# Chemical Reactions at the Cathode of the Leclanche Cell

In the dry cell, the overall reactions that occur are: $$\textbf{anode:}~~~ \ce{Zn -> Zn^{2+} + 2e^-}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$\textbf{cathode:} ~~\ce{2NH4+ + 2MnO2 + 2e- -> Mn2O3 + 2NH3 + H2O}$$ I have read that the overall reaction at the cathode actually consists of two separate reactions. However different sources have conflicting chemical equations for these two reactions.

Some sources says that the hydrogen is ammonium gets reduced first and then there is a separate redox reaction between the manganese dioxide and hydrogen gas: $$\ce{2NH4 + 2e- -> 2NH3 + H2}$$ $$\ce{2MnO2 + H2 -> 2Mn2O3 + H2O}$$ While other sources say that the manganese first gets reduced and then the produced hydroxide ions are neutralised by the ammonium: $$\ce{2MnO2 + H2O + e- -> Mn2O3 + 2OH-}$$ $$\ce{NH4+ + OH- -> NH3 + H2O}$$ Which of these reactions are correct?

Also, if the first set of equations are correct, does the redox reaction between the manganese dioxide and hydrogen gas also contribute to the voltage produced by the battery, or is it only the the redox reaction between the ammonium and zinc that contribute to the $1.5\ \mathrm{V}$ that is produced.