Question

Balance the following redox reactions by the ion-electron method in acidic medium.
$MnO_{4}^{-}(aq)+S{{O}_{2}}(g)\to M{{n}^{2+}}(aq)+HSO_{4}^{-}(aq)$

Answer 1

A redox reaction is a chemical reaction in which electrons are transferred between two reactants involved in it. The transfer of electrons from one reactant to another can be identified by observing the change in the oxidation states of the reacting atoms or molecules.
In the ion-electron method the redox chemical equation is separated into two individual half-equations. One half reaction is for oxidation reaction and other half is for reduction reaction. Each of the two half-reactions are balanced separately and then finally combined to give the balanced redox chemical equation.

Complete step by step answer:
In the question it is given that balances the reaction by the ion-electron method in an acidic medium. Means we have to balance the electrons on both sides of the reaction in an acidic medium.
The given reaction is
$MnO_{4}^{-}(aq)+S{{O}_{2}}(g)\to M{{n}^{2+}}(aq)+HSO_{4}^{-}(aq)$
In the above reaction first we have to identify the species which are undergoing oxidation and reduction.
Manganese (Mn) changes its oxidation state from +7 in $MnO_{4}^{-}$ to +2 in$M{{n}^{2+}}$, means Manganese undergoing reduction (gain of electrons).

& x+4(O)+K=0 \\\ & x+4(-2)+1=0 \\\ & x=7 \\\ \end{aligned}$$ Here x = oxidation state of Manganese Sulphur (S) changes its oxidation state from +4 in $$S{{O}_{2}}$$ to +6 in$$HSO_{4}^{-}$$, means Sulphur is undergoing oxidation. Oxidation state of sulphur in$$\text{S}{{\text{O}}_{\text{2}}}$$is as follows. $$\begin{aligned} & x+2(O)=0 \\\ & x+2(-2)=0 \\\ & x=4 \\\ \end{aligned}$$ Oxidation state of sulphur in $$HSO_{4}^{-}$$ is as follows. $$\begin{aligned} & x+4(O)+H=-1 \\\ & x+4(-2)+1=-1 \\\ & \text{ }x=6 \\\ \end{aligned}$$ Based on the above date we have to write the half reactions for oxidation and reduction. $$MnO_{4}^{-}(aq)+5{{e}^{-}}\to M{{n}^{2+}}(aq)$$ (Reduction Reaction) $$S{{O}_{2}}(g)\to HSO_{4}^{-}(aq)+2{{e}^{-}}$$(Oxidation reaction) Now, we have to balance the half reactions by making an equal number of electrons on both sides. $$\left( MnO_{4}^{-}(aq)+5{{e}^{-}}\to M{{n}^{2+}}(aq) \right)\times 2$$ $$\left( S{{O}_{2}}(g)\to HSO_{4}^{-}(aq)+2{{e}^{-}} \right)\times 5$$ Now write the total overall reaction. $$2MnO_{4}^{-}(aq)+5S{{O}_{2}}(g)+10{{e}^{-}}\to 2M{{n}^{2+}}(aq)+5HSO_{4}^{-}(aq)+10{{e}^{-}}$$ Ten electrons on both sides cancel each other and form the following chemical equation. $$2MnO_{4}^{-}(aq)+5S{{O}_{2}}(g)\to 2M{{n}^{2+}}(aq)+5HSO_{4}^{-}(aq)$$ To balance the number of oxygen atoms we have to add water on the left side means on the side of the reactants. $$2MnO_{4}^{-}(aq)+5S{{O}_{2}}(g)+2{{H}_{2}}O\to 2M{{n}^{2+}}(aq)+5HSO_{4}^{-}(aq)$$ To balance the number of hydrogens we have added $${{H}^{+}}$$on the reactants side. Now, the overall reaction is $$2MnO_{4}^{-}(aq)+5S{{O}_{2}}(g)+2{{H}_{2}}O+{{H}^{+}}\to 2M{{n}^{2+}}(aq)+5HSO_{4}^{-}(aq)$$ **Note:** If we are going to balance the equation in an acidic medium we have to add water and $${{H}^{+}}$$in the chemical reaction to balance oxygen and hydrogen. If we are going to balance the equation in an acidic medium we have to add $$O{{H}^{-}}$$ in the chemical reaction to balance oxygen and hydrogen.