Balancing redox reaction of zinc and hot diluted nitric acid

I need to balance the following reaction:

$$\ce{Zn + HNO3 -> Zn(NO3)2 + NH4NO3 + H2O}$$

I assigned all oxidation numbers:

$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{2+}{Zn}(\overset{+5}{N}\overset{-2}{O_3})2 +\overset{-3}{N}\overset{+1}{H_4}\overset{+5}{N}\overset{-2}{O_3} +\overset{+1}{H2}\overset{-2}{O}},$$

but I'm having troubles finding the half reactions. I know

$$\ce{Zn -> Zn^{2+} + 2 e-}$$

is one half reaction, but what about the other? Since ammonium nitrate has nitrogen atoms in two different oxidation states, what do I do with them? Do I add them to see if the nitrogen is or it is not balanced?

• Think a bit: none of the hydrogens or oxygens change oxidation numbers. But some (not all, obviously) of the nitrogens drop from +5 to -3 in oxidation numbers. Those are in the ammonium ions. – Ed V Apr 19 at 19:06
• You just need to know that 4*2=1*8, so 4 Zn for 1 ammonium. That simple. – Poutnik Apr 20 at 2:47

The oxidation state of $$\ce{Zn}$$ as calculated by you is incorrect. The correct oxidation state of $$\ce{Zn}$$ is $$+2$$ as it is associated with two mono negatively charged $$\ce{NO3-}$$ ions.

Thus the unbalanced reaction with correct oxidation states would be:

$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{+2}{Zn}(\overset{+5}{N}\overset{-2}{O_3})2 +\overset{-3}{N}\overset{+1}{H_4}\overset{+5}{N}\overset{-2}{O_3} +\overset{+1}{H2}\overset{-2}{O}}$$

Now the half-cell reactions are as follows: $$\ce{Zn -> Zn^{2+} + 2e-}\\ \ce{NO3- + 10H+ + 8e- -> NH4+ + 3H2O}\\$$

Thus the balanced iconic reaction would be:

$$\ce{4Zn + NO3- + 10H+ -> 4Zn^{2+} + NH4+ + 3H2O}\\\\$$

Now adding nine $$\ce{NO3-}$$ spectator ions to the ionic reaction gives us the standard molecular reaction.

$$\ce{4Zn + 10HNO3 -> 4Zn(NO3)2 + NH4NO3 + 3H2O}\\\\$$

What did you learn from this?

Always try to write the half-cell reaction in ionic form and only involve the ions which undergo oxidation state change. Don't bother yourself with the spectator ions at first. After you balance the equation in an ionic form adjust it to the molecular form by adding relevantly needed ions. Identify the oxidizing and the reducing species is the key.

• @EdV In defense of OP, they tried to assign oxidation numbers and made a scheme for the oxidation reaction, so there is definitely some effort. Judging from the comments, formatting also wasn't easy, but they got it pretty well done IMO. As for the answer, I don't see anything wrong with it: it is self-contatined and was posted after several hints in the comments. – andselisk Apr 19 at 19:54
• @EdV I've just corrected the question and cleaned up the comment section that was becoming a bit heated. You are probably correct, but сurrently the question and the answer look decent to me, and everything else is on the OP's conscience. – andselisk Apr 19 at 20:10
• @andselisk It is appreciated! As a moderator, you are always fair and unabashedly rigorous. I hope you keep being a moderator. I just see this one as a lost learning opportunity. End of story. – Ed V Apr 19 at 20:23
• Usually a half-equation does not contain an ionic substance containing two identical atoms but at a different oxidation number, like here $\ce{NH4NO3}$ . It is better to write it with ammonium ion only. The nitrate ion will come later on, as Nisarg have done. – Maurice Apr 19 at 21:02
• @Xetrez Downvoting is not necessarily related to violating rules ( closing votes are ). It is related - as hovering over hint suggests - to subjective evaluation the question or answer is not elaborated enough, unclear, not useful for community or otherwise inferior. The opposite for upvoting. It is preferred but not required habit to comment downvoting to give feedback for improvement. – Poutnik Apr 20 at 2:41