Question

Balance the following redox reaction in acidic medium :

$H_{2}S+{NO}_3^{-}⟶N{O}_2+S_8$

• A. ${8H}_{2}S+{16NO}_3^{-}+4H^{+}⟶16N{O}_2+S_8+8H_{2}O$
• B. ${8H}_{2}S+{16NO}_3^{-}+16H^{+}⟶16N{O}_2+S_8+16H_{2}O$
• C. ${8H}_{2}S+{16NO}_3^{-}+8H^{+}⟶16N{O}_2+S_8+8H_{2}O$
• D. None of these

Answer 1

Answer: (B)

${8H}_{2}S+{16NO}_3^{-}+16H^{+}⟶16N{O}_2+S_8+16H_{2}O$

The unbalanced redox equation is as follows:

$H_{2}S+{NO}_3^{-}⟶N{O}_2+S_8\left(acidic\right)$

Balance all atoms other than H and O.

$8H_{2}S+{NO}_3^{-}⟶N{O}_2+S_8\left(acidic\right)$

The oxidation number of S changes from -2 to 0. The change in the oxidation number per S atom is 2.

Total change in oxidation number for 8 S atoms is 16.
The oxidation number of N changes from 5 to 4. The change in the oxidation number is 1.

So, to balance the increase in the oxidation number with decrease in the oxidation number multiply ${NO}_3^{-}$ and $N{O}_2$ with 16.

$8H_{2}S+16{NO}_3^{-}⟶16N{O}_2+S_8\left(acidic\right)$

To balance O atoms add 16 water molecules on RHS and to balance H atoms, add 16 $H^{+}$ on LHS.

${8H}_{2}S+{16NO}_3^{-}+16H^{+}⟶16N{O}_2+S_8+16H_{2}O$

This is the balanced chemical equation.