Question

The standard reduction potentials for two half-cell reactions are given below. $\mathrm{Cd}^{2+}(aq) + 2e^- \longrightarrow \mathrm{Cd}(s); E^\circ = -0.40\mathrm{V}$ and $\mathrm{Ag}^{+}(aq) + e^- \longrightarrow \mathrm{Ag}(s); E^\circ = 0.80\mathrm{V}$. The standard free energy change for the reaction $2\mathrm{Ag}^{+}(aq) + \mathrm{Cd}(s) \longrightarrow 2\mathrm{Ag}(s) + \mathrm{Cd}^{2+}(aq)$ is given by

• A. $115.8\mathrm{kJ}$
• B. $-115.8\mathrm{kJ}$
• C. $-231.6\mathrm{kJ}$
• D. $231.6\mathrm{kJ}$

Hint:

$\Delta G^\circ = -nFE^\circ_{cell}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
$\Delta G^\circ = -nFE^\circ_{cell}$.
Step 2: Detailed Explanation:
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-0.40) = 1.20$ V. $\Delta G^\circ = -nFE^\circ = -2 \times 96500 \times 1.20 = -231600$ J = $-231.6$ kJ.
Step 3: Final Answer:
$\Delta G^\circ = -231.6$ kJ.