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Ammonium nitrate comproportionates into dinitrogen monoxide in the following reaction:

$\ce{NH4+ (aq) + NO3- (aq) -> N2O(g) + 2H2O (l)}$

I attempted to write the half-reactions by calculating the oxidation state for nitrogen in the ammonium and nitrate ions, and then finding the electrons needed for each half-reaction.

For example, for the ammonium half-reaction, I noticed that $\ce{N}$ is in a -3 oxidation state in $\ce{NH4+}$ and in a +1 oxidation state in $\ce{N2O}$. I concluded that four electrons must reduce ammonia and wrote the following preliminary equation:

$\ce{NH4+ (aq) + 4e- -> N2O (g)}$

Which I balanced the elements using standard acidic redox into:

$\ce{2NH4+ (aq) + H2O (l) + 4e- -> N2O(g) + 6H+ (aq)}$

However, I realized that my chemical equation is not balanced with respect to charge, indicating I have a flawed half-reaction.

What would be the correct half-reactions for this chemical reaction?

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    $\begingroup$ It is half-reaction(oxidation)$$\ce{2NH4+ (aq) + H2O (l) -> N2O(g) + 10H+ (aq) + 8e-}$$ $\endgroup$ – Adnan AL-Amleh Dec 11 '18 at 23:27
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    $\begingroup$ Ahhh. I had two mistakes: recognizing which species undergoes oxidation vs reduction and using the appropriate number of e-. $\endgroup$ – Ethiopius Dec 11 '18 at 23:51

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