Designing a galvanic cell from two half-reactions half-reaction Mn04 (aq)+2H,0)+

Question

Designing a galvanic cell from two half-reactions half-reaction Mn04 (aq)+2H,0)+3MnO(s)+4OH (aq) Cu".(aq)+e-? Cu-(a) standard reduction potential Ered = +0.59 V Ered +0.153 V Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that |L happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written Explanation Check

Explanation / Answer

For a spontaneous reaction

E°cell > 0

The balanced half reaction at cathode is reduction reaction

MnO4-(aq) + H2O(l) + 3e- = MnO2(s) + OH-(aq)

Ered = 0.59 V

The balanced half reaction at anode is oxidation reaction

Cu+(aq) = Cu2+ (aq) + 1e-

Multiply by 3

3Cu+(aq) = 3Cu2+(aq) + 3e-

Eox = - 0.153 V

Overall cell reaction

MnO4-(aq) + H2O(l) + 3Cu+(aq) = MnO2(s) + OH-(aq) + 3Cu2+(aq)

E°cell = Eox + Ered

= - 0.153 + 0.59

= 0.437 V

E°cell > 0

Reaction is spontaneous

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