Question

Balance the following redox reactions in basic medium:

$Al+{NO}_3^{-}⟶Al{\left(OH\right)}_4^{-}+{NH}_3$

• A. $8Al+3{NO}_3^{-}+8{OH}^{-}+18H_{2}O⟶8Al{\left(OH\right)}_4^{-}+{3NH}_3$
• B. $8Al+3{NO}_3^{-}+5{OH}^{-}+18H_{2}O⟶8Al{\left(OH\right)}_4^{-}+{3NH}_3$
• C. $8Al+3{NO}_3^{-}+6{OH}^{-}+9H_{2}O⟶8Al{\left(OH\right)}_4^{-}+{3NH}_3$
• D. None of these

Answer 1

The correct option is B $8Al+3{NO}_3^{-}+5{OH}^{-}+18H_{2}O⟶8Al{\left(OH\right)}_4^{-}+{3NH}_3$

The unbalanced redox equation is as follows:

$Al+{NO}_3^{-}⟶Al{\left(OH\right)}_4^{-}+{NH}_3\left(basic\right)$

All atoms other than H and O are balanced.
The oxidation number of Al changes from 0 to +3. The change in the oxidation number is 3.
The oxidation number of N changes from +5 to -3. The change in the oxidation number is 8.

To balance the increase in the oxidation number with decrease in the oxidation number multiply $Al$ and $Al{\left(OH\right)}_4^{-}$ with 8 and multiply ${NO}_3^{-}$ and $N{H}_3$ with 3..

$8Al+3{NO}_3^{-}⟶8Al{\left(OH\right)}_4^{-}+3{NH}_3\left(basic\right)$

To balance O atoms, add 23 water molecules on LHS.

$8Al+3{NO}_3^{-}+23H_{2}O⟶8Al{\left(OH\right)}_4^{-}+3{NH}_3\left(basic\right)$

To balance H atoms, add 5$H^{+}$ on RHS.

$8Al+3{NO}_3^{-}+23H_{2}O⟶8Al{\left(OH\right)}_4^{-}+3{NH}_3+5H^{+}\left(basic\right)$

Since the reaction occurs in basic medium, add 5 $O{H}^{-}$ ions on either side.

$8Al+3{NO}_3^{-}+23H_{2}O+5O{H}^{-}⟶8Al{\left(OH\right)}_4^{-}+3{NH}_3+5H^{+}+5O{H}^{-}\left(basic\right)$

On RHS, 5 $H^{+}$ ions combine with 5 $O{H}^{-}$ ions to form 5 water molecules which cancels with 5 water molecules on LHS.

$8Al+3{NO}_3^{-}+5{OH}^{-}+18H_{2}O⟶8Al{\left(OH\right)}_4^{-}+{3NH}_3$

This is the balanced chemical equation.