Question
The reaction of aqueous $$KMn{O_4}$$ with $${H_2}{O_2}$$ in acidic conditions gives
A.
$$M{n^{4 + }}\,{\text{and}}\,{O_2}$$
B.
$$M{n^{2 + }}\,{\text{and}}\,{O_2}$$
C.
$$M{n^{2 + }}\,{\text{and}}\,{O_3}$$
D.
$$M{n^{4 + }}\,{\text{and}}\,Mn{O_2}$$
Answer :
$$M{n^{2 + }}\,{\text{and}}\,{O_2}$$
Solution :
The reaction of aqueous $$KMn{O_4}$$ with $${H_2}{O_2}$$ in acidic medium is
$$3{H_2}S{O_4} + 2KMn{O_4} + 5{H_2}{O_2} \to $$ $$5{O_2} + 2MnS{O_4} + 8{H_2}O + {K_2}S{O_4}$$
In the above reaction, $$KMn{O_4}$$ oxidises $${H_2}{O_2}$$ to $${O_2}$$ and itself i.e. $$\left[ {MnO_4^ - } \right]$$ gets reduced to $$M{n^{2 + }}$$ $$ion$$ as $$MnS{O_4}.$$ Hence, aqueous solution of $$KMn{O_4}$$ with $${H_2}{O_2}$$ yields $$M{n^{2 + }}$$ and $${O_2}$$ in acidic conditions.