Question

Balance the following redox reactions in acidic medium :

$Zn+{NO}_3^{-}⟶{Zn}^{2+}+N_2$

• A. $5Zn+{2NO}_3^{-}+12H^{+}⟶{5Zn}^{2+}+N_2+3H_{2}O$
• B. $5Zn+{2NO}_3^{-}+12H^{+}⟶{5Zn}^{2+}+N_2+6H_{2}O$
• C. $5Zn+{2NO}_3^{-}+6H^{+}⟶{3Zn}^{2+}+N_2+6H_{2}O$
• D. None of these

Answer 1

The correct option is B $5Zn+{2NO}_3^{-}+12H^{+}⟶{5Zn}^{2+}+N_2+6H_{2}O$

The unbalanced redox equation is as follows:

$Zn+{NO}_3^{-}⟶{Zn}^{2+}+N_2$

Balance all atoms other than H and O.

$Zn+2{NO}_3^{-}⟶{Zn}^{2+}+N_2$

The oxidation number of Zn changes from 0 to +2. The change in the oxidation number is 2.
The oxidation number of N changes from +5 to 0. The change in the oxidation number per N atom is 5. For 2 N atoms, the change in the oxidation number is 10.

To balance the increase in the oxidation number with decrease in the oxidation number multiply $Zn$ and $Z{n}^{2+}$ with 5.

$5Zn+2{NO}_3^{-}⟶5{Zn}^{2+}+N_2$

To balance O atoms, add 6 water molecules on RHS.

$5Zn+2{NO}_3^{-}⟶5{Zn}^{2+}+N_2+6H_{2}O$

To balance H atoms, add 12$H^{+}$ on LHS.

$5Zn+2{NO}_3^{-}+12H^{+}⟶5{Zn}^{2+}+N_2+6H_{2}O$

This is the balanced chemical equation.