Question

Write the net ionic equations for the following reactions in basic solutions. ${H_2}{O_2}\left( {aq} \right) + C{l_2}{O_7}\left( {aq} \right) \to ClO_2^ - \left( {aq} \right) + {O_2}\left( g \right)$
A.$4{H_2}{O_2}\left( {aq} \right) + C{l_2}{O_7}\left( {aq} \right)2O{H^ - }\left( {aq} \right) \to 2ClO_2^ - \left( {aq} \right) + 4{O_2}\left( g \right) + 5{H_2}O$
B.$3{H_2}{O_2}\left( {aq} \right) + C{l_2}{O_7}\left( {aq} \right)2O{H^ - }\left( {aq} \right) \to 2ClO_2^ - \left( {aq} \right) + 4{O_2}\left( g \right) + 4{H_2}O$
C.$6{H_2}{O_2}\left( {aq} \right) + 2C{l_2}{O_7}\left( {aq} \right)2O{H^ - }\left( {aq} \right) \to 2ClO_2^ - \left( {aq} \right) + 4{O_2}\left( g \right) + 10{H_2}O$
D.None of these

Answer 1

To balance the given net ionic equation we can use any one of the two methods to balance the redox reactions. We have given the case of a basic solution so we have to consider it while balancing the reaction.

Complete step by step answer:
We will use the oxidation number method to balance the ionic equation. The oxidation number method is based on the change in the oxidation number of the reducing agent and the oxidizing agent. We will follow some steps.
Step $1$: We will write the skeletal ionic equation which is ${H_2}{O_2}\left( {aq} \right) + C{l_2}{O_7}\left( {aq} \right) \to ClO_2^ - \left( {aq} \right) + {O_2}\left( g \right)$

Step $2$: we will assign oxidation number for individual atom which changes oxidation number.
${H_2}\mathop {{O_2}}\limits^{ - 1} \left( {aq} \right) + \mathop {C{l_2}}\limits^{ + 7} {O_7}\left( {aq} \right) \to \mathop {Cl}\limits^{ + 3} O_2^ - \left( {aq} \right) + \mathop {{O_2}}\limits^0 \left( g \right)$
Step $3$: We have identified that the oxidation number $Cl$ decreases from $+ 7$ in $C{l_2}{O_7}$ to $+ 3$ in $ClO_2^ -$ and the oxidation number $O$ increases from $- 1$ in ${H_2}{O_2}$ to $0$ in ${O_2}$. Now we will multiply the increase or decrease in oxidation number with the number of atoms changing. So here for this ionic reaction, we have a reduction in $C{l_2}{O_7} = 2 \times 4 = 8$ and increment in the oxidation number $O = 2 \times 1 = 2$
Step $4$: Now we will multiply the formula with the suitable integer to equalize the increase and the decrease in the oxidation number. So we will multiply ${H_2}{O_2}$ by $4$ and $C{l_2}{O_7}$ by $1$. We have also balanced the other atoms $Cl$.
$4{H_2}{O_2}\left( {aq} \right) + C{l_2}{O_7}\left( {aq} \right) \to 2ClO_2^ - \left( {aq} \right) + 4{O_2}\left( g \right)$
Step $5$: Add water molecules to the side containing fewer $O$ atoms to balance the $O$ atoms. So we will add $3{H_2}O$ to the right side.
$4{H_2}{O_2}\left( {aq} \right) + C{l_2}{O_7}\left( {aq} \right) \to 2ClO_2^ - \left( {aq} \right) + 4{O_2}\left( g \right) + 3{H_2}O$.
Step $6$: In the basic medium proper number of ${H_2}O$ molecules are added to the side containing less number of $H$ atoms while an equal number $O{H^ - }$ is added to the other side. So here we will add $2{H_2}O$ with $2O{H^ - }$ to balance $H$ atoms.
$4{H_2}{O_2}\left( {aq} \right) + C{l_2}{O_7}\left( {aq} \right)2O{H^ - }\left( {aq} \right) \to 2ClO_2^ - \left( {aq} \right) + 4{O_2}\left( g \right) + 5{H_2}O$
This is the balanced reaction.
Therefore, the correct option is (A).

Note:
The alternative method to balance the ionic equation is the ion-electron method also known as the half-reaction method. It is based on splitting the redox reaction into two half-reactions one involving oxidation and the other involving reduction.
In an acidic medium, the proper number of ${H^ + }$ ions is added to the side containing less number of $H$ atoms