Question

Given below are half-cell reactions:
$MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O$,
$E°_{(Mn^{+2}/MnO_4^-)} = –1.510 V$
$(\frac{1}{2})O_2 + 2H^+ + 2e^- → H_2O,$
$E°(O_2/H_2O) = +1.223 V$
Will the permanganate ion, $MnO_4^-$ liberate $O_2$ from water in the presence of an acid?

• A. Yes, because E°cell = +0.287V
• B. No, because E°cell = -0.287V
• C. Yes, because E°cell = +2.733 V
• D. No, because E°cell = -2.733 V

Answer 1

Answer: (A)

Yes, because E°cell = +0.287V

Answer 2

The correct option is (A): Yes, because E°cell = +0.287V