Question

The complete balanced equation is:

$Cu+N{O}_3^{-}+....\to C{u}^{2+}+N{O}_2+.....$

• A. $Cu+4H^{+}+2{NO}_3^{-}\to C{u}^{2+}+2N{O}_2+2H_{2}O$
• B. $Cu+6H^{+}+{NO}_3^{-}\to C{u}^{2+}+2N{O}_2+2H_{2}O$
• C. $Cu+4H^{+}+2{NO}_3^{-}\to 2C{u}^{2+}+N{O}_2+2H_{2}O$
• D. none of these

Answer 1

The correct option is D none of these

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