Chapter Notes

REDOX REACTIONS

A redox reaction is a type of chemical reaction where both oxidation and reduction occur simultaneously. These reactions are fundamental to countless processes in chemistry, biology, and industry, from the burning of fuel to the functioning of batteries and even biological processes within our bodies. The core principle is simple: where there is oxidation, there is always reduction.

Classical Idea of Redox Reactions

The earliest understanding of redox reactions was based on the addition or removal of oxygen and hydrogen.

Oxidation

Originally, oxidation was defined as the addition of oxygen to a substance. This definition was later expanded.

Classical Definitions of Oxidation:

• Addition of oxygen: When magnesium burns, it combines with oxygen. $2 \text{Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{MgO(s)}$
• Addition of an electronegative element: Magnesium reacts with chlorine, an electronegative element. $\text{Mg(s)} + \text{Cl}_2\text{(g)} \rightarrow \text{MgCl}_2\text{(s)}$
• Removal of hydrogen: In this reaction, hydrogen is removed from hydrogen sulphide ($H_2S$). $2 \text{H}_2\text{S(g)} + \text{O}_2\text{(g)} \rightarrow 2 \text{S(s)} + 2 \text{H}_2\text{O(l)}$
• Removal of an electropositive element: Potassium (an electropositive element) is removed from potassium ferrocyanide. $2 \text{K}_4[\text{Fe(CN)}_6]\text{(aq)} + \text{H}_2\text{O}_2\text{(aq)} \rightarrow 2 \text{K}_3[\text{Fe(CN)}_6]\text{(aq)} + 2 \text{KOH(aq)}$

In summary, oxidation is the addition of oxygen or another electronegative element, or the removal of hydrogen or another electropositive element.

Reduction

Reduction was initially defined as the removal of oxygen. Like oxidation, this definition was also broadened over time.

Classical Definitions of Reduction:

• Removal of oxygen: Mercuric oxide decomposes, and oxygen is removed from it. $2 \text{HgO(s)} \xrightarrow{\Delta} 2 \text{Hg(l)} + \text{O}_2\text{(g)}$
• Removal of an electronegative element: Chlorine (an electronegative element) is removed from ferric chloride. $2 \text{FeCl}_3\text{(aq)} + \text{H}_2\text{(g)} \rightarrow 2 \text{FeCl}_2\text{(aq)} + 2 \text{HCl(aq)}$
• Addition of hydrogen: Hydrogen is added to ethene to form ethane. $\text{CH}_2=\text{CH}_2\text{(g)} + \text{H}_2\text{(g)} \rightarrow \text{H}_3\text{C}-\text{CH}_3$
• Addition of an electropositive element: Mercury (an electropositive element) is added to mercuric chloride. $2 \text{HgCl}_2\text{(aq)} + \text{SnCl}_2\text{(aq)} \rightarrow \text{Hg}_2\text{Cl}_2\text{(s)} + \text{SnCl}_4\text{(aq)}$

In summary, reduction is the removal of oxygen or another electronegative element, or the addition of hydrogen or another electropositive element.

Redox Reactions in Terms of Electron Transfer

A more modern and comprehensive way to understand redox reactions is by looking at the transfer of electrons between species.

• Oxidation is the loss of electron(s) by any species.
• Reduction is the gain of electron(s) by any species.
• An oxidising agent (or oxidant) is a substance that accepts electrons and gets reduced.
• A reducing agent (or reductant) is a substance that donates electrons and gets oxidised.

Let's look at the formation of sodium chloride ($NaCl$): $2 \text{Na(s)} + \text{Cl}_2\text{(g)} \rightarrow 2 \text{NaCl(s)}$

This reaction can be broken down into two half-reactions:

1. Oxidation Half-Reaction (loss of electrons): Sodium atoms lose electrons to become sodium ions. $2 \text{Na(s)} \rightarrow 2 \text{Na}^+\text{(g)} + 2 \text{e}^-$
2. Reduction Half-Reaction (gain of electrons): Chlorine molecules gain electrons to become chloride ions. $\text{Cl}_2\text{(g)} + 2 \text{e}^- \rightarrow 2 \text{Cl}^-\text{(g)}$

Here, sodium acts as the reducing agent because it donates electrons, and chlorine acts as the oxidising agent because it accepts electrons.

Given

• Reaction: $2 \text{Na(s)} + \text{H}_2\text{(g)} \rightarrow 2 \text{NaH(s)}$
• The product, sodium hydride ($NaH$), is an ionic compound represented as $\text{Na}^+\text{H}^-$.

To Find

Justify that the reaction is a redox reaction by showing the half-reactions.

Solution

We can split the reaction into two half-reactions based on the transfer of electrons.

Oxidation Half-Reaction: Sodium (Na) loses an electron to form the sodium ion ($\text{Na}^+$). $2 \text{Na(s)} \rightarrow 2 \text{Na}^+\text{(g)} + 2 \text{e}^-$

Reduction Half-Reaction: Hydrogen ($\text{H}_2$) gains an electron to form the hydride ion ($\text{H}^-$). $\text{H}_2\text{(g)} + 2 \text{e}^- \rightarrow 2 \text{H}^-\text{(g)}$

Since one species (sodium) is losing electrons (oxidation) and another species (hydrogen) is gaining electrons (reduction), the overall reaction is a redox change.

Competitive Electron Transfer Reactions

Some metals have a stronger tendency to lose electrons than others. This creates a competition for electron release, which determines the direction of a redox reaction.

• The zinc strip gets coated with reddish-brown copper metal.
• The blue color of the solution fades.

This happens because of the reaction: $\text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)}$

In this reaction, zinc has a stronger tendency to lose electrons than copper.

• Zinc is oxidised: It loses two electrons ($\text{Zn} \rightarrow \text{Zn}^{2+} + 2 \text{e}^-$).
• Copper ions are reduced: They gain two electrons ($\text{Cu}^{2+} + 2 \text{e}^- \rightarrow \text{Cu}$).

However, if you place a copper strip in a zinc sulfate solution, no reaction occurs. This tells us that zinc is a better reducing agent (electron donor) than copper.

Similarly, if you place a copper strip in a silver nitrate solution, the copper will be oxidised and silver ions will be reduced: $\text{Cu(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{Ag(s)}$ This shows that copper is a better reducing agent than silver.

From these experiments, we can establish an order of electron-releasing tendency, also known as the metal activity series: $\text{Zn} > \text{Cu} > \text{Ag}$ This series helps predict which redox reactions will occur spontaneously.

Oxidation Number

For covalent compounds where electrons are shared, not completely transferred, the concept of oxidation number is used to track electron shifts.

Oxidation number (or oxidation state) is the imaginary charge an atom would have if all its bonds to other atoms were completely ionic. This is a bookkeeping tool based on a set of rules.

Rules for Assigning Oxidation Number:

1. Free Elements: The oxidation number of an atom in its elemental form (e.g., $\text{H}_2$, $\text{O}_2$, $\text{Na}$, $\text{S}_8$) is zero.
2. Monatomic Ions: The oxidation number is equal to the charge on the ion (e.g., $\text{Na}^+$ is +1, $\text{Cl}^-$ is -1, $\text{Mg}^{2+}$ is +2).
3. Oxygen: The oxidation number of oxygen is usually -2.
   • Exceptions: In peroxides (like $\text{H}_2\text{O}_2$), it is -1. In superoxides (like $\text{KO}_2$), it is -1/2. When bonded to fluorine (e.g., $\text{OF}_2$), it is +2.
4. Hydrogen: The oxidation number of hydrogen is usually +1.
   • Exception: In metal hydrides (like $\text{NaH}$), it is -1.
5. Halogens: Fluorine is always -1. Other halogens (Cl, Br, I) are usually -1, unless bonded to oxygen or a more electronegative halogen, in which case they can have positive oxidation numbers.
6. Sum of Oxidation Numbers:
   • For a neutral compound, the sum of all oxidation numbers must be zero.
   • For a polyatomic ion, the sum must equal the charge of the ion.

Redox Reactions in Terms of Oxidation Number

• Oxidation is an increase in oxidation number.
• Reduction is a decrease in oxidation number.
• Oxidising Agent is a substance that contains an element whose oxidation number decreases.
• Reducing Agent is a substance that contains an element whose oxidation number increases.

Stock Notation

Stock notation is a way to name compounds of metals that can have multiple oxidation states. The oxidation number is written as a Roman numeral in parentheses after the metal's symbol.

• Aurous chloride ($\text{AuCl}$) is written as $\text{Au(I)Cl}$.
• Auric chloride ($\text{AuCl}_3$) is written as $\text{Au(III)Cl}_3$.
• Stannous chloride ($\text{SnCl}_2$) is written as $\text{Sn(II)Cl}_2$.
• Stannic chloride ($\text{SnCl}_4$) is written as $\text{Sn(IV)Cl}_4$.

Given

• Reaction: $2\text{Cu}_2\text{O(s)} + \text{Cu}_2\text{S(s)} \rightarrow 6\text{Cu(s)} + \text{SO}_2\text{(g)}$

To Find

• Justify it's a redox reaction.
• Identify the species oxidised and reduced.
• Identify the oxidant and reductant.

Solution

First, let's assign oxidation numbers to each element in the reactants and products.

• Copper (Cu): The oxidation number of Cu in both $\text{Cu}_2\text{O}$ and $\text{Cu}_2\text{S}$ is +1. In the product, elemental Cu has an oxidation number of 0. Since the oxidation number decreases from +1 to 0, copper is reduced.
• Sulphur (S): The oxidation number of S in $\text{Cu}_2\text{S}$ is -2. In the product $\text{SO}_2$, its oxidation number is +4. Since the oxidation number increases from -2 to +4, sulphur is oxidised.

Conclusion:

• Because both reduction (of copper) and oxidation (of sulphur) occur, this is a redox reaction.
• Oxidant (Oxidising Agent): $\text{Cu}_2\text{O}$ causes the oxidation of sulphur. The copper (Cu(I)) in it is the species that gets reduced.
• Reductant (Reducing Agent): $\text{Cu}_2\text{S}$ causes the reduction of copper. The sulphur (S(-II)) in it is the species that gets oxidised.

Types of Redox Reactions

Redox reactions can be classified into four main types.

1. Combination Reactions

These are reactions where two or more substances combine to form a single product. For it to be a redox reaction, at least one of the reactants must be in its elemental form. $\text{A} + \text{B} \rightarrow \text{C}$

• Carbon burning in oxygen: $\stackrel{0}{\text{C}}\text{(s)} + \stackrel{0}{\text{O}}_2\text{(g)} \rightarrow \stackrel{+4-2}{\text{CO}_2}\text{(g)}$
• Methane burning in oxygen: $\stackrel{-4+1}{\text{CH}_4}\text{(g)} + 2\stackrel{0}{\text{O}}_2\text{(g)} \rightarrow \stackrel{+4-2}{\text{CO}_2}\text{(g)} + 2\stackrel{+1-2}{\text{H}_2\text{O}}\text{(l)}$

2. Decomposition Reactions

These are reactions where a compound breaks down into two or more simpler substances. For it to be a redox reaction, at least one of the products must be in its elemental form. $\text{C} \rightarrow \text{A} + \text{B}$

• Decomposition of water: $2\stackrel{+1-2}{\text{H}_2\text{O}}\text{(l)} \rightarrow 2\stackrel{0}{\text{H}}_2\text{(g)} + \stackrel{0}{\text{O}}_2\text{(g)}$
• Decomposition of potassium chlorate: $2\stackrel{+1+5-2}{\text{KClO}_3}\text{(s)} \rightarrow 2\stackrel{+1-1}{\text{KCl}}\text{(s)} + 3\stackrel{0}{\text{O}}_2\text{(g)}$

3. Displacement Reactions

In these reactions, an atom or ion in a compound is replaced by an atom or ion of another element. $\text{X} + \text{YZ} \rightarrow \text{XZ} + \text{Y}$

• Metal Displacement: A more reactive metal displaces a less reactive metal from its salt solution. $\stackrel{+2+6-2}{\text{CuSO}_4}\text{(aq)} + \stackrel{0}{\text{Zn}}\text{(s)} \rightarrow \stackrel{0}{\text{Cu}}\text{(s)} + \stackrel{+2+6-2}{\text{ZnSO}_4}\text{(aq)}$
• Non-metal Displacement: This often involves the displacement of hydrogen or a halogen.
  • A reactive metal like sodium displaces hydrogen from water: $2\stackrel{0}{\text{Na}}\text{(s)} + 2\stackrel{+1-2}{\text{H}_2\text{O}}\text{(l)} \rightarrow 2\stackrel{+1-2+1}{\text{NaOH}}\text{(aq)} + \stackrel{0}{\text{H}}_2\text{(g)}$
  • A more reactive halogen like chlorine displaces a less reactive one like bromine: $\stackrel{0}{\text{Cl}_2}\text{(g)} + 2\text{K}\stackrel{-1}{\text{Br}}\text{(aq)} \rightarrow 2\text{K}\stackrel{-1}{\text{Cl}}\text{(aq)} + \stackrel{0}{\text{Br}_2}\text{(l)}$

4. Disproportionation Reactions

This is a special type of redox reaction where a single element in an intermediate oxidation state is simultaneously oxidised and reduced. The element must be able to exist in at least three different oxidation states.

• Decomposition of hydrogen peroxide: $2\text{H}_2\stackrel{-1}{\text{O}}_2\text{(aq)} \rightarrow 2\text{H}_2\stackrel{-2}{\text{O}}\text{(l)} + \stackrel{0}{\text{O}}_2\text{(g)}$ Here, oxygen in the -1 oxidation state is reduced to -2 (in $\text{H}_2\text{O}$) and oxidised to 0 (in $\text{O}_2$).

• Reaction of chlorine with a base: $\stackrel{0}{\text{Cl}_2}\text{(g)} + 2\text{OH}^-\text{(aq)} \rightarrow \stackrel{-1}{\text{Cl}}\text{O}^-\text{(aq)} + \stackrel{-1}{\text{Cl}}^-\text{(aq)} + \text{H}_2\text{O(l)}$ Here, chlorine in the 0 oxidation state is oxidised to +1 (in $\text{ClO}^-$) and reduced to -1 (in $\text{Cl}^-$).

The Paradox of Fractional Oxidation Number

Sometimes, calculating the oxidation number for an element in a compound yields a fraction, such as +4/3 for carbon in $\text{C}_3\text{O}_2$ or +2.5 for sulphur in $\text{Na}_2\text{S}_4\text{O}_6$.

This fractional value is not the actual charge on any single atom; it is the average oxidation state of all atoms of that element in the molecule. The real oxidation states of individual atoms are always whole numbers, which can be determined by looking at the molecule's structure.

• The two terminal sulphur atoms each have an oxidation state of +5.
• The two middle sulphur atoms each have an oxidation state of 0.

The average is $\frac{(+5) + (0) + (0) + (+5)}{4} = \frac{10}{4} = +2.5$.

Balancing of Redox Reactions

There are two main methods to balance redox equations.

1. Oxidation Number Method

This method focuses on making the total increase in oxidation number equal to the total decrease.

Steps:

1. Write the skeletal equation.
2. Assign oxidation numbers to identify which elements are oxidised and reduced.
3. Calculate the change in oxidation number per atom and for the whole molecule/ion. Use coefficients to make the total increase equal to the total decrease.
4. Balance charges. If the reaction is in an acidic solution, add $\text{H}^+$ ions. If in a basic solution, add $\text{OH}^-$ ions.
5. Balance atoms. Add $\text{H}_2\text{O}$ molecules to balance hydrogen and oxygen atoms.

Given

• Reactants: $\text{Cr}_2\text{O}_7^{2-}$ and $\text{SO}_3^{2-}$
• Products: $\text{Cr}^{3+}$ and $\text{SO}_4^{2-}$
• Medium: Acidic

To Find

The balanced net ionic equation.

Formula

Oxidation Number Method

Solution

Step 1: Skeletal Ionic Equation $\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + \text{SO}_3^{2-}\text{(aq)} \rightarrow \text{Cr}^{3+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}$

Step 2: Assign Oxidation Numbers $\stackrel{+6}{\text{Cr}}_2\text{O}_7^{2-}\text{(aq)} + \stackrel{+4}{\text{S}}\text{O}_3^{2-}\text{(aq)} \rightarrow \stackrel{+3}{\text{Cr}}^{3+}\text{(aq)} + \stackrel{+6}{\text{S}}\text{O}_4^{2-}\text{(aq)}$

• Chromium (Cr) is reduced from +6 to +3 (Decrease of 3 per atom).
• Sulphur (S) is oxidised from +4 to +6 (Increase of 2 per atom).

Step 3: Balance the Increase and Decrease in Oxidation Number

• Total decrease for $\text{Cr}_2\text{O}_7^{2-}$ (2 Cr atoms) = $2 \times 3 = 6$.
• Total increase for $\text{SO}_3^{2-}$ (1 S atom) = $2$.
• To make the changes equal (6), we need to multiply the sulphur species by 3. $\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 3\text{SO}_3^{2-}\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{SO}_4^{2-}\text{(aq)}$ (We also balance the Cr atoms on the right side).

Step 4: Balance Charges

• Charge on the left side: $(-2) + 3(-2) = -8$.
• Charge on the right side: $2(+3) + 3(-2) = 0$.
• To balance the charges in an acidic medium, we add $\text{H}^+$ to the more negative side (left). We need 8 $\text{H}^+$ ions. $\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 3\text{SO}_3^{2-}\text{(aq)} + 8\text{H}^+\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{SO}_4^{2-}\text{(aq)}$

Step 5: Balance Atoms (H and O)

• There are 8 H atoms on the left. We add $4\text{H}_2\text{O}$ to the right to balance them. $\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 3\text{SO}_3^{2-}\text{(aq)} + 8\text{H}^+\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{SO}_4^{2-}\text{(aq)} + 4\text{H}_2\text{O(l)}$
• A final check of oxygen atoms: Left side has $7 + (3 \times 3) = 16$ O atoms. Right side has $(3 \times 4) + 4 = 16$ O atoms. The equation is balanced.

Final Answer $\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 3\text{SO}_3^{2-}\text{(aq)} + 8\text{H}^+\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{SO}_4^{2-}\text{(aq)} + 4\text{H}_2\text{O(l)}$

2. Half-Reaction Method (Ion-Electron Method)

This method involves balancing the oxidation and reduction half-reactions separately before combining them.

Steps:

1. Write the unbalanced ionic equation.
2. Separate into two half-reactions: one for oxidation and one for reduction.
3. Balance atoms other than O and H in each half-reaction.
4. Balance O and H atoms. In acidic medium, add $\text{H}_2\text{O}$ to balance O and $\text{H}^+$ to balance H.
5. Balance charges by adding electrons ($\text{e}^-$) to the appropriate side.
6. Equalise the number of electrons in both half-reactions by multiplying them by suitable integers.
7. Add the two half-reactions and cancel out the electrons.
8. Verify that atoms and charges are balanced.

Given

• Reactants: $\text{MnO}_4^-$ and $\text{I}^-$
• Products: $\text{MnO}_2$ and $\text{I}_2$
• Medium: Basic

To Find

The balanced ionic equation.

Formula

Half-Reaction Method

Solution

Step 1: Skeletal Ionic Equation $\text{MnO}_4^-\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + \text{I}_2\text{(s)}$

Step 2: Separate into Half-Reactions

• Oxidation: $\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)}$
• Reduction: $\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)}$

Step 3 & 4: Balance Atoms (I, O, H)

• Oxidation Half-Reaction:

  • Balance I atoms: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)}$

• Reduction Half-Reaction:

  • Mn is balanced.
  • Balance O by adding $2\text{H}_2\text{O}$ to the right: $\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$
  • Balance H by adding $4\text{H}^+$ to the left: $\text{MnO}_4^-\text{(aq)} + 4\text{H}^+\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$
  • Since the medium is basic, add $4\text{OH}^-$ to both sides: $\text{MnO}_4^-\text{(aq)} + 4\text{H}^+\text{(aq)} + 4\text{OH}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)} + 4\text{OH}^-\text{(aq)}$
  • Combine $\text{H}^+$ and $\text{OH}^-$ to form $4\text{H}_2\text{O}$ on the left: $\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)} + 4\text{OH}^-\text{(aq)}$
  • Simplify by canceling $2\text{H}_2\text{O}$ from both sides: $\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$

Step 5: Balance Charges

• Oxidation: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)} + 2\text{e}^-$
• Reduction: $\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} + 3\text{e}^- \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$

Step 6: Equalise Electrons

• Multiply oxidation half-reaction by 3.
• Multiply reduction half-reaction by 2. $6\text{I}^-\text{(aq)} \rightarrow 3\text{I}_2\text{(s)} + 6\text{e}^-$ $2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} + 6\text{e}^- \rightarrow 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Step 7: Add Half-Reactions

• Cancel the $6\text{e}^-$ from both sides. $6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Final Answer The balanced equation is: $6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Redox Reactions and Electrode Processes

Redox reactions are the basis of electrochemistry, the study of how chemical energy can be converted into electrical energy and vice versa.

When a zinc rod is placed in a copper sulfate solution, electrons are transferred directly from zinc atoms to copper ions. This process releases heat. However, if we separate the reactants, we can force the electrons to travel through an external wire, creating an electric current. This setup is called a galvanic cell or Daniell cell.

Key Components of a Galvanic Cell:

• Two Half-Cells: Each half-cell contains a metal rod (the electrode) submerged in a solution of its own ions. For example, a zinc rod in zinc sulfate solution and a copper rod in copper sulfate solution.
• Redox Couple: The combination of the oxidised and reduced forms of a substance in a half-cell (e.g., $\text{Zn}^{2+}/\text{Zn}$ and $\text{Cu}^{2+}/\text{Cu}$).
• External Wire: Connects the two electrodes, allowing electrons to flow from the oxidation half-cell (anode) to the reduction half-cell (cathode).
• Salt Bridge: A U-shaped tube containing an electrolyte (like KCl) that connects the two solutions. It allows ions to migrate between the half-cells to maintain electrical neutrality, completing the circuit.

Electrode Potential

Each electrode has a potential, known as the electrode potential, which is a measure of its tendency to be oxidised or reduced. When measured under standard conditions (298 K, 1 M concentration, 1 atm pressure for gases), it is called the Standard Electrode Potential ($E^\ominus$).

• By convention, the standard electrode potential of the hydrogen electrode ($2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$) is defined as 0.00 volts.
• A negative $E^\ominus$ indicates that the redox couple is a stronger reducing agent than the $\text{H}^+/\text{H}_2$ couple.
• A positive $E^\ominus$ indicates that the redox couple is a weaker reducing agent than the $\text{H}^+/\text{H}_2$ couple.

The difference in electrode potentials between the two half-cells creates a voltage that drives the flow of electrons, generating electricity.