Q125 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Chemistry
What is standard decrease in Gibbs energy for the following cell?
$\text{Al}_{(s)} \mid \text{Al}^{+3}_{(1M)} \parallel \text{Cd}^{+2}_{(1M)} \mid \text{Cd}$
if $E^\circ\,\text{Al}^{+3}/\text{Al} = -1.66$ V, $E^\circ\,\text{Cd}^{+2}/\text{Cd} = -0.40$ V
$\text{Al}_{(s)} \mid \text{Al}^{+3}_{(1M)} \parallel \text{Cd}^{+2}_{(1M)} \mid \text{Cd}$
if $E^\circ\,\text{Al}^{+3}/\text{Al} = -1.66$ V, $E^\circ\,\text{Cd}^{+2}/\text{Cd} = -0.40$ V
- A$-12.36$ kJ
- B$-123.6$ kJ
- Correct answerC$-729.54$ kJ
- D$-119.274$ kJ
Solution
The standard cell potential is given by:
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
$E^\circ_{\text{cell}} = -0.40 - (-1.66) = 1.26 \text{ V}$
The half-cell reactions are:
Anode: $\text{Al} \rightarrow \text{Al}^{+3} + 3e^-$
Cathode: $\text{Cd}^{+2} + 2e^- \rightarrow \text{Cd}$
To balance the number of electrons, the overall reaction involves the transfer of $6$ electrons, so $n = 6$.
The standard Gibbs free energy change is:
$\Delta G^\circ = -n F E^\circ_{\text{cell}}$
$\Delta G^\circ = -6 \times 96500 \times 1.26$
$\Delta G^\circ = -729540 \text{ J} = -729.54 \text{ kJ}$
Answer: $-729.54$ kJ
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
$E^\circ_{\text{cell}} = -0.40 - (-1.66) = 1.26 \text{ V}$
The half-cell reactions are:
Anode: $\text{Al} \rightarrow \text{Al}^{+3} + 3e^-$
Cathode: $\text{Cd}^{+2} + 2e^- \rightarrow \text{Cd}$
To balance the number of electrons, the overall reaction involves the transfer of $6$ electrons, so $n = 6$.
The standard Gibbs free energy change is:
$\Delta G^\circ = -n F E^\circ_{\text{cell}}$
$\Delta G^\circ = -6 \times 96500 \times 1.26$
$\Delta G^\circ = -729540 \text{ J} = -729.54 \text{ kJ}$
Answer: $-729.54$ kJ
Hence, the correct answer is option: 'C'.