Ncert Exercises & Solutions

8.18 Balance the following redox reactions by ion – electron method :

(a) MnO4– (aq) + I– (aq) → MnO2 (s) + I2(s) (in basic medium)

(b) MnO4– (aq) + SO2 (g) → Mn2+ (aq) + HSO4– (aq) (in acidic solution)

(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)

(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)}

(b)Following the steps as in part (a), we have the oxidation half-reaction as:

S O_{2 (g)} + 2 H_{2} O_{(I)} \to H S {O^{-}}_{4 (a q)} + 3 {H^{+}}_{(a q)} + 2 {e^{-}}_{(a q)}

And the reduction half-reaction as:

{MnO}_{4}^{-}_{(aq)} + 8 {H^{+}}_{(aq)} + 5 e^{-} \to {Mn}^{2 +}_{(aq)} + 4 H_{2} O_{(I)}

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:

2 {MnO}_{4}^{-}_{(aq)} + 5 {SO}_{2 (g)} + 2 H_{2} O_{(g)} + 2 H_{2} O_{(I)} + {H^{+}}_{(aq)} \to 2 {Mn}^{2 +}_{(aq)} + 5 {HSO}_{4}^{-}_{(aq)}

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

And the reduction half reaction as:

H_{2} O_{2 (aq)} + 2 {H^{+}}_{(aq)} + 2 e^{-} \to 2 H_{2} O_{(I)}

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half-reaction, we have the net balanced redox reaction as:

H_{2} O_{2 (aq)} + 2 {Fe}^{2 +}_{(aq)} + 2 {H^{+}}_{(aq)} \to 2 {Fe}^{3 +}_{(aq)} + 2 H_{2} O_{(I)}

(d) Following the steps as in part (a), we have the oxidation half reaction as:

{SO}_{2 (g)} + 2 H_{2} O_{(l)} \to {SO}_{4}^{2 -}_{(aq)} + 4 {H^{+}}_{(aq)} + 2 e^{-}

And the reduction half reaction as:

{Cr}_{2} {O_{7}^{2 -}}_{(aq)} + 14 {H^{+}}_{(aq)} + 6 e^{-} \to 2 {Cr}^{3 +}_{(aq)} + 7 H_{2} O_{(l)}

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

{Cr}_{2} {O_{7}^{2 -}}_{(aq)} + 3 {SO}_{2 (g)} + 2 {H^{+}}_{(aq)} \to 2 {Cr}^{3 +}_{(aq)} + 3 {SO}_{4}^{2 -}_{(aq)} + H_{2} O_{(l)}

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