Having trouble understanding basic redox

I am doing a basic chemistry paper by correspondance over the holidays from my university.

I am unsure if I am going in the right direction for basic redox.

I can understand redox equations of the following form, I break these down into half equations and combine them.

$$\ce{MnO_4^- + C_2O_4^{2-} -> Mn^{2+} + CO_2}$$

Where I have issues is with the following (unbalanced) equation

$$\ce{H_2 + NO -> NH_3 + H_2O}$$

I am asked to show balanced half equations and the final combined equation.

I can see the following changes in oxidation states
reduction: $$\ce{ H -> H^{1+} + e^{1-} }$$ (for both the H in H2O and NH3)
oxidation: $$\ce{ N^{2+} + 5e^- -> N^{3-} }$$

My issue is then with balancing these and combining them, I am not sure if the half equation involving $\ce{H}$ should have $\ce{H2O}$ as the product or the $\ce{H3}$ from $\ce{NH3}$.

1. How should I go about breaking this into half equations?

2. Is it ok for both the half-equations to have molecules of the same compound in the product?

3. Should I ever have water as the product when trying to show a half equation (ie $\ce{H -> H_2O}$)?

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Assuming this reaction is taking place in aqueous phase, you can follow the conventional 7 steps for balancing redox reactions although this one does require some extra thoughts.
I will be writing the equation as it should be after each step.

$$\ce{H2 + NO -> NH3 + H2O}$$

Step 1: Ionise the required compounds and remove spectator ions

None of the compounds on either side are ionic other than $\small\ce{H2O}$ also we have no spectator ions as well. $$\small\ce{H2 + NO -> NH3 + H+ + OH-}$$

Step 2: Split into Oxidation and reduction halves

This is where the thought is required, if you decide of taking the oxidation half as $\ce{H2 -> NH3}$ (you can't simply take $\ce{H3^3-}$ as $\ce{NH3}$ is not ionic rather covalent) you would never be able to balance the nitrogen as the only source of nitrogen is $\ce{NO}$ if you observe the left hand side.
So it is decided that hydrogen cannot convert into ammonia, leaving us with only one option $\ce{H2 -> H2O}$.

$$\begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 -> H+ + OH-} & \ce{NO -> NH3} \end{array}$$

Step 3: Balance only those atoms undergoing redox

$$\begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 -> H+ + OH-} & \ce{NO -> NH3} \\ \text{(because Hydrogen is } & \text{(because only Nitrogen } \\ \text{undergoing redox but} & \text{is undergoing redox,} \\ \text{Oxygen is not.)} & \text{it's already balanced)} \\ \end{array}$$

Step 4: Balance Oxygen by adding water($\ \small\ce{H2O}\$)

$$\begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 + H2O -> H+ + OH-} & \ce{NO -> NH3 + H2O} \end{array}$$

Step 5: Balance Hydrogen by adding $\ \small\ce{H+}$

$$\begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 + H2O -> H+ + OH- + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\ \ce{H2 + H2O -> 3H+ + OH-} & \ce{NO + 5H+ -> NH3 + H2O} \end{array}$$

→ If medium is basic add equal amount of $\small\ce{OH-}$ to $\small\ce{H+}$ and combine them to make $\small\ce{H2O}$

Till the above steps mass has been balanced, now $\small\ce{H+}$ and $\small\ce{OH-}$ can be combined to form $\small\ce{H2O}$ and on further simplification oxidation half becomes $\small\ce{H2 -> 2H+}$

Step 6: Balance charge in each side of each half by adding electrons

$$\begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 -> 2H+ + 2e-} & \ce{NO + 5H+ + 5e- -> NH3 + H2O} \end{array}$$

→ Verify correctness by ensuring electrons are on right hand side in oxidation half and on left hand side in reduction half.

Step 7: Make net electrons produced equal to net electrons consumed

Multiply oxidation half by 5 and reduction half by 2 (this is basically like taking the LCM and multiplying both halves to make the number of electrons equal to the LCM, here 10.)

$$\begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{5H2 -> 10H+ + 10e-} & \ce{2NO + 10H+ + 10e- -> 2NH3 + 2H2O} \end{array}$$

Finally add the two halves and cancel out the common cmopounds and you get the balanced equation $$\begin{matrix} \ce{&5H2 + &2NO + &10H+ + &10e- &-> &10H+ + &10e- + &2NH3 + &2H2O} \\ \ce{&5H2 + &2NO & & &-> & & &2NH3 + &2H2O}\\ \end{matrix}$$

In this case the oxidation half could have been directly/trivially stated as $\small\ce{H2 -> 2H+}$

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 I guess StackExchange will start charging me if I post answers as big as this henceforth... – Aditya Sriram Jan 1 at 6:49 Well, we haven't yet started charging Terry yet (see his answers on Chem and Physics). And they didn't charge me for chemistry.stackexchange.com/users/83/terry-bollinger or physics.stackexchange.com/questions/19372/… , so I doub't they'll do anything about your tiny answer ;-) – ManishEarth♦ Jan 1 at 13:32 Right after posting that comment I went to this thread and thought to myself "Nope they won't charge me, If they did, these people from the above thread will go bankrupt". I came to know there was a maximum character limit of 30000. – Aditya Sriram Jan 1 at 14:55 Yeah, there is. Closest I've gotten is 27k characters, in this post of mine. [Read that if you have the time, it's relevant to everyone who uses SE] – ManishEarth♦ Jan 1 at 15:00 @AdityaSriram Thank you very much for your answer, it didn't feel correct to have H2O as the RHS for the oxidation of hydrogen, so I went on a wild goose chase. Thank you for your clear explanation. – cjh Jan 2 at 3:27