Question

For decolourisation of 1 mole of acidified $KMn{O}_4$ the moles of $H_2{O}_2$ required are:

• A. 1/2
• B. 3/2
• C. 5/2
• D. 7/2

Answer 1

The correct option is C 5/2

Reaction involved is:

$KMn{O}_4+H_2{O}_2+H^{+}\to K^{+}+M{n}^{2+}+H_{2}O+O_2$

In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.

Balancing a chemical reaction as:

Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:

$K\stackrel{+7}{Mn}{O}_4+H_2\stackrel{-1}{O_2}\to \stackrel{+2}{M{n}^{2+}}+\stackrel{0}{O_2}$

Step 2: Identify the atoms that are oxidized and those that are reduced as:

Reduction: $K\stackrel{+7}{Mn}{O}_4\to \stackrel{+2}{M{n}^{2+}}$

Oxidation: $H_2\stackrel{-1}{O_2}\to \stackrel{0}{O_2}$

Step 3: oxidation-number change is:

Reduction: : $K\stackrel{+7}{Mn}{O}_4\to \stackrel{+2}{M{n}^{2+}}$: Gain of 5 electrons total

Oxidation: $H_2\stackrel{-1}{O_2}\to \stackrel{0}{O_2}$: Loss of total 2 electrons

Step 4: Balance the total change in oxidation number as:

Reduction: : $K\stackrel{+7}{Mn}{O}_4\to \stackrel{+2}{M{n}^{2+}}\times 2$: Gain of 10 electrons

Oxidation: : $H_2\stackrel{-1}{O_2}\to \stackrel{0}{O_2}\times 5$: Loss of total 10 electrons

OR

Reduction: : $2K\stackrel{+7}{Mn}{O}_4\to 2\stackrel{+2}{M{n}^{2+}}$

Oxidation: : $5H_2\stackrel{-1}{O_2}\to 5\stackrel{0}{O_2}$

Step 5: Balance K, O atoms in reduction reaction by adding $H_{2}O$ and then balance H by $H^{+}$ as:

Reduction: : $2K\stackrel{+7}{Mn}{O}_4+16H^{+}\to 2\stackrel{+2}{M{n}^{2+}}+2K^{+}+8H_{2}O$

Oxidation: : $5H_2\stackrel{-1}{O_2}\to 5\stackrel{0}{O_2}+10H^{+}$

Thus overall Balanced reaction is:

$2K\stackrel{+7}{Mn}{O}_4+16H^{+}+5H_2\stackrel{-1}{O_2}\to 2\stackrel{+2}{M{n}^{2+}}+2K^{+}+8H_{2}O+5\stackrel{0}{O_2}+10H^{+}$

cancel $H^{+}$ from both sides:

$2KMn{O}_4+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+2K^{+}+8H_{2}O+5O_2$

Thus 2 mol of $KMn{O}_4$ require 5 mol of $H_2{O}_2$ for decolorization. Therefore for decolourisation of 1 mole of acidified $KMn{O}_4$ the moles of $H_2{O}_2$ required are 5/2.