Question

Balance the following equation by ion electron method (step by step).
$$\text{Cu + HN}{\text{O}}_3\to \text{ Cu(N}{\text{O}}_3{\text{)}}_2\text{ + N}{\text{O}}_2$$

Answer 1

Hint: Solve the reducing half reaction and oxidizing half reaction separately and then add both reactions to get a balanced reaction. The radical with higher oxidation number will be reduced and the one with lower oxidation number will get oxidized.

Complete step-by-step answer:
In the given reaction we know that $HN{O}_3$will be reduced to $N{O}_2$ with a change in oxidation state of 1. On the other side $Cu$will get oxidized to $$C{u}^{2+}$$.
We will now write the separate half reactions.
-Oxidizing half reaction:
$$Cu\to Cu{(N{O}_3)}_2+2e^{-}$$
-Reducing half reaction:
$HN{O}_3\to N{O}_2+1e^{-}$
-As we can see 2 moles of electrons are needed to oxidize one mole of $Cu$ ,we multiply the reducing half reaction by 2 to cancel out the electrons on both sides. We then add both the half reactions and once again check if all atoms are accounted for and add water molecules accordingly.
$$\text{Cu + 4HN}{\text{O}}_3\to \text{ Cu(N}{\text{O}}_3{\text{)}}_2\text{ + 2N}{\text{O}}_2+2{\text{H}}_2\text{O}$$
Therefore, this is the complete and balanced chemical equation.

Additional Information: The above question could be solved by directly substituting the products and then writing the required coefficients however this does not hold good for all reactions as sometimes the electron may not be balanced, in the case of comproportionation or disproportionation reaction. That is the reason why the half reaction method is reliable as we can see the electron transfer directly and can balance them first.

Note: In some chemical reactions the same reactant undergoes oxidation as well as reduction. In such cases write the reactant in oxidation half reaction as well as in reducing half reaction and then add the two half reactions to get a final balanced chemical reaction.