Is the role of potassium hydroxide ($\ce{KOH}$) in an alkaline battery to provide hydroxide for the reaction with zinc? Would the battery cease to work if $\ce{KOH}$ was removed and only $\ce{OH-}$ from autoionization ($\pu{10^{-7} M}$) remained?


The role of $\ce{KOH}$ is to provide enough ions for redox reactions. Remember that redox reactions come in pairs, for every reduction there must be oxidation. If $\ce{Zn}$ reacts with $\ce{OH-}$ to form a hydroxide, what would be a counter-reaction?

A battery also needs to have certain ionic conductivities to be functional. Could you ever get enough $\ce{OH-}$ ions through autoionization?

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  • $\begingroup$ Thanks. Yes I'm aware of the conductivity. The cathode reaction is MnO2 + H2O + 2e-. $\endgroup$ – Volta Jun 17 at 14:08
  • $\begingroup$ I think the reason I asked is because potassium hydroxide is labeled as the electrolyte, whereas it is actually the anode reaction, its the oxidation of hydroxide that releases electrons. The anode reaction, the O that reacts with Zn comes from 2 OH- --> O + H2O + 2 e-. $\endgroup$ – Volta Jun 17 at 15:24
  • $\begingroup$ The oxidation state of the zinc does not change. ZnO has 38 electrons, O has 8 and Zn(s) has 30, and the lewis structure of ZnO is analogous to Zn2+ (28 electrons) and O-2 (10 electrons), still 38 in total. $\endgroup$ – Volta Jun 17 at 15:34
  • $\begingroup$ Or, the oxidation state of zinc does change. But the total number of electrons in it (ZnO) remains the same, zinc is actually releasing its electrons into the oxygen it bonds with. The source of the electrons released from the anode is the hydroxide ions. $\endgroup$ – Volta Jun 17 at 15:52

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