Question

For the reaction between ${\rm{KMn}}{{\rm{O}}_{\rm{4}}}$ and ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$, the number of electrons transferred per mole of ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$are
(a) one
(b) two
(c) three
(d) four

Answer 1

Hint: We know that a reaction in which there is a change in the oxidation number (increase as well as decrease) of some reacting species involved in the reaction is termed as redox reaction. That means, in a redox reaction both oxidation and reduction takes place.

Complete step by step solution:
We know that manganate (VII) ion, ${\rm{Mn}}{{\rm{O}}_{\rm{4}}}^ - $ oxidizes hydrogen peroxide gas to oxygen gas. The reaction is done with potassium permanganate (VII) solution and hydrogen peroxide solution acidified with dilute sulphuric acid. Let’s first write the half reactions.
We know that magnate (VII) ions are reduced to manganese ions. So, the reaction is shown as ,
${\rm{Mn}}{{\rm{O}}_{\rm{4}}}^ - \to {\rm{M}}{{\rm{n}}^{2 + }}$
Now, we need to balance the equation.
As there are four oxygen atoms in the reactant side, we have to add four molecules of water to balance the oxygen atoms.
${\rm{Mn}}{{\rm{O}}_{\rm{4}}}^ - \to {\rm{M}}{{\rm{n}}^{2 + }} + 4{{\rm{H}}_{\rm{2}}}{\rm{O}}$
Now, to balance hydrogen atoms, we have to eight protons to the reactant side.
${\rm{Mn}}{{\rm{O}}_{\rm{4}}}^ - + 8{{\rm{H}}^ + } \to {\rm{M}}{{\rm{n}}^{2 + }} + 4{{\rm{H}}_{\rm{2}}}{\rm{O}}$
Now, we have balanced the charge on both sides. Now, at LHS charge is 7 and in RHS charge is +2. So, we have added 5 electrons to the LHS to make it +2.
${\rm{Mn}}{{\rm{O}}_{\rm{4}}}^ - + 8{{\rm{H}}^ + } + 5{e^ - } \to {\rm{M}}{{\rm{n}}^{2 + }} + 4{{\rm{H}}_{\rm{2}}}{\rm{O}}$…… (1)
Now, we write the oxidation reaction of ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$.
Hydrogen peroxide $\left( {{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}} \right)$ turns to oxygen \[{{\rm{O}}_{\rm{2}}}\].
${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}} \to {{\rm{O}}_{\rm{2}}}$
To balance hydrogen atoms in the above reaction we have to add 2 protons to the RHS.
${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}} \to {{\rm{O}}_{\rm{2}}} + 2{{\rm{H}}^ + }$
Now, we have to balance the charge. In LHS, charge is zero and in RHS charge is +2. So, we have to add 2 electrons to RHS.
${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}} \to {{\rm{O}}_{\rm{2}}} + 2{{\rm{H}}^ + } + 2{e^ - }$……. (2)
Now, we have to multiply equation (1) by 2 and equation (2) by 5.
${\rm{2Mn}}{{\rm{O}}_{\rm{4}}}^ - + 16{{\rm{H}}^ + } + 10{e^ - } \to 2{\rm{M}}{{\rm{n}}^{2 + }} + 8{{\rm{H}}_{\rm{2}}}{\rm{O}}$ …… (3)
${\rm{5}}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}} \to 5{{\rm{O}}_{\rm{2}}} + 10{{\rm{H}}^ + } + 10{e^ - }$…… (4)
From equation (4), we observe that 5 mole of ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$ transfers 10 electrons.
So, one mole of ${{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}$ transfers = $\dfrac{{10}}{5} = 2$ electrons.

Therefore, the correct answer is Option A.

Note: Increase of oxidation number indicates oxidation reaction and decrease of oxidation number indicates reduction reaction. Oxidizing agents are those that undergo reduction and reducing agents are those that undergo oxidation.