For decolourisation of 1 mole of acidified KMnO4 the moles of H2O2 required are:
A
1/2
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B
3/2
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C
5/2
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D
7/2
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Solution
The correct option is C 5/2 Reaction involved is: KMnO4+H2O2+H+→K++Mn2++H2O+O2
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: +7KMnO4+−1H2O2→+2Mn2++0O2
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction: +7KMnO4→+2Mn2+
Oxidation: −1H2O2→0O2
Step 3: oxidation-number change is:
Reduction: +7KMnO4→+2Mn2+: Gain of 5 electrons total
Oxidation: −1H2O2→0O2: Loss of total 2 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: +7KMnO4→+2Mn2+×2: Gain of 10 electrons
Oxidation: :−1H2O2→0O2×5: Loss of total 10 electrons
OR
Reduction: :+72KMnO4→+22Mn2+
Oxidation: :−15H2O2→02O2
Thus overall Balanced reaction is: +72KMnO4+16H++−15H2O2→+22Mn2++2K++8H2O+05O2+10H+
cancel H+ from both sides: 2KMnO4+6H++5H2O2→2Mn2++2K++8H2O+5O2
Thus 2 mol of KMnO4 require 5 mol of H2O2 for decolorization. Therefore for decolourisaiton of 1 mole of acidified KMnO4 the moles of H2O2 required are 5/2.