JEE Main · 2008 · 27 Apr 2008 · Offline · Redox Reactions and Electrochemistry

The standard cell e.m.f. is the cathode minus the anode standard potential, $(-0.42)-(-0.72)=0.30\mathrm{V}$, chromium being the anode. The overall reaction $2\mathrm{Cr}+3\mathrm{F}{\mathrm{e}}^{2+}\to 2\mathrm{C}{\mathrm{r}}^{3+}+3\mathrm{Fe}$ transfers six electrons, so the Nernst correction is $\frac{0.0591}{6}\log \frac{[\mathrm{C}{\mathrm{r}}^{3+}{]}^2}{[\mathrm{F}{\mathrm{e}}^{2+}{]}^3}=\frac{0.0591}{6}\log \frac{(0.1{)}^2}{(0.01{)}^3}=\frac{0.0591}{6}\log 10^4$. That term is $\frac{0.0591\times 4}{6}\approx 0.039\mathrm{V}$, and subtracting it from $0.30\mathrm{V}$ leaves about $0.26\mathrm{V}$.