Question
Balance the following equations by the oxidation number method.
$\text{I}_2+\text{NO}^-_3\xrightarrow{ \ \ \ \ \ \ \ }\text{NO}_2+\text{IO}^-_3$
$\text{I}_2+\text{NO}^-_3\xrightarrow{ \ \ \ \ \ \ \ }\text{NO}_2+\text{IO}^-_3$
✓
Answer
Total increase in O.N. = 5 × 2 = 10
Total decrease in O.N. = 1
To equalize O.N. multiply $\text{NO}^-_3$, by 10
$\text{I}_2+10\text{NO}^-_3\xrightarrow{ \ \ \ \ \ \ }10\text{NO}_2+\text{IO}^-_3$
Balancing atoms other than O and H
$\text{I}_2+10\text{NO}^-_3\xrightarrow{ \ \ \ \ \ \ \ }10\text{NO}_2+2\text{IO}_3^-$
Balanching O and H
$\text{I}_2+\text{IONO}^-_3+8\text{H}^+\xrightarrow{ \ \ \ \ \ \ \ }10\text{NO}_2+2\text{IO}^-_3+4\text{H}_2\text{O}$
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