JEE Advanced 2017 Paper 2 Chemistry: Paragraph 1

The cell is a concentration cell with the same electrode on both sides, so $E^\circ=0$ and $E=0.0592\log\dfrac{[\mathrm{Ag^+}]_{right}}{[\mathrm{Ag^+}]_{left}}$. On the left, saturation with $\mathrm{AgBr}$ in $0.100\ \mathrm{M}$ bromide gives $[\mathrm{Ag^+}]_{left}=K_{sp}/0.100$. Thus $0.610=0.0592\log\dfrac{0.100\times0.100}{K_{sp}}$, so $\log(0.0100/K_{sp})=10.30$, and $0.0100/K_{sp}=2.0\times10^{10}$, which gives $K_{sp}=5.0\times10^{-13}$. If the bromide concentration is ignored and $[\mathrm{Ag^+}]_{left}=\sqrt{K_{sp}}$ is used, the result would be about $2.5\times10^{-23}$, which is unrealistically small. Treating the right-hand side as $1\ \mathrm{M}$ gives $5.0\times10^{-12}$, since the factor of $0.100$ is dropped. The silver ion concentration in the left compartment is only about $5\times10^{-12}\ \mathrm{M}$, so the left electrode is the negative terminal, matching the stated polarity.

The $\mathrm{AgBr/Ag}$ electrode is the $\mathrm{Ag^+/Ag}$ electrode in a solution whose silver ion concentration is fixed by $K_{sp}$. At unit bromide activity, $[\mathrm{Ag^+}]=K_{sp}/[\mathrm{Br^-}]=5.0\times10^{-13}\ \mathrm{M}$. The Nernst equation for $\mathrm{Ag^+ + e^-\to Ag}$ gives $E=E^\circ_{\mathrm{Ag^+/Ag}}+0.0592\log[\mathrm{Ag^+}]$. Hence $E^\circ_{\mathrm{AgBr/Ag}}=0.80+0.0592\log(5.0\times10^{-13})=0.80+0.0592\times(-12.30)=0.80-0.728=+0.07\ \mathrm{V}$. The potential is lower than $0.80\ \mathrm{V}$ because precipitation lowers the silver ion concentration, which makes Ag a stronger reducing agent. The value $-0.07\ \mathrm{V}$ would result from a sign error in the log term, and $+1.53\ \mathrm{V}$ from adding rather than subtracting the correction.