How will the reduction potential of a hydrogen electrode change when the pH of its solution changes from 0 to 7 at 298 K?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Decrease by 0.413 V
For the hydrogen electrode \( \ce{H^+(aq) + e^- -> 1/2H2(g)} \), E° = 0 V. With hydrogen gas at 1 bar, the Nernst equation at 298 K gives:
\( \displaystyle E=0-0.059\log\frac{1}{[\ce{H^+}]}=-0.059\,\mathrm{pH} \)
At pH 0: \( E=-0.059\times0=0\,\mathrm{V} \)
At pH 7: \( E=-0.059\times7=-0.413\,\mathrm{V} \)
The reduction potential falls from 0 V to −0.413 V, a decrease of 0.413 V.