Question

The $\Delta G^\circ$ for the reaction, $\text{Cd}^{2+}(aq) + \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + \text{Cd}(s)$ is: (Given $E_{\text{Cd}^{2+}/\text{Cd}}^\circ = -403\text{ V}$, $E_{\text{Zn}^{2+}/\text{Zn}}^\circ = -763\text{ V}$)}

• A. $-44.5\text{ kJ}$
• B. $-50\text{ kJ}$
• C. $-72.2\text{ kJ}$
• D. $-69.5\text{ kJ}$

Hint:

A spontaneous redox process always has a positive cell potential ($E_{\text{cell}}^\circ > 0$) and a corresponding negative free energy change ($\Delta G^\circ < 0$). Calculating $0.763 - 0.403 = 0.36\text{ V}$ quickly leads to $-2 \times 96500 \times 0.36 \approx -69.5\text{ kJ}$.

Answer 1

The Correct Option is D

Understanding the Concept: The standard Gibbs free energy change ($\Delta G^\circ$) of an electrochemical cell reaction is related to its standard cell potential ($E_{\text{cell}}^\circ$) by the relationship: \[ \Delta G^\circ = -nFE_{\text{cell}}^\circ \] where $n$ is the number of moles of electrons transferred, and $F$ is Faraday's constant ($96500\text{ C mol}^{-1}$). The standard cell potential is calculated using: \[ E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \]
Step 1: Identify the half-reactions and calculate the standard potential ($E_{\text{cell}}^\circ$).
From the overall cell equation:
$\text{Zn}$ is oxidized: $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$ (Anode)
$\text{Cd}^{2+}$ is reduced: $\text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s)$ (Cathode)
This shows that $n = 2$ electrons are transferred. Let's find $E_{\text{cell}}^\circ$: \[ E_{\text{cell}}^\circ = E_{\text{Cd}^{2+}/\text{Cd}}^\circ - E_{\text{Zn}^{2+}/\text{Zn}}^\circ = (-0.403\text{ V}) - (-0.763\text{ V}) \] \[ E_{\text{cell}}^\circ = -0.403 + 0.763 = +0.360\text{ V} \]
Step 2: Substitute values into the Gibbs free energy relation.
\[ \Delta G^\circ = -2 \times 96500\text{ C mol}^{-1} \times 0.360\text{ V} \] \[ \Delta G^\circ = -193000 \times 0.360 = -69480\text{ J} \] Converting Joules to kilojoules ($\text{kJ}$): \[ \Delta G^\circ = -\frac{69480}{1000} = -69.48\text{ kJ} \approx -69.5\text{ kJ} \]