A galvanic cell is built as Ag(s) | AgBr(s) | KBr(0.010 M) \| AgNO3(0.10 M) | Ag(s) at 298 K. Given Ksp(AgBr)=5.0×10^-13 and (2.303RT)/(F)=0.0591 V…

The left half-cell is a silver electrode in contact with $\mathrm{AgBr}$ and $0.010\ \mathrm{M}$ bromide, so the silver ion concentration there is fixed by the solubility product: $[\mathrm{Ag^+}]_{L}=\dfrac{K_{sp}}{[\mathrm{Br^-}]}=\dfrac{5.0\times10^{-13}}{0.010}=5.0\times10^{-11}\ \mathrm{M}$. The right half-cell has $[\mathrm{Ag^+}]_R=0.10\ \mathrm{M}$. Both electrodes are silver, so $E^\circ=0$ and this is a concentration cell with $n=1$, with the more dilute side as anode. Then $E=0.0591\log\dfrac{[\mathrm{Ag^+}]_R}{[\mathrm{Ag^+}]_L}=0.0591\log\dfrac{0.10}{5.0\times10^{-11}}=0.0591\log(2.0\times10^{9})$. Since $\log(2\times10^{9})=9.301$, $E=0.0591\times9.301=0.5497\approx0.55\ \mathrm{V}$. Because the right-hand silver ion concentration is far larger, the right electrode is the cathode where $\mathrm{Ag^+}$ is reduced, and the left electrode is the anode where silver is oxidised and immediately precipitates as $\mathrm{AgBr}$ in the bromide medium, so the left half-cell stays saturated and its silver ion level does not drift.