Question

Balance the following equation in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
$C{l_2}{O_7}(g) + {H_2}{O_2}(aq) \to ClO_2^ - (aq) + {O_2}(g) + {H^ + }$

Answer 1

First, we will note the oxidising and reducing agent by their change in oxidation number. We should first know the ion-electron method. Then after we will balance it. Once we have done that we will move to the oxidation number method. And we will again balance it with it. Then we will match the both balanced reactions to know if we are right.

Complete step by step solution:
Step1: The oxidation number in Oxygen is increased from $- 1$ to $0$ whereas, the oxidation number of chlorine is decreased from $+ 7$ to $+ 3$. From here we can conclude that the $C{l_2}{O_7}$ is the oxidising agent and the ${H_2}{O_2}$ is a reducing agent.
Step2: The ion-electron method also called as the half-reaction method, in it the redox reaction is divided in two halves. One is for oxidation and the other is for reduction. Both are balanced separately and then combined later to make a complete balance reaction.
Step3: Ion electron method for balancing
We write the oxidation part of equation first
${H_2}{O_2}(aq) \to {O_2}(g)$
As there is a difference of 2 in Oxidation number, we add 2 electrons
${H_2}{O_2}(aq) \to {O_2}(g) + 2{e^ - }$
2 more hydroxide ions are added to make it charge neutral
${H_2}{O_2}(aq) + 2O{H^ - } \to {O_2}(g) + 2{e^ - }$
Now the hydrogen and oxygen molecules are more on one side so we add 2 water molecules.
${H_2}{O_2}(aq) + 2O{H^ - } \to {O_2}(g) + 2{e^ - } + 2{H_2}O(aq)$
Now we write the reduction half
$C{l_2}{O_7} \to ClO_2^ - (aq)$
The Chlorine is already balanced so we add 8 electrons to balance the oxidation number
$C{l_2}{O_7} + 8{e^ - } \to ClO_2^ - (aq)$
To balance the charge 6 hydride ions are added.
$C{l_2}{O_7} + 8{e^ - } \to ClO_2^ - (aq) + 6O{H^ - }(aq)$
Now the oxidation half is multiplied by 4 to eliminate the electrons and is added to the reduction half.
The balanced equation is
$C{I_2}{O_7}(g) + 4{H_2}{O_2}(aq) + 2O{H^ - } \to ClO_2^ - (aq) + 4{O_2}(g) + 5{H_2}O(l)$
Step4: Oxidation number method is a way of balancing the equation while watching the electrons in a redox reaction. It is based on the idea that electrons are transferred between charged atoms.
Step5: Oxidation number method
Gain in oxidation number of ${H_2}{O_2}$ is 2. While the decrease in oxidation number of $C{l_2}{O_7}$ is 8.
${H_2}{O_2}$ and ${O_2}$ are multiplied by 4 and $ClO_2^ -$ by 2
$C{l_2}{O_7}(g) + 4{H_2}{O_2}(aq) \to 2ClO_2^ - (aq) + 4{O_2}(g)$
The oxygen atoms are balanced by adding 3 water molecules and Hydrogen atoms are balanced by adding 2 hydroxide ions and 2 more water molecules. We get a balanced reaction in the end.
$C{l_2}{O_7}(g) + 4{H_2}{O_2}(aq) + 2O{H^ - } \to 2ClO_2^ - (aq) + 4{O_2}(g) + 5{H_2}O(l)$
Since we can see that we get the same equation using both methods.

Note: The balancing of equations is necessary because it follows the Law of Conservation of Mass. This is one of the guiding principles. The balanced reaction helps us to calculate the amount of reactant needed to produce a particular amount of product.