Question

For decolourisation of 1 mole of acidified \[KMn{{O}_{4}}\] the moles of a.\[{{H}_{2}}{{O}_{2}}\]required are:
b.\[\dfrac{1}{2}\]
c.\[\dfrac{3}{2}\]
d.\[\dfrac{5}{2}\]
e.\[\dfrac{7}{2}\]

Answer 1

Hint: \[KMn{{O}_{4}}\] or potassium permanganate is a inorganic chemical compound which is also known as Condy’s crystals or the permanganate of potash. On dissolution of potassium permanganate crystals in water the formation of purple solution takes place. It is a strong oxidising agent which does not tend to produce toxic products.

Complete step-by-step answer:So the reaction which is involved on reacting potassium permanganate with hydrogen peroxide is the following:
\[KMn{{O}_{4}}+{{H}_{2}}{{O}_{2}}+{{H}^{+}}\to {{K}^{+}}+M{{n}^{2+}}+{{H}_{2}}O+{{O}_{2}}\]
Here the equation given above is not balanced as the number of atoms present on both reactants and products side is not the same. For the equation to be balanced we should have the same number of atoms on both sides.
So now we will balance the above equation. So firstly we will find the oxidation states of each of the atoms in the equation. So in the \[KMn{{O}_{4}}\]the oxidation number of manganese is +7. In hydrogen peroxide the oxidation number of oxygen is -1. In the product side the oxidation number of manganese is +2 and that of oxygen is 0 as it is a neutral molecule.
Now we will determine the substance which has undergone reduction and oxidation. So in reduction the positive oxidation number on the atom decreases while in oxidation the negative oxidation number of the atom increases.
So here in reduction we see that manganese in potassium permanganate has the oxidation of +7 but it reduces to +2 in the product side as \[M{{n}^{2+}}\]. See the following equation for clarity: \[KMn{{O}_{4}}(+7)\to M{{n}^{2+}}(+2)\]. The gain of five electrons takes place in the reaction.
Here the hydrogen peroxide oxidises to oxygen as the -1 charge increases to 0 in this reaction of oxidation. So the loss of total 2 electrons takes place. See the following equation for clarity:
\[{{H}_{2}}{{O}_{2}}(-1)\to {{O}_{2}}(0)\]. Here the loss of 2 electrons takes place.
Now the charge in both of the equations is not the same or the total change in the oxidation number is not same so we need to balance it. To balance the change of oxidation number we will multiply the reduction reaction by 2 and the reaction of oxidation by 5.
So now the reaction of reduction which we have now is :
\[2KMn{{O}_{4}}\to 2M{{n}^{2+}}\]
The reaction of oxidation now is :
\[5{{H}_{2}}{{O}_{2}}\to 5{{O}_{2}}\]
Now we need to balance the potassium and oxygen atoms in the reduction reaction by the addition of water molecule and then balancing this water molecule by addition of hydrogen ion as:
\[2KMn{{O}_{4}}+6{{H}^{+}}+5{{H}_{2}}{{O}_{2}}\to 2M{{n}^{2+}}+2{{K}^{+}}+8{{H}_{2}}O+5{{O}_{2}}\]
Now balancing the oxidation reaction:
\[5{{H}_{2}}{{O}_{2}}\to 5{{O}_{2}}+10{{H}^{+}}\]
Now the overall balanced reaction is:
\[2KMn{{O}_{4}}+16{{H}^{+}}+5{{H}_{2}}{{O}_{2}}\to 2M{{n}^{2+}}+2{{K}^{+}}+8{{H}_{2}}O+5{{O}_{2}}+10{{H}^{+}}\]
Now we will cancel the hydrogen ions from both product and reactant side and get the following final reaction equation as:
\[2KMn{{O}_{4}}+6{{H}^{+}}+5{{H}_{2}}{{O}_{2}}\to 2M{{n}^{2+}}+2{{K}^{+}}+8{{H}_{2}}O+5{{O}_{2}}\]
Now in the above reaction we see that 2 moles of \[KMn{{O}_{4}}\]need 5 moles of \[{{H}_{2}}{{O}_{2}}\]for the decolourisation. So I mole of \[KMn{{O}_{4}}\]will require \[\dfrac{5}{2}\]moles of \[{{H}_{2}}{{O}_{2}}\]for the decolourisation.

so the correct answer is option (C).

Note: Potassium permanganate is used as an oxidant in the wide spectrum of the chemical reactions. On reaction of the potassium permanganate with hydrochloric acid the production of chlorine takes place. It is also used as the regeneration chemical in the well water treatment. It is also used as the disinfectant.