
Balance the following equation by ion electron method (step by step).
\[\text{Cu + HN}{{\text{O}}_{3}}\text{ }\to \text{ Cu(N}{{\text{O}}_{3}}{{\text{)}}_{2}}\text{ + N}{{\text{O}}_{2}}\]
Answer
564.6k+ views
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:
\[Cu\to Cu{{(N{{O}_{3}})}_{2}}+2{{e}^{-}}\]
-Reducing half reaction:
$HN{{O}_{3}}\,\to \,\,N{{O}_{2}}\,+\,1{{e}^{-}}$
-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.
\[\text{Cu + 4HN}{{\text{O}}_{3}}\text{ }\to \text{ Cu(N}{{\text{O}}_{3}}{{\text{)}}_{2}}\text{ + 2N}{{\text{O}}_{2}}\,+\,2{{\text{H}}_{2}}\text{O}\]
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.
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:
\[Cu\to Cu{{(N{{O}_{3}})}_{2}}+2{{e}^{-}}\]
-Reducing half reaction:
$HN{{O}_{3}}\,\to \,\,N{{O}_{2}}\,+\,1{{e}^{-}}$
-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.
\[\text{Cu + 4HN}{{\text{O}}_{3}}\text{ }\to \text{ Cu(N}{{\text{O}}_{3}}{{\text{)}}_{2}}\text{ + 2N}{{\text{O}}_{2}}\,+\,2{{\text{H}}_{2}}\text{O}\]
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

