Question

Balance the following below equation.

$Au+{NO}_3^{-}+C{l}^{-}+H^{+}\to {AuCl}_4^{-}+N{O}_2+H_{2}O$

• A. $Au+3{NO}_3^{-}+4C{l}^{-}+6H^{+}\to {AuCl}_4^{-}+3N{O}_2+3H_{2}O$
• B. $Au+2{NO}_3^{-}+5C{l}^{-}+8H^{+}\to {AuCl}_4^{-}+6N{O}_2+3H_{2}O$
• C. $Au+4{NO}_3^{-}+2C{l}^{-}+6H^{+}\to {AuCl}_4^{-}+5N{O}_2+3H_{2}O$
• D. None of these

Answer 1

Answer: (A)

$Au+3{NO}_3^{-}+4C{l}^{-}+6H^{+}\to {AuCl}_4^{-}+3N{O}_2+3H_{2}O$

The unbalanced redox equation is as follows:

$Au+{NO}_3^{-}+C{l}^{-}+H^{+}\to {AuCl}_4^{-}+N{O}_2+H_{2}O$

Balance all atoms other than H and O.

$Au+{NO}_3^{-}+4C{l}^{-}+H^{+}\to {AuCl}_4^{-}+N{O}_2+H_{2}O$

The oxidation number of gold changes from 0 to +3. The change in the oxidation number of gold is 3.
The oxidation number of nitrogen changes from +5 to +4. The change in the oxidation number of nitrogen is 1.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $N{O}_3^{-}$ and $N{O}_2$ with 3.

$Au+3{NO}_3^{-}+4C{l}^{-}+H^{+}\to {AuCl}_4^{-}+3N{O}_2+H_{2}O$

O atoms are balanced by adding 12 $H_{2}O$ molecules on RHS.

$Au+3{NO}_3^{-}+4C{l}^{-}+H^{+}\to {AuCl}_4^{-}+3N{O}_2+3H_{2}O$

Hydrogen atoms are balanced by adding 5 $H^{+}$ on LHS.

$Au+3{NO}_3^{-}+4C{l}^{-}+6H^{+}\to {AuCl}_4^{-}+3N{O}_2+3H_{2}O$

This is the balanced chemical equation.