Question

Calculate the cell e.m.f. and $\Delta G$ for the cell reaction at 298 K for the cell.

Zn (s) | $Zn^{2\oplus}$ (0.0004M) || $Cd^{2\oplus}$ (0.2M) | Cd(s)

Given, $E_{Zn^{2\oplus}|Zn}^o$ = - 0.763 V; $E_{Cd^{2\oplus}|Cd}^o$ = - 0.403 V at 298 K. F = 96500 C $mol^{-1}$.

Answer 1

E_{cell} = 0.440 V, \Delta G = -84.90 kJ mol^{-1}

Answer 2

1. Identify Anode and Cathode, and write Half-Cell Reactions:

The given standard electrode potentials are: $E_{Zn^{2\oplus}|Zn}^o = -0.763 \text{ V}$ $E_{Cd^{2\oplus}|Cd}^o = -0.403 \text{ V}$

The species with the more negative standard reduction potential undergoes oxidation (anode). Comparing the two, Zn has a more negative potential than Cd. Therefore, Zn acts as the anode, and $Cd^{2+}$ acts as the cathode.

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

The overall cell reaction is: $\text{Zn(s)} + \text{Cd}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cd(s)}$ From the overall reaction, the number of electrons transferred (n) is 2.

2. Calculate the Standard Cell Potential ($E_{cell}^o$): $E_{cell}^o = E_{cathode}^o - E_{anode}^o$ $E_{cell}^o = E_{Cd^{2\oplus}|Cd}^o - E_{Zn^{2\oplus}|Zn}^o$ $E_{cell}^o = (-0.403 \text{ V}) - (-0.763 \text{ V})$ $E_{cell}^o = -0.403 \text{ V} + 0.763 \text{ V}$ $E_{cell}^o = 0.360 \text{ V}$

3. Calculate the Reaction Quotient (Q): For the cell reaction $\text{Zn(s)} + \text{Cd}^{2+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cd(s)}$, the reaction quotient Q is: $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]}$ Given concentrations: $[\text{Zn}^{2+}] = 0.0004 \text{ M}$ and $[\text{Cd}^{2+}] = 0.2 \text{ M}$ $Q = \frac{0.0004}{0.2} = \frac{4 \times 10^{-4}}{2 \times 10^{-1}} = 2 \times 10^{-3} = 0.002$

4. Calculate the Cell e.m.f. ($E_{cell}$) using the Nernst Equation: The Nernst equation at 298 K (25°C) is: $E_{cell} = E_{cell}^o - \frac{0.0592}{n} \log_{10} Q$ Substitute the calculated values: $E_{cell} = 0.360 \text{ V} - \frac{0.0592}{2} \log_{10} (0.002)$ $E_{cell} = 0.360 \text{ V} - 0.0296 \times \log_{10} (2 \times 10^{-3})$ $E_{cell} = 0.360 \text{ V} - 0.0296 \times (\log_{10} 2 + \log_{10} 10^{-3})$ $E_{cell} = 0.360 \text{ V} - 0.0296 \times (0.3010 - 3)$ $E_{cell} = 0.360 \text{ V} - 0.0296 \times (-2.699)$ $E_{cell} = 0.360 \text{ V} + 0.0798904 \text{ V}$ $E_{cell} = 0.4398904 \text{ V}$ Rounding to three decimal places, $E_{cell} = 0.440 \text{ V}$.

5. Calculate the Gibbs Free Energy Change ($\Delta G$): The relationship between Gibbs free energy change and cell potential is: $\Delta G = -nFE_{cell}$ Where: n = 2 (number of electrons transferred) F = 96500 C $mol^{-1}$ (Faraday constant) $E_{cell}$ = 0.4398904 V (calculated cell e.m.f.)

$\Delta G = -2 \times 96500 \text{ C mol}^{-1} \times 0.4398904 \text{ V}$ $\Delta G = -84900.70768 \text{ J mol}^{-1}$ Converting to kJ/mol: $\Delta G = -84.9007 \text{ kJ mol}^{-1}$ Rounding to two decimal places, $\Delta G = -84.90 \text{ kJ mol}^{-1}$.