The standard electrode potential (reduction) of the Ag+|Ag electrode is 0.800 V at 25°C. What is its electrode potential in a solution containing 10−3 M Ag+ ions? (2.303RT/F = 0.059 V at 298 K)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.623 V
The Nernst equation gives the potential of an electrode at any ion concentration. For the electrode reaction
\( \displaystyle \ce{Ag^+(aq) + e^- -> Ag(s)} \qquad n=1 \)
the concentration of solid silver is taken as unity, so at 298 K:
\( \displaystyle E=E^{\circ}-\frac{0.059}{n}\log\frac{1}{[\ce{Ag^+}]} \)
Substituting E° = 0.800 V, n = 1 and [Ag+] = 10−3 M:
\( \displaystyle E=0.800-0.059\log\frac{1}{10^{-3}}=0.800-0.059\times3 \)
\( \displaystyle E=0.800-0.177=0.623\,\mathrm{V} \)
A lower \( \ce{Ag^+} \) concentration makes the reduction less favourable, so the potential is less than E°.