If standard emf of cell $\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Cu}_{(1 \mathrm{M})}^{+2} \| \mathrm{Ag}_{(1 \mathrm{M})}^{+1}\right| \mathrm{Ag}_{(\mathrm{s})} \quad$ is 1.55 V. What is electrical work done under standard conditions?

The galvanic cell notation $\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}_{(1\mathrm{M})}^{+2} \| \mathrm{Ag}_{(1\mathrm{M})}^{+1}\right|\mathrm{Ag}_{(\mathrm{s})}$ indicates zinc oxidation and silver ion reduction. The balanced cell reaction is $\mathrm{Zn}_{(\mathrm{s})} + 2\mathrm{Ag}^{+1}_{(\mathrm{aq})} \rightarrow \mathrm{Zn}^{+2}_{(\mathrm{aq})} + 2\mathrm{Ag}_{(\mathrm{s})}$, with $n = 2$ electrons transferred.

The maximum electrical work is given by $W_{\text{elec}} = -\Delta G^\circ$, where $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, yielding $W_{\text{elec}} = -nFE^\circ_{\text{cell}}$.

Using $E^\circ_{\text{cell}} = 1.55\text{ V}$ and $F = 96500\text{ C/mol}$:
$W_{\text{elec}} = -(2)(96500)(1.55) = -299150\text{ J}$.

Converting to kilojoules gives $W_{\text{elec}} = -299.15\text{ kJ}$, corresponding to option C.