The solubility product of $\mathrm{AgBr}(\mathrm{s})$ is $5 \times 10^{-13}$ at 298 K . If the standard reduction potential of the half-cell, $\mathrm{E}_{\mathrm{Ag\mid AgBr}(\mathrm{s}) \mathrm{Br}^{-}}^{0}$ is 0.07 V , the standard reduction potential $\mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^{0}$ (in volts) is ____ .
Answer
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.
Explanation
Answer: ≈ 0.80 V — E°(AgBr/Ag) = E°(Ag⁺/Ag) + 0.0591 log Ksp, so E°(Ag⁺/Ag) = 0.07 + 0.0591 × 12.30 = 0.797 V.