Question

What is standard decrease in Gibbs energy for the following cell?
\(\text{Al}_{(s)}|\text{Al}_{(1M)}^{+3}\parallel \text{Cd}_{(1M)}^{+2}|\text{Cd}\)
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
• C. \(-729.54\) kJ
• D. \(-119.274\) kJ

Hint:

Find the cell EMF, the number of electrons transferred (6), and use the change in Gibbs energy equals minus nFE.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(\Delta G^{\circ} = -nFE^{\circ}_{cell}\), with \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\).

Step 2: Cell reaction
Al is oxidised at the anode and \(\text{Cd}^{2+}\) is reduced at the cathode.
\(2\text{Al} + 3\text{Cd}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Cd}\), so \(n = 6\).
\(E^{\circ}_{cell} = -0.40 - (-1.66) = 1.26\) V.

Step 3: Calculation
\(\Delta G^{\circ} = -6\times 96500\times 1.26 = -729540\) J \(= -729.54\) kJ.

Final Answer:
The standard Gibbs energy change is -729.54 kJ. \[ \boxed{\text{(C) }-729.54\text{ kJ}} \]