Balance the following redox reactions by the ion-electron method in basis medium

$M n O_{4}^{-} (a q) + I^{-} (a q) \to M n O_{2} (s) + I_{2} (s)$

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Solution

The unbalanced chemical equation is

$M n O_{4}^{-} (a q) + I^{-} (a q) \to M n O_{2} (s) + I_{2} (s)$
The oxidation half reaction is

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

The reduction half reaction is

$M n O_{4}^{-} (a q) \to M n O_{2} (a q)$
Balance I atoms and charges in the oxidation half reaction.

$2 I^{-} (a q) \to I_{2} (s) + 2 e^{-}$
In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.

$M n O_{4}^{-} (a q) + 3 e^{-} \to M n O_{2} (a q)$
Balance charge in the reduction half reaction by adding 4 hydroxide
ions to product side.

$M n O_{4}^{-} (a q) + 3 e^{-} \to M n O_{2} (a q) + 4 O H^{-}$
To balance O atoms, add 2 water molecules to reactant side.

$M n O_{4}^{-} (a q) + 3 e^{-} + 2 H_{2} O \to M n O_{2} (a q) + 4 O H^{-}$
To equalize the number of electrons, multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2.

$6 I^{-} (a q) \to 3 I_{2} (s) + 6 e^{-}$

$2 M n O_{4}^{-} (a q) + 6 e^{-} + 4 H_{2} O \to 2 M n O_{2} (a q) + 8 O H^{-}$
Add two half cell reactions to obtain the balanced equation.

$2 M n O_{4}^{-} (a q) + 6 I^{-} (a q) + 4 H_{2} O_{2} (l) \to 2 M n O_{2} (s) + 3 I_{2} (s) + 8 O H^{-}$

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Similar questions

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

M n O_{4}^{⊖} (a q) + I^{⊖} (a q) \to M n O_{2} (s) + I_{2} (s)

I^{-}

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

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)

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

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