For decolourisation of 1 mole of acidified $K M n O_{4}$ the moles of $H_{2} O_{2}$ required are:

1/2

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3/2

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5/2

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7/2

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Solution

The correct option is C 5/2
Reaction involved is:
$K M n O_{4} + H_{2} O_{2} + H^{+} \to K^{+} + M n^{2 +} + H_{2} O + O_{2}$

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 I: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atoms as:
$+ 7 K M n O_{4} + - 1 H_{2} O_{2} \to + 2 M n^{2 +} +_{2}$
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: $+ 7 K M n O_{4} \to + 2 M n^{2 +}$
Oxidation: $- 1 H_{2} O_{2} \to_{2}$
Step 3: oxidation-number change is:
Reduction: $+ 7 K M n O_{4} \to + 2 M n^{2 +} :$ Gain of 5 electrons total
Oxidation: $- 1 H_{2} O_{2} \to_{2} :$ Loss of total 2 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: $+ 7 K M n O_{4} \to + 2 M n^{2 +} \times 2 :$ Gain of 10 electrons
Oxidation: $: - 1 H_{2} O_{2} \to_{2} \times 5 :$ Loss of total 10 electrons
OR
Reduction: $: + 7 2 K M n O_{4} \to + 2 2 M n^{2 +}$
Oxidation: $: - 1 5 H_{2} O_{2} \to 0 2 O_{2}$
Thus overall Balanced reaction is:
$+ 7 2 K M n O_{4} + 16 H^{+} + - 1 5 H_{2} O_{2} \to + 2 2 M n^{2 +} + 2 K^{+} + 8 H_{2} O + 0 5 O_{2} + 10 H^{+}$
cancel $H^{+}$ from both sides:
$2 K M n O_{4} + 6 H^{+} + 5 H_{2} O_{2} \to 2 M n^{2 +} + 2 K^{+} + 8 H_{2} O + 5 O_{2}$

Thus 2 mol of $K M n O_{4}$ require 5 mol of $H_{2} O_{2}$ for decolorization. Therefore for decolourisaiton of 1 mole of acidified $K M n O_{4}$ the moles of $H_{2} O_{2}$ required are 5/2.

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