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Oxidation Number Method

Balance the f...

Question

Balance the following redox equations by oxidation number method.
occur in acidic medium
$S n_{(a q)}^{2 +} + I O_{4 (a q)}^{-} \to S n_{(a q)}^{4 +} + I_{(a q)}^{-}$

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Solution

${S n}_{(a q)}^{2 +} + {I O}_{4 (a q)}^{-} ⟶ {S n}_{(a q)}^{4 +} + I_{(a q)}^{-}$
1) Separating into half reactions,
${S n}^{2 +} ⟶ {S n}^{4 +} + {2 e}^{-}$ (oxidation)
$+ 7$ $- 1$
${8 e}^{-} + {I O}_{4}^{-} ⟶ I^{-}$ (Reduction)
2) Balancing the atoms & charge,
${S n}^{2 +} ⟶ {S n}^{4 +} + {2 e}^{-}$
${I O}_{4}^{-} + {8 e}^{-} + 8 H^{+} ⟶ I^{-}$
3) Balancing oxygen atoms
${S n}^{2 +} ⟶ {S n}^{4 +} + {2 e}^{-} ⟶ (1)$
${I O}_{4}^{-} + {8 e}^{-} + {8 H}^{+} ⟶ I^{-} + 4 H_{2} O ⟶ (2)$
4) Make electron gain equivalent to electron lost.
$(E q n (1) \times 4) + (E q n (2) \times 1)$
${I O}_{4}^{-} + {8 e}^{-} + 8 H^{+} ⟶ I^{-} + 4 H_{2} O - ------------------------------------ -$
${4 S n}^{2 +} + {I O}_{4}^{-} + 8 H^{+} ⟶ {4 S n}^{4 +} + I^{-} + 4 H_{2} O$

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