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Ion Electron Method

Balance the f...

Question

Balance the following redox reactions by ion-electron method:
$M n O_{4}^{-} (a q) + I^{-} (a q) \to M n O_{2} (s) + I_{2} (s)$ (in basic medium)

$M n O_{4}^{-} (a q) + S O_{2} (g) \to M n^{2 +} (a q) + (a q) H S O_{4 (a q)}^{-}$ (in acidic solution)

$H_{2} O_{2} (a q) + F e^{2 +} (a q) \to F e^{3 +} (a q) + H_{2} O (l)$ (in acidic solution)

$C r_{2} O_{7}^{2 -} + S O_{2} (g) \to C r^{3 +} (a q) + S O_{4}^{2 -}$ (aq) (in acidic solution)

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Solution

Step 1: The two half reactions involved in the given reaction are:
Oxidation half reaction: ${- 1 I}_{(a q)} ⟶ {0 I}_{2 (s)}$
Reduction halfreaction: $+ 7 M n O_{4 (a q)}^{-} ⟶ + 4 M n O_{2 (a q)}$

Step 2:
Balancing I in the oxidation half-reaction, we have:
$2 I_{(a q)}^{-} ⟶ I_{2 (s)}$
Now, to balance the charge, we add 2 $e^{-}$ to the RHS of the reaction.
$2 I_{P}^{-} (a q) ⟶ I_{2 (s)} + 2 e^{-}$
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.
$M n O_{4 (a q)}^{-} + 3 e^{-} ⟶ M n O_{2 (a q)}$
Now, to balance the charge, we add 4 $O H^{-}$ ions to the RHS of the reaction as the reaction is taking place in a basic medium.
$M n O_{4 (a q)}^{-} + 3 e^{-} ⟶ M n O_{2 (a q)} + 4 O H^{-}$
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.
$M n O_{4 (a q)}^{-} + 2 H_{2} O + 3 e^{-} ⟶ M n o_{2 (a q)} + 4 O H^{-}$
Step 5:
Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
$6 I_{(a q)}^{-} ⟶ 3 I_{2 (s)} + 6 e^{-} 2 M n O_{4 (a q)}^{-} + 4 H_{2} O + 6 e^{-} ⟶ 2 M n O_{2 (s)} + 8 O H_{(a q)}^{-}$ Answer: $6 I_{(a q)}^{-} + 2 M n O_{4 (a q)}^{-} + 4 H_{2} O ⟶ 3 I_{2 (s)} + 2 M n O_{2 (s)} + 8 O H_{(a q)}^{-}$

Following the steps as in part (a), we have the oxidation half reaction as:
$S O_{2 (g)} + 2 H_{2} O_{(I)} ⟶ H S O_{4 (a q)}^{-} + 3 H_{(a q)}^{+} + 2 e_{(a q)}^{-}$
And the reduction half reaction as:
$M n O_{4 (a q)}^{-} + 8 H_{(a q)}^{+} + 5 e^{-} ⟶ M n_{(a q)}^{2 +} + 4 H_{2} O_{(I)}$
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:
$2 M n O_{4 (a q)}^{-} + 5 S O_{s (g)} + 2 H_{2} O_{(I)} + H_{(a q)}^{+} ⟶ 2 M n_{(a q)}^{2 +} + 5 H S O_{4 (a q)}^{-}$

Following the steps as in part (a), we have the oxidation half reaction as:
$F e_{(a q)}^{2 +} ⟶ F e_{(a q)}^{3 +} + e^{-}$
And the reduction half reaction as:
$H_{2} O_{2 (a q)} + 2 H_{(a q)}^{+} + 2 e^{-} ⟶ 2 H_{2} O_{(I)}$
Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
$H_{2} O_{2 (a q)} + 2 F e_{(a q)}^{2 +} + 2 H_{(a q)}^{+} ⟶ 2 F e_{(a q)}^{3 +} + 2 H_{2} O_{(I)}$

$S O_{2 (g)} + 2 H_{2} O_{(I)} \to S O_{(a q)}^{2 -} + 4 H_{(a q)}^{+} + 2 e^{-}$
And the reduction half reaction as:
$C r_{2} O_{7 (a q)}^{2 -} + 14 H_{(a q)}^{+} + 6 e^{-} ⟶ 2 C r_{(a q)}^{3 +} + 7 H_{2} O_{(I)}$
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
$C r_{2} O_{7 (a q)}^{2 -} + 3 S O_{2 (g)} + 2 H_{(a q)}^{+} ⟶ 2 C r_{(a q)}^{3 +} + 3 S O_{4 (a q)}^{2 -} + H_{2} O_{(I)}$

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Similar questions

Q. Balance the following redox reactions by ion-electron method.

H_{2} O_{2} (a q) + F e^{2 +} (a q) \to F e^{3 +} (a q) + H_{2} O (l)

(in acidic solution)

Q. Balance the following redox reactions by ion electron method.

C r_{2} O_{7}^{2 -} + S O_{2} (g) \to C r^{3 +} (a q) + S O_{4}^{2 -} (a q)

(in acidic solution)

Q. Balance the following redox reactions by the ion-electron method in acidic medium.

M n O_{4}^{-} (a q) + S O_{2} (g) \to M n^{2 +} (a q) + H S O_{4}^{-} (a q)

Q. Balance the redox reaction by Half reaction method.

{F e}^{2 +} (a q) + {C r}_{2} O_{7}^{2 -} (a q) (acid medium) - ------- \to {F e}^{3 +} (a q) + {C r}^{3 +} (a q)

Balance the following redox reactions:

S^{2 -} (a q) + I_{2} (s) ⟶ {S O}_{4}^{2 -} (a q) + I^{-} (a q)

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Ion Electron Method

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