Question

Balance the following equation.

$C{u}_{2}O+H^{+}+{NO}_3^{-}\to C{u}^{2+}+NO+H_{2}O$

• A. $3C{u}_{2}O+14H^{+}+2{NO}_3^{-}\to 6C{u}^{2+}+2NO+7H_{2}O$
• B. $C{u}_{2}O+14H^{+}+6{NO}_3^{-}\to 6C{u}^{2+}+3NO+7H_{2}O$
• C. $3C{u}_{2}O+7H^{+}+4{NO}_3^{-}\to 6C{u}^{2+}+6NO+7H_{2}O$
• D. None of these

Answer 1

The correct option is A $3C{u}_{2}O+14H^{+}+2{NO}_3^{-}\to 6C{u}^{2+}+2NO+7H_{2}O$

The unbalanced redox equation is as follows:

$C{u}_{2}O+H^{+}+{NO}_3^{-}\to C{u}^{2+}+NO+H_{2}O$

Balance all atoms other than H and O.

$C{u}_{2}O+H^{+}+{NO}_3^{-}\to 2C{u}^{2+}+NO+H_{2}O$

The oxidation number of copper changes from +1 to +2. The change in the oxidation number of one atom of copper is 1.
Total change in the oxidation number for 2 Cu atoms is 2.
The oxidation number of nitrogen changes from +5 to +2. The change in the oxidation number of nitrogen is 3.

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $C{u}_{2}O$ and $C{u}^{2+}$ with 3 and by multiplying $N{O}_3^{-}$ and NO with 2.

$3C{u}_{2}O+H^{+}+2{NO}_3^{-}\to 6C{u}^{2+}+2NO+H_{2}O$

O atoms are balanced by adding 6 water molecules on RHS.

$3C{u}_{2}O+H^{+}+2{NO}_3^{-}\to 6C{u}^{2+}+2NO+7H_{2}O$

Hydrogen atoms are balanced by adding 13 $H^{+}$ on LHS.

$3C{u}_{2}O+14H^{+}+2{NO}_3^{-}\to 6C{u}^{2+}+2NO+7H_{2}O$

This is the balanced chemical equation.