Question

Balance the following redox reaction:

$Zn+{NO}_3^{-}+{OH}^{-}⟶{NH}_3+Zn{O}_2^{2-}$

• A. $4Zn+{NO}_3^{-}+{7OH}^{-}⟶{NH}_3+4Zn{O}_2^{2-}+2H_{2}O$
• B. $4Zn+{NO}_3^{-}+{7OH}^{-}⟶{NH}_3+3Zn{O}_2^{2-}+2H_{2}O$
• C. $3Zn+{NO}_3^{-}+{7OH}^{-}⟶{NH}_3+4Zn{O}_2^{2-}+2H_{2}O$
• D. None of these

Answer 1

The correct option is A $4Zn+{NO}_3^{-}+{7OH}^{-}⟶{NH}_3+4Zn{O}_2^{2-}+2H_{2}O$

The unbalanced redox equation is as follows:
$Zn+{NO}_3^{-}+{OH}^{-}⟶{NH}_3+Zn{O}_2^{2-}$
All atoms other than H and O are balanced.
The oxidation number of Zn changes from 0 to 2. The change in the oxidation number of Zn is 2.
The oxidation number of N changes from 5 to -3. The change in the oxidation number is 8.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying Zn and $Zn{O}_2^{2-}$ with 4.
$4Zn+{NO}_3^{-}+{OH}^{-}⟶{NH}_3+4Zn{O}_2^{2-}$
O atoms are balanced by adding 4 water molecules on LHS.
$4Zn+{NO}_3^{-}+{OH}^{-}+4H_{2}O⟶{NH}_3+4Zn{O}_2^{2-}$
To balance H atoms, add 6 $H^{+}$ on RHS.
$4Zn+{NO}_3^{-}+{OH}^{-}+4H_{2}O⟶{NH}_3+4Zn{O}_2^{2-}+6H^{+}$
Since the reaction occurs in basic medium, add 6 hydroxide ions on both sides of the equation.
$4Zn+{NO}_3^{-}+4H_{2}O+7O{H}^{-}⟶{NH}_3+4Zn{O}_2^{2-}+6H^{+}+6O{H}^{-}$
On RHS, 6 $H^{+}$ ions combine with 6 $O{H}^{-}$ ions to form 6 water molecules out of which 4 cancel out with 4 water molecules on LHS.
$4Zn+{NO}_3^{-}+7O{H}^{-}⟶{NH}_3+4Zn{O}_2^{2-}+2H_{2}O$
This is the balanced chemical equation.