Question

Balance the following equation by ion electron method-
$C{u}_{2}O+H^{+}+N{O}_3^{-}\to C{u}^{2+}+NO+H_{2}O$

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

Answer 1

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

The unbalanced redox equation is as follows:
$C{u}_{2}O+H^{+}+N{O}_3^{-}\to C{u}^{2+}+NO+H_{2}O$

Assign the oxidation number to each element present in the reaction.
$\stackrel{+1}{C{u}_2}O+H^{+}+\stackrel{+5}{N}{O}_3^{-}\to \stackrel{+2}{C{u}^{2+}}+\stackrel{+2}{N}O+H_{2}O$

Split the reaction into two halves i.e. the oxidation half and the reduction half.
Balance the atom other than oxygen and hydrogen atom in both half reactions.
Oxidation half equation:
$\stackrel{+1}{C{u}_2}O\to 2C{u}^{2+}$
Reduction half equation:
$\stackrel{+5}{N}{O}_3^{-}\to \stackrel{+2}{N}O$
$Cu$ and $N$ are balanced.

Now, balance $O$ atoms by adding water molecules on the side containing less number of $O$ atoms and balance hydrogen atoms by adding $H^{+}$ ions to the side containing less number of $H$ atoms.
Oxidation half equation- $C{u}_{2}O+2H^{+}\to 2C{u}^{2+}+H_{2}O$---(i)
Reduction half equation- $4H^{+}+N{O}_3^{-}\to NO+2H_{2}O$---(ii)

Equation (i) and (ii) are balanced atom wise. Now balance both equations charge wise. To balance the charge, add electrons to the electrically positive side.
Oxidation half equation-
$C{u}_{2}O+2H^{+}\to 2C{u}^{2+}+H_{2}O+2e^{-}$
Reduction half reaction-
$4H^{+}+N{O}_3^{-}+3e^{-}\to NO+2H_{2}O$

Multiply the half equations by suitable factors to equalize the number of electrons in the two half equations.
Oxidation half reaction- $3C{u}_{2}O+6H^{+}\to 6C{u}^{2+}+3H_{2}O+6e^{-}$
Reduction half reaction- $8H^{+}+2N{O}_3^{-}+6e^{-}\to 2NO+4H_{2}O$

Add the two half equations and cancel the number of electrons on both sides.
$3C{u}_{2}O+14H^{+}+2N{O}_3^{-}\to 6C{u}^{2+}+2NO+7H_{2}O$