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Question: For decolourisation of 1 mole of acidified \[KMn{O_4}\] the mole of \[{H_2}{O_2}\] required are: A...

For decolourisation of 1 mole of acidified KMnO4KMn{O_4} the mole of H2O2{H_2}{O_2} required are:
A. 12\dfrac{1}{2}
B. 32\dfrac{3}{2}
C. 52\dfrac{5}{2}
D. 72\dfrac{7}{2}

Explanation

Solution

In the given question KMnO4KMn{O_4} is present in acidic condition so the reactant includes hydrogen ion. To find the moles of hydrogen peroxide we need to balance the equation. In this reaction a redox reaction is taking place.

Complete answer:
The reaction involved in the decolourization of acidified KMnO4KMn{O_4}with hydrogen peroxide H2O2{H_2}{O_2} is shown below.
KMnO4+H2O2+H+→K++Mn2++H2O+O2KMn{O_4} + {H_2}{O_2} + {H^ + } \to {K^ + } + M{n^{2 + }} + {H_2}O + {O_2}
For balancing the equation, first assign the oxidation state of each atom.
KMn+7O4+H2O2−1→Mn2++O20KM{n^{ + 7}}{O_4} + {H_2}O_2^{ - 1} \to M{n^{2 + }} + O_2^0
In this reaction, redox reaction is taking place.
The redox reaction is defined as the reaction where transfer of electrons takes place during the reaction. In this reaction reduction and oxidation occur simultaneously.
The reduction half of the reaction is shown below.
KMn+7O4+5e−→Mn2+KM{n^{ + 7}}{O_4} + 5{e^ - } \to M{n^{2 + }}
Here manganese (VII) is reduced by gaining 5 electrons to Mn2+M{n^{2 + }}
The oxidation half of the reaction is shown below.
H2O2−1→O2+2e−{H_2}O_2^{ - 1} \to {O_2} + 2{e^ - }
Here oxygen is oxidized by losing 2 electrons.
To balance the total change in oxidation state, multiply the reduction half with 2 and oxidation half with 5.
KMn+7O4+5e−→Mn2+×2=10e−KM{n^{ + 7}}{O_4} + 5{e^ - } \to M{n^{2 + }} \times 2 = 10{e^ - }
H2O2−1→O2+2e−×5=10e−{H_2}O_2^{ - 1} \to {O_2} + 2{e^ - } \times 5 = 10{e^ - }
Balance the potassium atom and oxygen atom in the reduction reaction by adding water H2O{H_2}O
.2KMn+7O4+16H+→2Mn2++2K++8H2O2KM{n^{ + 7}}{O_4} + 16{H^ + } \to 2M{n^{2 + }} + 2{K^ + } + 8{H_2}O
Balance the oxidation half as shown below.
H2O2−1→5O20+10H+{H_2}O_2^{ - 1} \to 5O_2^0 + 10{H^ + }
The overall balanced equation is given as shown below.
2KMn+7O4+5H2O2−1+16H+→2K++2Mn2++8H2O+5O20+10H+2KM{n^{ + 7}}{O_4} + 5{H_2}O_2^{ - 1} + 16{H^ + } \to 2{K^ + } + 2M{n^{2 + }} + 8{H_2}O + 5O_2^0 + 10{H^ + }
Cancel the hydrogen ion on both side of the reaction to get the final equation.
2KMn+7O4+5H2O2−1+6H+→2K++2Mn2++8H2O+5O202KM{n^{ + 7}}{O_4} + 5{H_2}O_2^{ - 1} + 6{H^ + } \to 2{K^ + } + 2M{n^{2 + }} + 8{H_2}O + 5O_2^0
By the above given equation we can see that for two mole of KMnO4KMn{O_4}
, five mole of hydrogen peroxide H2O2{H_2}{O_2}is required for decolourization.
Thus for decolorization of 1 mole acidified KMnO4KMn{O_4} 5/2 mole of H2O2{H_2}{O_2} is required.
Therefore, the correct option is C.

Note:
Oxidation is defined as the gain of hydrogen and reduction is defined as loss of hydrogen. The balanced equation follows the law of conservation of mass which means that the number of moles one the left side of reaction is equal to the number of moles on the right side of reaction.