Ncert Exercises & Solutions

The stoichiometry of iodide anion after balancing the following redox reaction by ion – electron method in basic medium is:

MnO₄^–(aq) + I^– (aq) → MnO₂ (s) + I₂(s)

1. Three (3)
2. Five (5)
3. Six (6)
4. Two (2)

Hint: In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4.

(a) Step 1: The two half-reactions involved in the given reaction are:
Oxidation half

{\overset{- I}{I}}_{(a q)} \to {\overset{0}{I}}_{2 (s)}

reaction:

\overset{+ 7}{M} n {O_{4}}^{-}_{(a q)} \to \overset{+ 4}{M} n O_{2 (a q)}

Reduction half reaction:

Step 2:

Balancing I in the oxidation half reaction, we have:

2 {I^{-}}_{(a q)} \to I_{2 (s)}

Now, to balance the charge, we add 2

e^{-}

to the RHS of the reaction.

2 {I^{-}}_{(a q)} \to I_{2 (s)} + 2 e^{-}

Step 3:
In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4.
Thus, 3 electrons are added to the LHS of the reaction.

{MnO}_{4}^{-}_{(aq)} + 3 e^{-} \to {MnO}_{2}_{(aq)}

Now, to balance the charge, we add 4

{OH}^{-}

ions to the RHS of the reaction as the reaction is taking place in a basic medium.

{MnO}^{-}_{4 (aq)} + 3 e^{-} \to {MnO}_{2 (aq)} + 4 {OH}^{-}

Step 4:
In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.

{MnO}_{4}^{-}_{(aq)} + 2 H_{2} O + 3 e^{-} \to {MnO}_{2 (aq)} + 4 {OH}^{-}

Step 5:
Equalizing the number of electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, we have:

6 {I^{-}}_{(aq)} \to 3 I_{2 (s)} + 6 e^{-}

2 {MnO}^{-}_{4 (aq)} + 4 H_{2} O + 6 e^{-} \to 2 {MnO}_{2 (s)} + 8 {OH}^{-}_{(aq)}

Step 6:
Adding the two half-reactions, we have the net balanced redox reaction as:

6 {I^{-}}_{(aq)} + 2 {MnO}^{-}_{4 (aq)} + 4 H_{2} O_{(I)} \to 3 I_{2 (s)} + 2 {MnO}_{2 (s)} + 8 {OH}^{-}_{(aq)}