An electrochemical cell is based on the following two half-reactions: Oxidation: Pb(s) → Pb2+(aq, 0.25M) + 2e−, E° = -0.13 V Reduction: MnO4−(aq, 1.80M) + 4H+(aq, 2.4M) + 3e− → MnO2(s) + 2H2O(l), E° = 1.68 V Compute the cell potential at 25 °C.
Added by Jonathan H.
Step 1
To balance the redox reaction, we need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the number of electrons transferred: 3Pb(s) → 3Pb²⁺(aq, 0.25M) + 6e⁻ 2MnO₄⁻(aq, 1.80M) + 8H⁺(aq, 2.4M) + 6e⁻ → 2MnO₂(s) + Show more…
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