Question

In the reaction: $Mn{O_4}^ - + x{H^ + } + n{e^ - } \to M{n^{ + 2}} + y{H_2}O$
What is the value of n?

Answer 1

In this question it is simple asking us to balance the given redox reaction. First we’ll find out which element undergoes oxidation and which is being reduced. Reduction is gain of electrons and Oxidation is loss of electrons.

Complete answer:
From the given reaction to find out what is being oxidized or reduced we’ll have to find out the Oxidation number of Mn. Let us assume the oxidation number of Mn in $MnO_4^ - = x$ and the Oxidation number of oxygen will always be $- 2$
Therefore, the value of x will be $x + 4( - 2) = - 1$
Hence, $x = - 1 + 8 = + 7$
The oxidation state of Mn in the reactant side is $+ 7$ and that on the product side is $+ 2$. The change in the oxidation state is $+ 5$ which means $5{e^ - }$ have been gained.
The initial redox reaction can be written as: $MnO_4^ - \to M{n^{ + 2}}$
First step to balance the reaction is to balance the number of oxygen atoms. For that we add water molecules on the oxygen deficient side.
$MnO_4^ - \to M{n^{ + 2}} + 4{H_2}O$
To balance the number of hydrogens, add ${H^ + }$ on the deficient side
$MnO_4^ - + 8{H^ + } \to M{n^{ + 2}} + 4{H_2}O$
In a redox reaction the overall charge should be balanced on the RHS and LHS. Taking the number of electrons gained to be n, we get the charge on both sides to be:
$- 1 + 8( + 1) + n( - 1) = + 2$
Hence the value of n will be: $n = - 1 + 8 - 2 = + 5$
Therefore, the final balanced redox equation will be: $MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O$
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O$

Note:
The given redox reaction is carried out in acidic medium, hence we’re adding${H^ + }$ ions. If the reaction is in basic medium the process of balancing includes the same steps as that of acidic medium but also addition of the same number of $O{H^ - }$ions as that of ${H^ + }$.