A primary battery has the reaction: $$\ce{2MnO2 + Zn + 2NH4Cl -> Mn2O3 + Zn(NH3)2Cl2 + H2O}$$ Write the half-cell reactions for the given equation.

I am having trouble understanding how you know which will be the oxidation reaction and which will be the reduction reaction. From what I have understood I should assign the oxidation states of each reactant and product and from that I assume that the reactions should be:

Ox: $\ce{Zn + 2NH4Cl -> [Zn(NH3)2]^{2+}}$ + ...

Red: $\ce{2MnO2 + ... -> Mn2O3 + H2O}$ + ...

And by balancing out the hydrogens and charges I get that:

Ox: $\ce{Zn + 2NH4Cl -> [Zn(NH3)2]^{2+} + 2H+ + 2 Cl- + 2e-}$

Red: $\ce{2MnO2 + 2H+ + 2e- -> Mn2O3 + H2O}$

Is this correct? Or should the oxidation reaction only be: $\ce{Zn -> Zn^{2+} + 2e-}$ and then the reduction reaction would consist of the remaining compounds?

All help is appreciated.

  • $\begingroup$ Your equation in the question is not balanced by mass, neither oxidation half-reaction. $\endgroup$ – Mathew Mahindaratne Oct 4 '20 at 14:44
  • $\begingroup$ @Poutnik Oh, I completely missed the Cl$_2$. So then my oxidation reaction would be: Zn+$2$NH$_4$Cl⟶[Zn(NH$_3$)$_2$]$^{+2}$+$2$H$^+$+$2$e$^−$ + $2$Cl$^-$ $\endgroup$ – katara Oct 4 '20 at 14:47
  • $\begingroup$ @Poutnik Thank you, I didn't know about that notation. But is what I wrote correct? $\endgroup$ – katara Oct 4 '20 at 14:53
  • $\begingroup$ @ Katara. Now your equation written with the $2$ chloride ions is correct $\endgroup$ – Maurice Oct 4 '20 at 15:15
  • $\begingroup$ Please fix the equations in your original question, not in the comments. $\endgroup$ – Karl Oct 4 '20 at 16:45

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