s})+4 \mathrm{NO}_{2}(\mathrm{~g})+ \mathrm{O}_{2}(\mathrm{~g})$

Analysis: A single compound, Lead(II) nitrate, breaks down into multiple simpler substances. Let's check oxidation states. In $\mathrm{Pb(NO_3)_2}$ , N is +5 and O is -2. In $\mathrm{NO_2}$ , N is +4. In $\mathrm{O_2}$ , O is 0. Since the oxidation states of Nitrogen (from +5 to +4) and Oxygen (from -2 to 0) change, it is a redox reaction.
Classification: This is a decomposition redox reaction.

(c) $\mathrm{NaH}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})$

Analysis: In this reaction, the hydride ion ( $\mathrm{H}^{-}$ ) from NaH displaces the hydrogen (with oxidation state +1) from water to form dihydrogen gas ( $\mathrm{H}_{2}$ , oxidation state 0). The oxidation state of H in NaH changes from -1 to 0 (oxidation), and the oxidation state of H in $\mathrm{H_2O}$ changes from +1 to 0 (reduction). This is a displacement reaction.
Classification: This is a displacement redox reaction (specifically, non-metal displacement).

(d) $2 \mathrm{NO}_{2}(\mathrm{~g})+2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}_{2}^{-}(\mathrm{aq})+ \mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Analysis: Let's check the oxidation state of Nitrogen. In the reactant $\mathrm{NO}_{2}$ , the oxidation state of N is +4. In the products, the oxidation state of N is +3 in $\mathrm{NO}_{2}^{-}$ and +5 in $\mathrm{NO}_{3}^{-}$ . The same element, Nitrogen, is being simultaneously reduced (from +4 to +3) and oxidized (from +4 to +5).
Classification: This is a disproportionation redox reaction.

Q7In-text Problems

Why do the following reactions proceed differently ? $\mathrm{Pb}_{3} \mathrm{O}_{4}+8 \mathrm{HCl} \rightarrow 3 \mathrm{PbCl}_{2}+\mathrm{Cl}_{2}+4 \mathrm{H}_{2} \mathrm{O}$ and $\mathrm{Pb}_{3} \mathrm{O}_{4}+4 \mathrm{HNO}_{3} \rightarrow 2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{PbO}_{2}+ 2 \mathrm{H}_{2} \mathrm{O}$

Solution

Understanding the Reactant $\mathrm{Pb}_{3} \mathrm{O}_{4}$ : First, it is important to understand that

\mathrm{Pb}_{3} \mathrm{O}_{4}

(red lead) is a mixed oxide, best represented as a stoichiometric mixture of 2 moles of Lead(II) oxide (PbO) and 1 mole of Lead(IV) oxide (

\mathrm{PbO}_{2}

). So, we can write it as

2\mathrm{PbO} \cdot \mathrm{PbO}_{2}

.

In PbO, lead has an oxidation state of +2. In

\mathrm{PbO}_{2}

, lead has an oxidation state of +4.

Reaction with HCl: Hydrochloric acid (HCl) acts as both an acid and a reducing agent (due to the

\mathrm{Cl}^{-}

ion).

Acid-Base Reaction: The basic component, PbO, reacts with the acid HCl. $2\mathrm{PbO} + 4\mathrm{HCl} \rightarrow 2\mathrm{PbCl}_{2} + 2\mathrm{H}_{2}O$
Redox Reaction: The oxidizing component, $\mathrm{PbO}_{2}$ (with Pb in +4 state), oxidizes the chloride ion ( $\mathrm{Cl}^{-}$ ) from HCl to chlorine gas ( $\mathrm{Cl}_{2}$ ). In this process, Pb(+4) is reduced to Pb(+2). $\mathrm{PbO}_{2} + 4\mathrm{HCl} \rightarrow \mathrm{PbCl}_{2} + \mathrm{Cl}_{2} + 2\mathrm{H}_{2}O$

Combining these two reactions gives the overall equation:

(2\mathrm{PbO} + \mathrm{PbO}_{2}) + 8\mathrm{HCl} \rightarrow 3\mathrm{PbCl}_{2} + \mathrm{Cl}_{2} + 4\mathrm{H}_{2}O

Reaction with $\mathrm{HNO}_{3}$ : Nitric acid (

\mathrm{HNO}_{3}

) is an acid and also a strong oxidizing agent.

Acid-Base Reaction: The basic component, PbO, reacts with the acid $\mathrm{HNO}_{3}$ . $2\mathrm{PbO} + 4\mathrm{HNO}_{3} \rightarrow 2\mathrm{Pb(NO_3)_2} + 2\mathrm{H}_{2}O$
Interaction with $\mathrm{PbO}_{2}$ : The other component, $\mathrm{PbO}_{2}$ , also contains an oxidizing agent (Pb in +4 state). Since nitric acid is also an oxidizing agent, no redox reaction occurs between $\mathrm{PbO}_{2}$ and $\mathrm{HNO}_{3}$ . $\mathrm{PbO}_{2}$ is passive and does not react with dilute nitric acid.

Combining the reacting and non-reacting parts gives the overall equation:

(2\mathrm{PbO} + \mathrm{PbO}_{2}) + 4\mathrm{HNO}_{3} \rightarrow 2\mathrm{Pb(NO_3)_2} + \mathrm{PbO}_{2} + 2\mathrm{H}_{2}O

Conclusion: The reactions proceed differently because HCl can act as a reducing agent, reducing

\mathrm{PbO}_{2}

to

\mathrm{PbCl}_{2}

. In contrast,

\mathrm{HNO}_{3}

is an oxidizing agent and cannot reduce

\mathrm{PbO}_{2}

; it only acts as an acid, reacting with the basic PbO component of the mixed oxide.

Q8In-text Problems

Write the net ionic equation for the reaction of potassium dichromate(VI), $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ with sodium sulphite, $\mathrm{Na}_{2} \mathrm{SO}_{3}$ , in an acid solution to give chromium(III) ion and the sulphate ion.

Solution

Method: Oxidation Number Method

Step 1: Write the skeletal ionic equation. The reactants are dichromate ion (

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}

) and sulphite ion (

\mathrm{SO}_{3}^{2-}

). The products are chromium(III) ion (

\mathrm{Cr}^{3+}

) and sulphate ion (

\mathrm{SO}_{4}^{2-}

).

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+\mathrm{SO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})

Step 2: Assign oxidation numbers to the elements undergoing change.

In $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ , Cr has an oxidation number of +6.
In $\mathrm{SO}_{3}^{2-}$ , S has an oxidation number of +4.
In $\mathrm{Cr}^{3+}$ , Cr has an oxidation number of +3.
In $\mathrm{SO}_{4}^{2-}$ , S has an oxidation number of +6. $\stackrel{+6}{\mathrm{Cr}}_{2} \mathrm{O}_{7}^{2-} + \stackrel{+4}{\mathrm{SO}}_{3}^{2-} \rightarrow \stackrel{+3}{\mathrm{Cr}}^{3+} + \stackrel{+6}{\mathrm{SO}}_{4}^{2-}$

Step 3: Calculate the change in oxidation number and balance the atoms.

Reduction: The oxidation number of each Cr atom decreases by 3 (from +6 to +3). Since there are two Cr atoms in $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ , the total decrease is $2 \times 3 = 6$ .
Oxidation: The oxidation number of the S atom increases by 2 (from +4 to +6). The total increase is 2.

To make the total increase and decrease equal, we must multiply the change for S by 3. This means we need 3 atoms of S. We place a coefficient of 3 in front of

\mathrm{SO}_{3}^{2-}

and

\mathrm{SO}_{4}^{2-}

. We also balance the Cr atoms by placing a 2 in front of

\mathrm{Cr}^{3+}

.

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) + 3\mathrm{SO}_{3}^{2-}(\mathrm{aq}) \rightarrow 2\mathrm{Cr}^{3+}(\mathrm{aq}) + 3\mathrm{SO}_{4}^{2-}(\mathrm{aq})

Step 4: Balance the ionic charges. The reaction takes place in an acidic solution, so we use

\mathrm{H}^{+}

ions to balance the charge.

Charge on the left side (LHS): $(-2) + 3(-2) = -8$ .
Charge on the right side (RHS): $2(+3) + 3(-2) = +6 - 6 = 0$ . To balance the charge, we need to add 8 positive charges to the LHS. We add $8\mathrm{H}^{+}$ . $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) + 3\mathrm{SO}_{3}^{2-}(\mathrm{aq}) + 8\mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2\mathrm{Cr}^{3+}(\mathrm{aq}) + 3\mathrm{SO}_{4}^{2-}(\mathrm{aq})$

Step 5: Balance the hydrogen and oxygen atoms. Balance the H atoms by adding water (

\mathrm{H}_{2} \mathrm{O}

) molecules.

H atoms on LHS: 8.
H atoms on RHS: 0. We add $4\mathrm{H}_{2} \mathrm{O}$ to the RHS. $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) + 3\mathrm{SO}_{3}^{2-}(\mathrm{aq}) + 8\mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2\mathrm{Cr}^{3+}(\mathrm{aq}) + 3\mathrm{SO}_{4}^{2-}(\mathrm{aq}) + 4\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Now, check the oxygen atoms to verify:

O atoms on LHS: $7 + (3 \times 3) = 16$ .
O atoms on RHS: $(3 \times 4) + 4 = 12 + 4 = 16$ . The equation is now fully balanced.

Final Answer: The net ionic equation is:

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) + 3\mathrm{SO}_{3}^{2-}(\mathrm{aq}) + 8\mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2\mathrm{Cr}^{3+}(\mathrm{aq}) + 3\mathrm{SO}_{4}^{2-}(\mathrm{aq}) + 4\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

Q9In-text Problems

Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

Solution

Method: Oxidation Number Method

Step 1: Write the skeletal ionic equation. The reactants are permanganate ion (

\mathrm{MnO}_{4}^{-}

) and bromide ion (

\mathrm{Br}^{-}

). The products are manganese dioxide (

\mathrm{MnO}_{2}

) and bromate ion (

\mathrm{BrO}_{3}^{-}

).

\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + \mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s}) + \mathrm{BrO}_{3}^{-}(\mathrm{aq})

Step 2: Assign oxidation numbers.

In $\mathrm{MnO}_{4}^{-}$ , Mn is +7.
In $\mathrm{Br}^{-}$ , Br is -1.
In $\mathrm{MnO}_{2}$ , Mn is +4.
In $\mathrm{BrO}_{3}^{-}$ , Br is +5. $\stackrel{+7}{\mathrm{Mn}}\mathrm{O}_{4}^{-} + \stackrel{-1}{\mathrm{Br}}^{-} \rightarrow \stackrel{+4}{\mathrm{Mn}}\mathrm{O}_{2} + \stackrel{+5}{\mathrm{Br}}\mathrm{O}_{3}^{-}$

Step 3: Balance the change in oxidation numbers.

Reduction: The oxidation number of Mn decreases by 3 (from +7 to +4). Total decrease = 3.
Oxidation: The oxidation number of Br increases by 6 (from -1 to +5). Total increase = 6.

To equalize the change, we multiply the Mn species by 2.

2\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + \mathrm{Br}^{-}(\mathrm{aq}) \rightarrow 2\mathrm{MnO}_{2}(\mathrm{~s}) + \mathrm{BrO}_{3}^{-}(\mathrm{aq})

Now, the total decrease is

2 \times 3 = 6

, which equals the total increase of 6.

Step 4: Balance the ionic charges. The reaction is in a basic medium, so we use

\mathrm{OH}^{-}

ions.

Charge on LHS: $2(-1) + (-1) = -3$ .
Charge on RHS: $0 + (-1) = -1$ . To balance the charges, we need to make the LHS more positive or the RHS more negative. We add $2\mathrm{OH}^{-}$ to the RHS to make the charge on both sides -3. $2\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + \mathrm{Br}^{-}(\mathrm{aq}) \rightarrow 2\mathrm{MnO}_{2}(\mathrm{~s}) + \mathrm{BrO}_{3}^{-}(\mathrm{aq}) + 2\mathrm{OH}^{-}(\mathrm{aq})$

Step 5: Balance the hydrogen and oxygen atoms. Balance the H atoms by adding water (

\mathrm{H}_{2} \mathrm{O}

) molecules.

H atoms on LHS: 0.
H atoms on RHS: 2. We add $1\mathrm{H}_{2} \mathrm{O}$ to the LHS. $2\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + \mathrm{Br}^{-}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2\mathrm{MnO}_{2}(\mathrm{~s}) + \mathrm{BrO}_{3}^{-}(\mathrm{aq}) + 2\mathrm{OH}^{-}(\mathrm{aq})$

Check the oxygen atoms to verify:

O atoms on LHS: $(2 \times 4) + 1 = 9$ .
O atoms on RHS: $(2 \times 2) + 3 + 2 = 4 + 3 + 2 = 9$ . The equation is fully balanced.

Final Answer: The balanced ionic equation is:

2\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + \mathrm{Br}^{-}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2\mathrm{MnO}_{2}(\mathrm{~s}) + \mathrm{BrO}_{3}^{-}(\mathrm{aq}) + 2\mathrm{OH}^{-}(\mathrm{aq})

Q10In-text Problems

Permanganate(VII) ion, $\mathrm{MnO}_{4}^{-}$ in basic solution oxidises iodide ion, $\mathrm{I}^{-}$ to produce molecular iodine ( $\mathrm{I}_{2}$ ) and manganese (IV) oxide ( $\mathrm{MnO}_{2}$ ). Write a balanced ionic equation to represent this redox reaction.

Solution

Method: Half-Reaction (Ion-Electron) Method

Step 1: Write the skeletal ionic equation.

\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + \mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s}) + \mathrm{I}_{2}(\mathrm{~s})

Step 2: Separate the equation into two half-reactions.

Oxidation Half-Reaction: Iodide is oxidized to iodine. $\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_{2}(\mathrm{~s})$
Reduction Half-Reaction: Permanganate is reduced to manganese dioxide. $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s})$

Step 3: Balance atoms other than O and H.

Oxidation half: Balance I atoms. $2\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_{2}(\mathrm{~s})$
Reduction half: Mn atoms are already balanced. $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s})$

Step 4: Balance O and H atoms for a basic medium.

Reduction Half-Reaction:
- Balance O atoms by adding $\mathrm{H}_{2} \mathrm{O}$ . There are 4 O on the left and 2 O on the right. Add $2\mathrm{H}_{2} \mathrm{O}$ to the right. $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
- Balance H atoms by adding $\mathrm{H}_{2} \mathrm{O}$ and $\mathrm{OH}^{-}$ . There are 4 H on the right. Add $4\mathrm{H}_{2} \mathrm{O}$ to the left and $4\mathrm{OH}^{-}$ to the right. This simplifies to adding $2\mathrm{H}_{2} \mathrm{O}$ to the left. $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{MnO}_{2}(\mathrm{~s}) + 4\mathrm{OH}^{-}(\mathrm{aq})$

Step 5: Balance the charges by adding electrons (e⁻).

Oxidation Half-Reaction: Charge on LHS is -2. Charge on RHS is 0. Add 2 e⁻ to the right. $2\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}_{2}(\mathrm{~s}) + 2\mathrm{e}^{-}$
Reduction Half-Reaction: Charge on LHS is -1. Charge on RHS is -4. Add 3 e⁻ to the left. $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 2\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) + 3\mathrm{e}^{-} \rightarrow \mathrm{MnO}_{2}(\mathrm{~s}) + 4\mathrm{OH}^{-}(\mathrm{aq})$

Step 6: Equalize the number of electrons in both half-reactions. The number of electrons lost must equal the number of electrons gained. The least common multiple of 2 and 3 is 6.

Multiply the oxidation half-reaction by 3. $6\mathrm{I}^{-}(\mathrm{aq}) \rightarrow 3\mathrm{I}_{2}(\mathrm{~s}) + 6\mathrm{e}^{-}$
Multiply the reduction half-reaction by 2. $2\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 4\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) + 6\mathrm{e}^{-} \rightarrow 2\mathrm{MnO}_{2}(\mathrm{~s}) + 8\mathrm{OH}^{-}(\mathrm{aq})$

Step 7: Add the two half-reactions together and cancel common species.

6\mathrm{I}^{-}(\mathrm{aq}) + 2\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 4\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) + 6\mathrm{e}^{-} \rightarrow 3\mathrm{I}_{2}(\mathrm{~s}) + 6\mathrm{e}^{-} + 2\mathrm{MnO}_{2}(\mathrm{~s}) + 8\mathrm{OH}^{-}(\mathrm{aq})

Cancel the

6\mathrm{e}^{-}

from both sides.

Final Answer: The balanced ionic equation is:

2\mathrm{MnO}_{4}^{-}(\mathrm{aq}) + 6\mathrm{I}^{-}(\mathrm{aq}) + 4\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2\mathrm{MnO}_{2}(\mathrm{~s}) + 3\mathrm{I}_{2}(\mathrm{~s}) + 8\mathrm{OH}^{-}(\mathrm{aq})

Redox ReactionsClass 11 Chemistry NCERT Solutions

Assign oxidation number to the underlined elements in each of the following species:

Solution

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?

Solution

Justify that the following reactions are redox reactions:

Solution

Fluorine reacts with ice and results in the change: H2O(s)+F2( g)→HF( g)+HOF( g)\mathrm{H}_{2} \mathrm{O}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{~g})+\mathrm{HOF}(\mathrm{~g})H2​O(s)+F2​( g)→HF( g)+HOF( g) Justify that this reaction is a redox reaction.

Solution

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5,Cr2O72−\mathrm{H}_{2} \mathrm{SO}_{5}, \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}H2​SO5​,Cr2​O72−​ and NO3−\mathrm{NO}_{3}^{-}NO3−​. Suggest structure of these compounds. Count for the fallacy.

Solution

Write formulas for the following compounds:

Solution

Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

Solution

While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?

Solution

Consider the reactions:

Solution

The compound AgF2\mathrm{AgF}_{2}AgF2​ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?

Solution

Solution

How do you count for the following observations ?

Solution

Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

Solution

Solution

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Solution

Solution

Consider the reactions:

Solution

Balance the following redox reactions by ion - electron method :

Solution

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

Solution

Solution

The Mn3+\mathrm{Mn}^{3+}Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+\mathrm{Mn}^{2+}Mn2+, MnO2\mathrm{MnO}_{2}MnO2​, and H+\mathrm{H}^{+}H+ ion. Write a balanced ionic equation for the reaction.

Solution

Consider the elements: Cs, Ne, I and F

Solution

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Solution

Refer to the periodic table given in your book and now answer the following questions:

Solution

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

Solution

Using the standard electrode potentials given in the Table 7.1, predict if the reaction between the following is feasible:

Solution

Predict the products of electrolysis in each of the following:

An aqueous solution of AgNO3\mathrm{AgNO}_{3}AgNO3​ with silver electrodes

An aqueous solution AgNO3\mathrm{AgNO}_{3}AgNO3​ with platinum electrodes

A dilute solution of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4}H2​SO4​ with platinum electrodes

An aqueous solution of CuCl2\mathrm{CuCl}_{2}CuCl2​ with platinum electrodes.

Solution

Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn .

Solution

Solution

Depict the galvanic cell in which the reaction Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s)\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s) takes place, Further show:

which of the electrode is negatively charged,

the carriers of the current in the cell, and

individual reaction at each electrode.

Solution

In the reactions given below, identify the species undergoing oxidation and reduction:

H2 S( g)+Cl2( g)→2HCl( g)+S(s)\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{~g})+\mathrm{S}(\mathrm{s})H2​ S( g)+Cl2​( g)→2HCl( g)+S(s)

3Fe3O4( s)+8Al(s)→9Fe(s)+4Al2O3( s)3 \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+8 \mathrm{Al}(\mathrm{s}) \rightarrow 9 \mathrm{Fe}(\mathrm{s}) +4 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})3Fe3​O4​( s)+8Al(s)→9Fe(s)+4Al2​O3​( s)

2Na(s)+H2( g)→2NaH(s)2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaH}(\mathrm{s})2Na(s)+H2​( g)→2NaH(s)

Solution

Justify that the reaction: 2Na(s)+H2( g)→2NaH(s)2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaH}(\mathrm{s})2Na(s)+H2​( g)→2NaH(s) is a redox change.

Solution

Solution

Solution

Solution

Suggest a scheme of classification of the following redox reactions

Solution

Solution

Write the net ionic equation for the reaction of potassium dichromate(VI), K2Cr2O7\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}K2​Cr2​O7​ with sodium sulphite, Na2SO3\mathrm{Na}_{2} \mathrm{SO}_{3}Na2​SO3​, in an acid solution to give chromium(III) ion and the sulphate ion.

Solution

Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

Solution

Fluorine reacts with ice and results in the change: $\mathrm{H}_{2} \mathrm{O}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{~g})+\mathrm{HOF}(\mathrm{~g})$ Justify that this reaction is a redox reaction.

Calculate the oxidation number of sulphur, chromium and nitrogen in $\mathrm{H}_{2} \mathrm{SO}_{5}, \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ and $\mathrm{NO}_{3}^{-}$ . Suggest structure of these compounds. Count for the fallacy.

The compound $\mathrm{AgF}_{2}$ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?

The $\mathrm{Mn}^{3+}$ ion is unstable in solution and undergoes disproportionation to give $\mathrm{Mn}^{2+}$ , $\mathrm{MnO}_{2}$ , and $\mathrm{H}^{+}$ ion. Write a balanced ionic equation for the reaction.

An aqueous solution of $\mathrm{AgNO}_{3}$ with silver electrodes

An aqueous solution $\mathrm{AgNO}_{3}$ with platinum electrodes

A dilute solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$ with platinum electrodes

An aqueous solution of $\mathrm{CuCl}_{2}$ with platinum electrodes.

Depict the galvanic cell in which the reaction $\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ takes place, Further show:

$\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{~g})+\mathrm{S}(\mathrm{s})$

$3 \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+8 \mathrm{Al}(\mathrm{s}) \rightarrow 9 \mathrm{Fe}(\mathrm{s}) +4 \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})$

$2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaH}(\mathrm{s})$

Justify that the reaction: $2 \mathrm{Na}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NaH}(\mathrm{s})$ is a redox change.

Write the net ionic equation for the reaction of potassium dichromate(VI), $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ with sodium sulphite, $\mathrm{Na}_{2} \mathrm{SO}_{3}$ , in an acid solution to give chromium(III) ion and the sulphate ion.

Permanganate(VII) ion, $\mathrm{MnO}_{4}^{-}$ in basic solution oxidises iodide ion, $\mathrm{I}^{-}$ to produce molecular iodine ( $\mathrm{I}_{2}$ ) and manganese (IV) oxide ( $\mathrm{MnO}_{2}$ ). Write a balanced ionic equation to represent this redox reaction.