A galvanic cell is Zn(s) | Zn^2+(0.10 M) \| Cu^2+(0.010 M) | Cu(s) at 298 K. Given E^°(Zn^2+/Zn)=-0.76 V, E^°(Cu^2+/Cu)=+0.34 V, F=96500 C mol^-1 and…

The cell reaction is $\mathrm{Zn+Cu^{2+}\to Zn^{2+}+Cu}$ with $n=2$ and $E^\circ_\mathrm{cell}=0.34-(-0.76)=1.10\ \mathrm{V}$. The Nernst equation gives $E=E^\circ-\tfrac{0.0592}{2}\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}=1.10-0.0296\log10=1.0704\ \mathrm{V}$, so (A) is correct. (B) $\Delta G=-nFE=-2(96500)(1.0704)=-206.6\ \mathrm{kJ\,mol^{-1}}$, i.e. about $-207$ kJ: correct. (C) A smaller $[\mathrm{Zn^{2+}}]$ makes the reaction quotient $Q$ smaller, so $\log Q$ falls and $E$ rises; dilution of the zinc half-cell increases the emf: wrong. (D) At equilibrium $E=0$ and $\log K=nE^\circ/0.0592=2(1.10)/0.0592=37.2$, so $K=1.5\times10^{37}$: correct. Note that $\Delta G$ at the stated concentrations uses $E$, not $E^\circ$; using $E^\circ$ would give $-212$ kJ, a typical distractor.