The standard Gibbs energy for the given cell reaction in kJ\ mol-1 at 298 K is: {Zn(s) + Cu^{2+}(aq) Zn^{2+}(aq) +…

Strategy: Energy change $\Delta G^\circ$ is directly proportional to the cell potential $E^\circ$ via the Faraday constant ($\Delta G^\circ = -nFE^\circ$).

Step 1: Identify parameters Reaction: $\ce{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$, $n = 2$. $E^\circ = 2 \text{ V}$. $F = 96000 \text{ C/mol}$ (as given).

Step 2: Calculate $\Delta G^\circ$ $\Delta G^\circ = -2 \times 96000 \times 2 = -384,000 \text{ J/mol}$ $\Delta G^\circ = -384 \text{ kJ/mol}$

Spontaneous reactions ($E^\circ > 0$) always have negative $\Delta G^\circ$.

$\boxed{\text{Answer: (d) -384}}$