ElectrochemistryHard

Question

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O,
If H+ concentration is decreased from 1 M to 10-4 M at 25oC, where as concentration of Mn2+ and MnO4- remain 1 M.

Options

A.the potential decreases by 0.38 V with decrease in oxidising power
B.the potential increases by 0.38 V with increase in oxidising power
C.the potential decreases by 0.25 V with decrease in oxidising power
D.the potential decreases by 0.38 V without affecting oxidising power

Solution

MnO4- + 8H+ + 5e- →  Mn2+ + 4H2O
E1 = Eo -
E2 = Eo - = - × 32 = - 0.37824
E1 - E2 = 0.38 Volt.

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