At the cathode or anode, there will be a reduction reaction by H2O MnO2 which is an electrolyte and will accept electrons, and become Mn2O3 and OH - which OH - will be brought back into use at the anode again, so the equation.At the cathode; 2MnO2 (s) + H2O (L) + 2E --- > Mn2O3 (s) + 2OH - (AQ).
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