= Solution
With $n=2$, the standard reaction <Gibbs free energy> and dimensionless <equilibrium constant> are
$$
\Delta_rG^\circ=-nFE^\circ=-2(96500)(0.304)
=\boxed{-58.672\ \mathrm{kJ\,mol^{-1}}},
\qquad
K=\exp\!\left(\frac{nFE^\circ}{RT}\right)
=\boxed{1.93\times10^{10}}.
$$
For the reaction as written, $Q=a_{\rm Fe^{2+}}/a_{\rm Sn^{2+}}$, and the <Nernst equation> gives $E=E^\circ-(RT/2F)\ln Q$. Consequently \b[zero cell potential requires $[\mathrm{Fe^{2+}}]/[\mathrm{Sn^{2+}}]=K\simeq1.93\times10^{10}$] under the specified unit-activity-coefficient assumption. Reversing that ratio reverses the conclusion. This mathematical ratio does not by itself guarantee experimentally attainable <concentrations> without precipitation or other chemistry.
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