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In an acidic solution, MnO2 acts as an oxidizing agent according to the half-reaction: MnO2 + 4H+ + 2e− → Mn2+ + 2H2O If the pH is lowered by 1 unit, by how much will the electrode potential of the Pt | MnO2, Mn2+ half-cell change?

1. 0.236 V
2. −0.236 V
3. −0.118 V
4. 0.118 V

Correct answer: 0.118 V

Solution

E = E0 - (0.059/2) log([Mn2+]/[H+]^4) = E0 + (0.236/2) log[H+] = E0 - 0.118*pH. Lowering pH by 1 unit increases the electrode potential by 0.118 V.

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