Question

Balance the following equation by oxidation number method.
$Cu+HN{O}_3\to Cu(N{O}_3{)}_2+N{O}_2+H_{2}O$.

Answer 1

Writing the oxidation numbers of all the atoms.
$\stackrel{0}{Cu}+\stackrel{+1}{H}\stackrel{+5}{N}\stackrel{-2}{O_3}\to \stackrel{+2}{Cu}(\stackrel{+5}{N}\stackrel{-2}{O_3}{)}_2+\stackrel{+4}{N}\stackrel{-2}{O_2}+\stackrel{+1}{H_2}\stackrel{-2}{O}$
Change in Ox. no. has occurred in copper and nitrogen.
$\stackrel{0}{Cu}\to \stackrel{+2}{Cu}(N{O}_3{)}_2$
$H\stackrel{+5}{N}{O}_3\to \stackrel{+4}{N}{O}_2$
Increase in Ox. no. of copper $=2$ units per molecule Cu.
Decrease in Ox. no. of nitrogen$=1$ unit per molecule $HN{O}_3$.
To make increase and decrease equal, eq. (ii) is multiplied by $2$
$Cu+2HN{O}_3\to Cu(N{O}_3{)}_2+2N{O}_2+H_{2}O$
Balancing nitrate ions, hydrogen and oxygen, the following equation is obtained:
$Cu+4HN{O}_3\to Cu(N{O}_3{)}_2+2N{O}_2+2H_{2}O$
This is the balanced equation.