Calculate the value of the equilibrium constant for the reaction 2MnO4- + 3Mn2+ + 2H2O ↔ 5MnO2(s) + 4H+ using the following half-reactions: MnO4- + 4H+ + 3e- ↔ MnO2(s) + 2H2O E°=+1.692V MnO2(s) + 4H+ + 2e- ↔ Mn2+ + 2H2O E°=+1.230V

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