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Calculate the EMF (in V) of the following electrochemical cell at 25 deg C: Pt | H2(1 atm) | H2SO4(0.05 M) || H+(1 M), MnO4-(0.1 M), Mn2+(0.1 M) | Pt Given: E_red of (MnO4-/Mn2+) = +1.51 V; 2.303RT/F = 0.06 V.

1. 1.39 V
2. 1.57 V
3. 1.52 V
4. -1.42 V

Correct answer: 1.57 V

Solution

At the cathode, Nernst gives E_cathode = 1.51 - (0.06/5)*log([Mn2+]/([MnO4-][H+]⁸)) = 1.51 V (since [H+]=1 M, all concentrations cancel). At the anode, [H+] = 2*0.05 = 0.1 M, so E_anode = 0 - (0.06/2)*log(1/(0.1)²) = -0.06 V (as a reduction). E_cell = 1.51 - (-0.06) = 1.57 V.

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