Question

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O; \( E^\circ = 1.51 \, \text{V} \) MnO4- + 4H+ + 2e- → Mn2+ + 2H2O; \( E^\circ = 1.23 \, \text{V} \)

• A. 1.70 V
• B. 0.91 V
• C. 1.37 V
• D. 0.548 V

Hint:

To calculate the cell potential, subtract the reduction potential of the cathode half-reaction from the anode half-reaction.

Answer 1

The Correct Option is B

The overall cell potential can be found by subtracting the reduction potentials of the half-reactions. The value is the difference between the two given potentials, which results in 0.91 V.