Question

The $\Delta_r G^\circ$ of the galvanic cell in which the following cell reaction takes place, \[ 2\mathrm{Cr(s)} + 3\mathrm{Cd^{2+}(aq)} \rightarrow 2\mathrm{Cr^{3+}(aq)} + 3\mathrm{Cd(s)} \] is ($E^\circ_{\mathrm{Cr^{3+}/Cr}} = -0.74$ V and $E^\circ_{\mathrm{Cd^{2+}/Cd}} = -0.40$ V)

• A. $-196.86\ \text{kJ mol}^{-1}$
• B. $+196.86\ \text{kJ mol}^{-1}$
• C. $-96.50\ \text{kJ mol}^{-1}$
• D. $+96.50\ \text{kJ mol}^{-1}$
• E. $+98.12\ \text{kJ mol}^{-1}$

Hint:

For electrochemistry: - Always identify cathode and anode correctly - Use $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ - $\Delta G^\circ = -nFE^\circ$

Answer 1

The Correct Option is A

Concept:
• Cell emf: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
• Gibbs free energy change: \[ \Delta_r G^\circ = -nFE^\circ_{\text{cell}} \] where $n$ = number of electrons transferred, $F = 96500\ \text{C mol}^{-1}$

Step 1: Identify oxidation and reduction.
\[ \mathrm{Cr \rightarrow Cr^{3+} + 3e^-} \quad (\text{oxidation}) \] \[ \mathrm{Cd^{2+} + 2e^- \rightarrow Cd} \quad (\text{reduction}) \]

Step 2: Balance electrons.
LCM of $3$ and $2$ is $6$: \[ 2\mathrm{Cr} \rightarrow 2\mathrm{Cr^{3+}} + 6e^- \] \[ 3\mathrm{Cd^{2+}} + 6e^- \rightarrow 3\mathrm{Cd} \]

Step 3: Determine $n$.
\[ n = 6 \]

Step 4: Calculate cell emf.
Cathode: Cd$^{2+}$/Cd = $-0.40$ V
Anode: Cr$^{3+}$/Cr = $-0.74$ V \[ E^\circ_{\text{cell}} = (-0.40) - (-0.74) = +0.34\ \text{V} \]

Step 5: Calculate $\Delta_r G^\circ$.
\[ \Delta_r G^\circ = -nFE^\circ = -6 \times 96500 \times 0.34 \] \[ = -196860\ \text{J mol}^{-1} = -196.86\ \text{kJ mol}^{-1} \]