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

$$\ce{MnO4^- + C2O4^{2-} -> Mn^{2+} + CO2}$$

Where I have issue is with the following (unbalanced) equation:

$$\ce{H2 + NO -> NH_3 + H2O}$$

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 $\ce{H}$ in water and ammonia)
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 $\ce{H -> H_2O}$)?

2 Answers 2


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.* $$\small\ce{H2 + NO -> NH3 + H+ + OH-}$$ *-Spectator ions are ions which don't participate in the reaction or retain their oxidation state.

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. Also, since you know that Nitrogen is being reduced (+2 -> -3) the other element, Hydrogen must get oxidised (0 -> +1). 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 5a: Balance Hydrogen by adding $\ \small\ce{H+}$

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

Step 5b: (Only in case of basic medium) Adding $\ \small\ce{OH-}$ to $\ \small\ce{H+}$

This step must follow Step 5a and means nothing on it's own. $$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline \ce{H2 \qquad -> 2H+} & \ce{NO + 5H+ \qquad \qquad -> NH3 + H2O \qquad \qquad} \\ \ce{H2 + 2OH- -> 2H+ + 2OH-} & \ce{NO + 5H+ + 5OH- -> NH3 + H2O + 5OH-} \\ \ce{H2 + 2OH- -> [2H+ + 2OH- ]} & \ce{NO + [5H+ + 5OH- ]-> NH3 + H2O + 5OH-} \\ \ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 5H2O \qquad -> NH3 + H2O + 5OH-} \\ \hline \ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 4H2O \qquad -> NH3 + 5OH- \quad \qquad }\\ \hline \end{array} $$

The medium being acidic or basic merely determines whether $\ \small\ce{H+}$ or $\ \small\ce{OH-}$ will be seen in the final redox equation.

Till the above steps mass has been balanced, now charge balancing.

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

$$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline (acidic)\quad \ce{H2 -> 2H+ + 2e-} & \ce{NO + 5H+ + 5e- -> NH3 + H2O} \\ \hline (basic) \quad \ce{H2 + 2OH- -> 2H2O + 2e-} & \ce{NO + 4H2O + 5e- -> NH3 + 5OH-} \\ \hline \end{array} $$

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

→ Also verify if net change in oxidation state is equal to the number of electrons appearing in each half. eg. Hydrogen is going from 0 to +1 oxidation state and there are 2 Hydrogen atoms in the left side of the oxidation half undergoing oxidation(don't count $\small\ce{H}$ from $\small\ce{OH-}$), meaning a net change of '2', which is equal to the number of electrons. Similarly Nitrogen is going from +2 to -3 and there is just one Nitrogen atom in the left side of the reduction half meaning a net change of '5' which is equal to the number of electrons appearing in the equation. (This is what goes on to be later called as the 'n-factor' or the 'valency-factor')

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.)

$$ \begin{array}{l|r} \text{Oxidation Half} & \text{Reduction half} \\ \hline (acidic) \ce{5H2 -> 10H+ + 10e-} & \ce{2NO + 10H+ + 10e- -> 2NH3 + 2H2O}\\ \hline (basic) \ce{5H2 + 10OH- -> 10H2O + 10e-} & \ce{2NO + 8H2O + 10e- -> 2NH3 + 10OH-} \\ \hline \end{array} $$

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

Basic Medium $$\begin{matrix} \ce{&5H2 + &10OH- + &2NO + &8H2O + &10e- &-> &10H2O + &10e- + &2NH3- + &10OH-} \\ \ce{&5H2 + & &2NO & & &-> &2H2O + & &2NH3}\\ \end{matrix}$$

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

You can also carry on $\small\ce{H+}$ ion till the end net reaction and then perform Step 5b which is less tedious.


I sympathise with your problem using electron half equations when one of the reactants produces two different products. For this reason and the fact that (as the previous answer points out) there are no ionic compounds invoved, I prefer to use oxidation number half equations.

Oxidation half equation: $\ce{H2 -> 2H+}$

Hydrogen oxidation numer goes up 2

Reduction half equation: $\ce{NO -> NH3}$
Nitrogen's oxidation number goes down by 5 (+2 to -3)

Oxygen is unchanged (-2)

Therefor to balance the oxidation numbers $5~\ce{H2}$s require $2~\ce{NO}$s and the equation becomes: $$\ce{5H2 + 2NO -> 2NH3 + 2H2O}$$


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