ElectrochemistryClass 12 Chemistry NCERT Solutions

The ability of a metal to displace another metal from its salt solution depends on its position in the electrochemical series. A metal with a lower standard reduction potential (more negative) is more reactive and can displace a metal with a higher standard reduction potential (less negative or positive) from its salt solution.

The standard electrode potentials ($E^\circ$) of the given metals are:

The order of increasing standard reduction potential is: Mg < Al < Zn < Fe < Cu

This is also the order of decreasing reactivity. Therefore, the order in which these metals displace each other from their salt solutions is from the most reactive to the least reactive.

Final Answer: The order is Mg, Al, Zn, Fe, Cu.

The reducing power of a metal is its tendency to lose electrons, i.e., to get oxidized. A metal with a lower (more negative) standard reduction potential ($E^\circ$) has a greater tendency to get oxidized and is therefore a stronger reducing agent.

The standard electrode potentials ($E^\circ$) of the given metals are:

To arrange the metals in increasing order of their reducing power, we need to arrange them in order of decreasing standard reduction potential (from most positive to most negative).

Order of decreasing $E^\circ$ values: Ag ($+0.80 \text{ V}$) < Hg ($+0.79 \text{ V}$) < Cr ($-0.74 \text{ V}$) < Mg ($-2.37 \text{ V}$) < K ($-2.93 \text{ V}$)

This corresponds to the increasing order of reducing power.

Final Answer: The increasing order of reducing power is Ag < Hg < Cr < Mg < K.

The galvanic cell for the given reaction is depicted as follows:

Cell Representation: $\text{Zn(s)} | \text{Zn}^{2+}\text{(aq)} || \text{Ag}^{+}\text{(aq)} | \text{Ag(s)}$

A zinc electrode is dipped in a solution of zinc sulphate, and a silver electrode is dipped in a solution of silver nitrate. The two solutions are connected by a salt bridge, and the electrodes are connected by an external wire.

(i) Which of the electrode is negatively charged? In this reaction, zinc is oxidized ($\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$), and silver ions are reduced ($\text{Ag}^+ + e^- \rightarrow \text{Ag}$). The electrode where oxidation occurs is the anode. By convention, in a galvanic cell, the anode is the negative electrode because it is the source of electrons flowing into the external circuit. Answer: The zinc electrode (anode) is negatively charged.

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

(iii) Individual reaction at each electrode. The overall reaction is split into two half-reactions:

Half-reactions: Anode (Oxidation): $\text{Cr(s)} \rightarrow \text{Cr}^{3+}\text{(aq)} + 3e^-$ Cathode (Reduction): $\text{Cd}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Cd(s)}$

Given Standard Potentials (from standard tables): $E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \text{ V}$ $E^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \text{ V}$

Standard Cell Potential ($E^\circ_{\text{cell}}$): $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $E^\circ_{\text{cell}} = (-0.40 \text{ V}) - (-0.74 \text{ V}) = 0.34 \text{ V}$

Standard Gibbs Energy ($\Delta_r G^\circ$): For the overall reaction, the number of electrons transferred, $n=6$. $\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$ $\Delta_r G^\circ = -6 \times (96487 \text{ C mol}^{-1}) \times (0.34 \text{ V})$ $\Delta_r G^\circ = -196833.48 \text{ J mol}^{-1} = -196.83 \text{ kJ mol}^{-1}$

Equilibrium Constant ($K_c$): $E^\circ_{\text{cell}} = \frac{2.303 RT}{nF} \log K_c = \frac{0.0591}{n} \log K_c$ at 298 K $0.34 = \frac{0.0591}{6} \log K_c$ $\log K_c = \frac{0.34 \times 6}{0.0591} = 34.517$ $K_c = \text{antilog}(34.517) = 3.29 \times 10^{34}$

Half-reactions: Anode (Oxidation): $\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + e^-$ Cathode (Reduction): $\text{Ag}^{+}\text{(aq)} + e^- \rightarrow \text{Ag(s)}$

Given Standard Potentials (from Table 2.1): $E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \text{ V}$ $E^\circ_{\text{Ag}^{+}/\text{Ag}} = +0.80 \text{ V}$

Standard Cell Potential ($E^\circ_{\text{cell}}$): $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $E^\circ_{\text{cell}} = (+0.80 \text{ V}) - (+0.77 \text{ V}) = 0.03 \text{ V}$

Standard Gibbs Energy ($\Delta_r G^\circ$): For the overall reaction, the number of electrons transferred, $n=1$. $\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$ $\Delta_r G^\circ = -1 \times (96487 \text{ C mol}^{-1}) \times (0.03 \text{ V})$ $\Delta_r G^\circ = -2894.61 \text{ J mol}^{-1} = -2.895 \text{ kJ mol}^{-1}$

Equilibrium Constant ($K_c$): $E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_c$ $0.03 = \frac{0.0591}{1} \log K_c$ $\log K_c = \frac{0.03}{0.0591} = 0.5076$ $K_c = \text{antilog}(0.5076) = 3.22$

The Nernst equation for a cell at 298 K is: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q$

(i) $Mg(s)|Mg^{2+}(0.001 M) || Cu^{2+}(0.0001 M)|Cu(s)$ Cell Reaction: $\text{Mg(s)} + \text{Cu}^{2+}\text{(aq)} \rightarrow \text{Mg}^{2+}\text{(aq)} + \text{Cu(s)}$ Here, $n=2$. $E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.36 \text{ V}$, $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V}$. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-2.36) = 2.70 \text{ V}$. Nernst Equation: $E_{\text{cell}} = 2.70 - \frac{0.0591}{2} \log \frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}$ EMF Calculation: $E_{\text{cell}} = 2.70 - \frac{0.0591}{2} \log \frac{0.001}{0.0001} = 2.70 - \frac{0.0591}{2} \log(10)$ $E_{\text{cell}} = 2.70 - 0.02955 = 2.67 \text{ V}$

(ii) $Fe(s)|Fe^{2+}(0.001 M) || H^+(1 M)|H_2(g)(1bar)|Pt(s)$ Cell Reaction: $\text{Fe(s)} + 2\text{H}^{+}\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$ Here, $n=2$. $E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \text{ V}$, $E^\circ_{\text{H}^{+}/\text{H}_2} = 0.00 \text{ V}$. $E^\circ_{\text{cell}} = 0.00 - (-0.44) = 0.44 \text{ V}$. Nernst Equation: $E_{\text{cell}} = 0.44 - \frac{0.0591}{2} \log \frac{[\text{Fe}^{2+}] \cdot p_{\text{H}_2}}{[\text{H}^+]^2}$ EMF Calculation: $E_{\text{cell}} = 0.44 - \frac{0.0591}{2} \log \frac{0.001 \times 1}{(1)^2} = 0.44 - \frac{0.0591}{2} \log(10^{-3})$ $E_{\text{cell}} = 0.44 - \frac{0.0591}{2} \times (-3) = 0.44 + 0.0887 = 0.5287 \text{ V}$

(iii) $Sn(s)|Sn^{2+}(0.050 M) || H^+(0.020 M)|H_2(g)(1bar)|Pt(s)$ Cell Reaction: $\text{Sn(s)} + 2\text{H}^{+}\text{(aq)} \rightarrow \text{Sn}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$ Here, $n=2$. $E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \text{ V}$, $E^\circ_{\text{H}^{+}/\text{H}_2} = 0.00 \text{ V}$. $E^\circ_{\text{cell}} = 0.00 - (-0.14) = 0.14 \text{ V}$. Nernst Equation: $E_{\text{cell}} = 0.14 - \frac{0.0591}{2} \log \frac{[\text{Sn}^{2+}] \cdot p_{\text{H}_2}}{[\text{H}^+]^2}$ EMF Calculation: $E_{\text{cell}} = 0.14 - \frac{0.0591}{2} \log \frac{0.050}{(0.020)^2} = 0.14 - \frac{0.0591}{2} \log(125)$ $E_{\text{cell}} = 0.14 - 0.02955 \times (2.0969) = 0.14 - 0.062 = 0.078 \text{ V}$

(iv) $Pt(s)|Br^-(0.010 M)|Br_2(l) || H^+(0.030 M)|H_2(g)(1 bar)|Pt(s)$ Cell Reaction: $2\text{Br}^{-}\text{(aq)} + 2\text{H}^{+}\text{(aq)} \rightarrow \text{Br}_2\text{(l)} + \text{H}_2\text{(g)}$ Here, $n=2$. $E^\circ_{\text{Br}_2/\text{Br}^-} = +1.09 \text{ V}$, $E^\circ_{\text{H}^{+}/\text{H}_2} = 0.00 \text{ V}$. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.00 - 1.09 = -1.09 \text{ V}$. Nernst Equation: $E_{\text{cell}} = -1.09 - \frac{0.0591}{2} \log \frac{p_{\text{H}_2}}{[\text{Br}^-]^2 [\text{H}^+]^2}$ EMF Calculation: $E_{\text{cell}} = -1.09 - \frac{0.0591}{2} \log \frac{1}{(0.010)^2 (0.030)^2} = -1.09 - \frac{0.0591}{2} \log \frac{1}{9 \times 10^{-8}}$ $E_{\text{cell}} = -1.09 - \frac{0.0591}{2} \log(1.11 \times 10^7) = -1.09 - 0.02955 \times (7.045)$ $E_{\text{cell}} = -1.09 - 0.208 = -1.298 \text{ V}$

The overall reaction can be split into two half-reactions:

Anode (Oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2e^-$; $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V}$

Cathode (Reduction): $\text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} + 2e^- \rightarrow 2\text{Ag(s)} + 2\text{OH}^{-}\text{(aq)}$; $E^\circ_{\text{Ag}_2\text{O}/\text{Ag}} = +0.344 \text{ V}$

Calculation of $E^\circ$ for the reaction: The standard cell potential ($E^\circ_{\text{cell}}$) is calculated as: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $E^\circ_{\text{cell}} = (+0.344 \text{ V}) - (-0.76 \text{ V})$ $E^\circ_{\text{cell}} = 1.104 \text{ V}$

Calculation of $\Delta_r G^\circ$ for the reaction: The standard Gibbs energy change ($\Delta_r G^\circ$) is related to the standard cell potential by the equation: $\Delta_r G^\circ = -nFE^\circ_{\text{cell}}$

Here, the number of moles of electrons transferred, $n=2$. Faraday's constant, $F = 96487 \text{ C mol}^{-1}$.

$\Delta_r G^\circ = -2 \times (96487 \text{ C mol}^{-1}) \times (1.104 \text{ V})$ $\Delta_r G^\circ = -213043.3 \text{ J mol}^{-1}$ $\Delta_r G^\circ = -213.04 \text{ kJ mol}^{-1}$

Conductivity ($\kappa$): Conductivity of a solution is defined as the conductance of a solution of 1 cm length with an area of cross-section of 1 cm$^2$. It is the inverse of resistivity ($\rho$). $\kappa = \frac{1}{\rho} = \frac{1}{R} \frac{l}{A}$ Its SI unit is Siemens per meter ($\text{S m}^{-1}$), but it is often expressed in $\text{S cm}^{-1}$.

Molar Conductivity ($\Lambda_m$): Molar conductivity of a solution at a given concentration is the conductance of the volume ($V$) of the solution containing one mole of the electrolyte, kept between two electrodes with a unit area of cross-section and at a distance of unit length. It is related to conductivity by the equation: $\Lambda_m = \frac{\kappa}{c}$ where $c$ is the molar concentration. If $\kappa$ is in $\text{S cm}^{-1}$ and $c$ is in mol L$^{-1}$, the expression is: $\Lambda_m (\text{S cm}^2 \text{mol}^{-1}) = \frac{\kappa \times 1000}{c}$

Given: Concentration of KCl solution, $c = 0.20 \text{ M} = 0.20 \text{ mol L}^{-1}$ Conductivity, $\kappa = 0.0248 \text{ S cm}^{-1}$

To Find: Molar conductivity, $\Lambda_m$

Formula: The molar conductivity ($\Lambda_m$) is related to conductivity ($\kappa$) and molar concentration ($c$) by the formula: $\Lambda_m = \frac{1000 \times \kappa}{c}$

Calculation: Substituting the given values into the formula: $\Lambda_m = \frac{1000 \text{ cm}^3 \text{ L}^{-1} \times 0.0248 \text{ S cm}^{-1}}{0.20 \text{ mol L}^{-1}}$ $\Lambda_m = \frac{24.8}{0.20} \text{ S cm}^2 \text{mol}^{-1}$ $\Lambda_m = 124 \text{ S cm}^2 \text{mol}^{-1}$

Final Answer: The molar conductivity of the 0.20 M KCl solution is $124 \text{ S cm}^2 \text{mol}^{-1}$.

Given: Concentration of KCl solution, $c = 0.001 \text{ M}$ Resistance of the solution, $R = 1500 \Omega$ Conductivity of the solution, $\kappa = 0.146 \times 10^{-3} \text{ S cm}^{-1}$

To Find: The cell constant, $G^*$

Formula: The relationship between conductivity ($\kappa$), resistance ($R$), and the cell constant ($G^*$) is given by: $\kappa = \frac{G^*}{R}$ Rearranging the formula to solve for the cell constant: $G^* = \kappa \times R$

Calculation: Substituting the given values into the formula: $G^* = (0.146 \times 10^{-3} \text{ S cm}^{-1}) \times (1500 \Omega)$ Since $S = \Omega^{-1}$, the units cancel out correctly. $G^* = 0.146 \times 1.5 \text{ cm}^{-1}$ $G^* = 0.219 \text{ cm}^{-1}$

Final Answer: The cell constant of the conductivity cell is $0.219 \text{ cm}^{-1}$.

Step 1: Data Tabulation and Unit Conversion We need to calculate molar conductivity ($\Lambda_m$) and the square root of concentration ($c^{1/2}$). It is convenient to work with units of S cm$^2$ mol$^{-1}$ for $\Lambda_m$ and (mol L$^{-1}$)$^{1/2}$ for $c^{1/2}$. First, convert conductivity $\kappa$ from S m$^{-1}$ to S cm$^{-1}$. $1 \text{ S m}^{-1} = 10^{-2} \text{ S cm}^{-1}$ So, $\kappa (\text{S cm}^{-1}) = \kappa (\text{S m}^{-1}) \times 10^{-2}$.

The given values are for $10^2 \times \kappa (\text{S m}^{-1})$, so $\kappa (\text{S m}^{-1}) = \frac{\text{given value}}{100}$. Therefore, $\kappa (\text{S cm}^{-1}) = \frac{\text{given value}}{100} \times 10^{-2} = \text{given value} \times 10^{-4}$.

The formula for molar conductivity is $\Lambda_m = \frac{1000 \times \kappa (\text{S cm}^{-1})}{c (\text{mol L}^{-1})}$.

Step 3: Plotting $\Lambda_m$ vs $c^{1/2}$ A plot of $\Lambda_m$ (y-axis) against $c^{1/2}$ (x-axis) is drawn. For a strong electrolyte like NaCl, this should yield a straight line according to the Debye-Hückel-Onsager equation: $\Lambda_m = \Lambda_m^\circ - A c^{1/2}$.

Step 4: Finding the value of $\Lambda_m^\circ$ The value of $\Lambda_m^\circ$ (limiting molar conductivity) is the intercept of the line on the y-axis, which corresponds to $c^{1/2} = 0$ (infinite dilution).

By extrapolating the straight line graph of the calculated points to $c^{1/2} = 0$, we get the y-intercept. From the plot, the intercept is found to be approximately $124.0 \text{ S cm}^2 \text{mol}^{-1}$.

Alternatively, we can calculate the slope 'A' and then find the intercept. Slope $A = \frac{\Delta \Lambda_m}{\Delta c^{1/2}} = \frac{118.5 - 123.7}{0.1000 - 0.0316} = \frac{-5.2}{0.0684} \approx -76$ Using the point (0.0316, 123.7) and the equation $y = mx+c$: $123.7 = (-76)(0.0316) + \Lambda_m^\circ$ $123.7 = -2.4 + \Lambda_m^\circ$ $\Lambda_m^\circ = 126.1 \text{ S cm}^2 \text{mol}^{-1}$ (Note: Graphical extrapolation gives a more accurate value)

Final Answer: The calculated values for $\Lambda_m$ are 123.7, 118.5, 115.8, 111.1, and 106.7 S cm$^2$ mol$^{-1}$. By plotting $\Lambda_m$ vs $c^{1/2}$ and extrapolating to zero concentration, the value of limiting molar conductivity is found to be $\Lambda_m^\circ \approx 124.0 \text{ S cm}^2 \text{mol}^{-1}$.

Given: Concentration of acetic acid, $c = 0.00241 \text{ M}$ Conductivity, $\kappa = 7.896 \times 10^{-5} \text{ S cm}^{-1}$ Limiting molar conductivity, $\Lambda_m^\circ = 390.5 \text{ S cm}^2 \text{mol}^{-1}$

Formula: $\Lambda_m = \frac{1000 \times \kappa}{c}$

Calculation: $\Lambda_m = \frac{1000 \text{ cm}^3 \text{ L}^{-1} \times (7.896 \times 10^{-5} \text{ S cm}^{-1})}{0.00241 \text{ mol L}^{-1}}$ $\Lambda_m = \frac{0.07896}{0.00241} \text{ S cm}^2 \text{mol}^{-1}$ $\Lambda_m = 32.76 \text{ S cm}^2 \text{mol}^{-1}$

Step A: Calculate the degree of dissociation ($\alpha$) Formula: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$

Calculation: $\alpha = \frac{32.76 \text{ S cm}^2 \text{mol}^{-1}}{390.5 \text{ S cm}^2 \text{mol}^{-1}}$ $\alpha = 0.0839$

Step B: Calculate the dissociation constant ($K_a$) Formula: For a weak electrolyte like acetic acid ($\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$): $K_a = \frac{c\alpha^2}{1-\alpha}$

Calculation: $K_a = \frac{(0.00241) \times (0.0839)^2}{1 - 0.0839}$ $K_a = \frac{0.00241 \times 0.007039}{0.9161}$ $K_a = \frac{1.696 \times 10^{-5}}{0.9161}$ $K_a = 1.85 \times 10^{-5}$

The charge on one mole of electrons is equal to one Faraday ($F$), which is approximately $96500 \text{ C mol}^{-1}$. We need to determine the number of moles of electrons required for each reduction.

(i) 1 mol of $Al^{3+}$ to Al The reduction half-reaction is: $\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$ From the stoichiometry, the reduction of 1 mole of $\text{Al}^{3+}$ ions requires 3 moles of electrons. Charge required = $3 \times F = 3 \times 96500 \text{ C} = 289500 \text{ C}$.

(ii) 1 mol of $Cu^{2+}$ to Cu The reduction half-reaction is: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ From the stoichiometry, the reduction of 1 mole of $\text{Cu}^{2+}$ ions requires 2 moles of electrons. Charge required = $2 \times F = 2 \times 96500 \text{ C} = 193000 \text{ C}$.

(iii) 1 mol of $MnO_4^-$ to $Mn^{2+}$ First, we determine the change in the oxidation state of Manganese (Mn). In $\text{MnO}_4^-$, let the oxidation state of Mn be $x$. Then, $x + 4(-2) = -1$, which gives $x = +7$. In $\text{Mn}^{2+}$, the oxidation state is +2. The change in oxidation state is from +7 to +2, which means a gain of 5 electrons. The reduction half-reaction (in acidic medium) is: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ From the stoichiometry, the reduction of 1 mole of $\text{MnO}_4^-$ ions requires 5 moles of electrons. Charge required = $5 \times F = 5 \times 96500 \text{ C} = 482500 \text{ C}$.

Step 1: Write the electrode reaction. Calcium is produced by the reduction of $\text{Ca}^{2+}$ ions at the cathode. $\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca(s)}$ This equation shows that 1 mole of Ca (atomic mass = 40.0 g/mol) is produced by 2 moles of electrons, which is equivalent to 2 Faradays (2F) of electricity.

Step 2: Calculate the moles of Ca to be produced. Moles of Ca = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{20.0 \text{ g}}{40.0 \text{ g/mol}} = 0.5 \text{ mol}$

Step 3: Calculate the required electricity in Faradays. From stoichiometry, 1 mol Ca requires 2 F of electricity. Therefore, 0.5 mol Ca will require: Electricity = $0.5 \text{ mol} \times 2 \frac{\text{F}}{\text{mol}} = 1 \text{ F}$

Final Answer for (i): 1 Faraday of electricity is required.

Step 1: Write the electrode reaction. Aluminium is produced by the reduction of $\text{Al}^{3+}$ ions at the cathode. $\text{Al}^{3+} + 3e^- \rightarrow \text{Al(s)}$ This equation shows that 1 mole of Al (atomic mass = 27.0 g/mol) is produced by 3 moles of electrons, which is equivalent to 3 Faradays (3F) of electricity.

Step 2: Calculate the moles of Al to be produced. Moles of Al = $\frac{\text{Given mass}}{\text{Molar mass}} = \frac{40.0 \text{ g}}{27.0 \text{ g/mol}} = 1.481 \text{ mol}$

Step 3: Calculate the required electricity in Faradays. From stoichiometry, 1 mol Al requires 3 F of electricity. Therefore, 1.481 mol Al will require: Electricity = $1.481 \text{ mol} \times 3 \frac{\text{F}}{\text{mol}} = 4.443 \text{ F}$

Final Answer for (ii): 4.443 Faradays of electricity are required.

Step 1: Write the balanced oxidation half-reaction. The oxidation of water produces oxygen gas. $2\text{H}_2\text{O(l)} \rightarrow \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4e^-$

Step 2: Determine electrons transferred per mole of reactant. The balanced equation shows that the oxidation of 2 moles of $\text{H}_2\text{O}$ produces 4 moles of electrons. Therefore, the oxidation of 1 mole of $\text{H}_2\text{O}$ will produce $\frac{4}{2} = 2$ moles of electrons.

Step 3: Calculate the charge in coulombs. The charge of 1 mole of electrons is 1 Faraday ($F$), which is 96500 C. Charge required = (moles of electrons) $\times F$ Charge required = $2 \text{ mol} \times 96500 \text{ C/mol} = 193000 \text{ C}$

Final Answer for (i): 193000 C of electricity is required.

Step 1: Determine the change in oxidation state. In FeO, the oxidation state of Fe is +2. In $\text{Fe}_2\text{O}_3$, the oxidation state of Fe is +3. The oxidation process is the conversion of $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$.

Step 2: Write the oxidation half-reaction. $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + 1e^-$

Step 3: Determine electrons transferred per mole of reactant. 1 mole of FeO contains 1 mole of $\text{Fe}^{2+}$. The oxidation of 1 mole of $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ involves the loss of 1 mole of electrons.

Step 4: Calculate the charge in coulombs. Charge required = (moles of electrons) $\times F$ Charge required = $1 \text{ mol} \times 96500 \text{ C/mol} = 96500 \text{ C}$

Final Answer for (ii): 96500 C of electricity is required.

Given: Current, $I = 5 \text{ A}$ Time, $t = 20 \text{ minutes} = 20 \times 60 \text{ seconds} = 1200 \text{ s}$ Electrolyte: $\text{Ni(NO}_3)_2$, which provides $\text{Ni}^{2+}$ ions. Molar mass of Ni, $M = 58.7 \text{ g/mol}$

To Find: Mass of Ni deposited, $w$

Step 1: Calculate the total charge passed ($Q$). Formula: $Q = I \times t$ Calculation: $Q = 5 \text{ A} \times 1200 \text{ s} = 6000 \text{ C}$

Step 2: Write the cathode reaction. Nickel ions are reduced at the cathode to form nickel metal. $\text{Ni}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Ni(s)}$

Step 3: Relate charge to the mass of Ni deposited. The reaction shows that 2 moles of electrons are required to deposit 1 mole of Ni. Charge for 1 mole of Ni = $2 \times F = 2 \times 96500 \text{ C} = 193000 \text{ C}$ So, 193000 C of charge deposits 1 mole of Ni, which is 58.7 g.

Step 4: Calculate the mass of Ni deposited by 6000 C. Mass deposited, $w = \frac{\text{Molar mass} \times Q}{n \times F}$ $w = \frac{58.7 \text{ g/mol} \times 6000 \text{ C}}{2 \times 96500 \text{ C/mol}}$ $w = \frac{352200}{193000} \text{ g}$ $w = 1.825 \text{ g}$

Final Answer: The mass of Ni deposited at the cathode is 1.825 g.

Given: Current, $I = 1.5 \text{ A}$ Mass of silver deposited in cell B, $w_{\text{Ag}} = 1.45 \text{ g}$ Molar mass of Ag, $M_{\text{Ag}} = 108 \text{ g/mol}$ Molar mass of Zn, $M_{\text{Zn}} = 65.4 \text{ g/mol}$ Molar mass of Cu, $M_{\text{Cu}} = 63.5 \text{ g/mol}$

Step 1: Calculate the moles of Ag deposited. Moles of Ag = $\frac{w_{\text{Ag}}}{M_{\text{Ag}}} = \frac{1.45 \text{ g}}{108 \text{ g/mol}} = 0.01343 \text{ mol}$

Step 2: Determine the moles of electrons passed. The reaction in cell B is: $\text{Ag}^+\text{(aq)} + 1e^- \rightarrow \text{Ag(s)}$ So, 1 mole of electrons deposits 1 mole of Ag. Therefore, moles of electrons passed = 0.01343 mol.

Step 3: Calculate the total charge ($Q$). $Q = (\text{moles of electrons}) \times F = 0.01343 \text{ mol} \times 96500 \text{ C/mol} = 1295.6 \text{ C}$

Step 4: Calculate the time ($t$). $t = \frac{Q}{I} = \frac{1295.6 \text{ C}}{1.5 \text{ A}} = 863.7 \text{ s}$ Time in minutes = $\frac{863.7}{60} = 14.4 \text{ min}$

Since the cells are connected in series, the same amount of charge (1295.6 C) and the same moles of electrons (0.01343 mol) pass through all three cells.

Mass of Zinc (Zn) deposited in cell A: Reaction: $\text{Zn}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Zn(s)}$ 2 moles of electrons deposit 1 mole of Zn. Moles of Zn deposited = $\frac{\text{moles of electrons}}{2} = \frac{0.01343}{2} = 0.006715 \text{ mol}$ Mass of Zn deposited, $w_{\text{Zn}} = (\text{moles of Zn}) \times M_{\text{Zn}} = 0.006715 \text{ mol} \times 65.4 \text{ g/mol} = 0.439 \text{ g}$

Mass of Copper (Cu) deposited in cell C: Reaction: $\text{Cu}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Cu(s)}$ 2 moles of electrons deposit 1 mole of Cu. Moles of Cu deposited = $\frac{\text{moles of electrons}}{2} = \frac{0.01343}{2} = 0.006715 \text{ mol}$ Mass of Cu deposited, $w_{\text{Cu}} = (\text{moles of Cu}) \times M_{\text{Cu}} = 0.006715 \text{ mol} \times 63.5 \text{ g/mol} = 0.426 \text{ g}$

A redox reaction is feasible (spontaneous) if the standard cell potential ($E^\circ_{\text{cell}}$) for the reaction is positive. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$.

(i) $Fe^{3+}(aq)$ and $I^-(aq)$ Possible reduction: $\text{Fe}^{3+}\text{(aq)} + e^- \rightarrow \text{Fe}^{2+}\text{(aq)}$; $E^\circ = +0.77 \text{ V}$ (Cathode) Possible oxidation: $2\text{I}^{-}\text{(aq)} \rightarrow \text{I}_2\text{(s)} + 2e^-$; $E^\circ = +0.54 \text{ V}$ (Anode) $E^\circ_{\text{cell}} = 0.77 - 0.54 = +0.23 \text{ V}$. Since $E^\circ_{\text{cell}}$ is positive, the reaction is feasible.

(ii) $Ag^+(aq)$ and $Cu(s)$ Possible reduction: $\text{Ag}^{+}\text{(aq)} + e^- \rightarrow \text{Ag(s)}$; $E^\circ = +0.80 \text{ V}$ (Cathode) Possible oxidation: $\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2e^-$; $E^\circ = +0.34 \text{ V}$ (Anode) $E^\circ_{\text{cell}} = 0.80 - 0.34 = +0.46 \text{ V}$. Since $E^\circ_{\text{cell}}$ is positive, the reaction is feasible.

(iii) $Fe^{3+}(aq)$ and $Br^-(aq)$ Possible reduction: $\text{Fe}^{3+}\text{(aq)} + e^- \rightarrow \text{Fe}^{2+}\text{(aq)}$; $E^\circ = +0.77 \text{ V}$ (Cathode) Possible oxidation: $2\text{Br}^{-}\text{(aq)} \rightarrow \text{Br}_2\text{(aq)} + 2e^-$; $E^\circ = +1.09 \text{ V}$ (Anode) $E^\circ_{\text{cell}} = 0.77 - 1.09 = -0.32 \text{ V}$. Since $E^\circ_{\text{cell}}$ is negative, the reaction is not feasible.

(iv) $Ag(s)$ and $Fe^{3+}(aq)$ Possible oxidation: $\text{Ag(s)} \rightarrow \text{Ag}^{+}\text{(aq)} + e^-$; $E^\circ = +0.80 \text{ V}$ (Anode) Possible reduction: $\text{Fe}^{3+}\text{(aq)} + e^- \rightarrow \text{Fe}^{2+}\text{(aq)}$; $E^\circ = +0.77 \text{ V}$ (Cathode) $E^\circ_{\text{cell}} = 0.77 - 0.80 = -0.03 \text{ V}$. Since $E^\circ_{\text{cell}}$ is negative, the reaction is not feasible.

(v) $Br_2(aq)$ and $Fe^{2+}(aq)$ Possible reduction: $\text{Br}_2\text{(aq)} + 2e^- \rightarrow 2\text{Br}^{-}\text{(aq)}$; $E^\circ = +1.09 \text{ V}$ (Cathode) Possible oxidation: $\text{Fe}^{2+}\text{(aq)} \rightarrow \text{Fe}^{3+}\text{(aq)} + e^-$; $E^\circ = +0.77 \text{ V}$ (Anode) $E^\circ_{\text{cell}} = 1.09 - 0.77 = +0.32 \text{ V}$. Since $E^\circ_{\text{cell}}$ is positive, the reaction is feasible.