Indicate whether the following conversions represent an oxidation, a reduction or none (neither oxidation nor reduction).
(i) $H C l O_{3}$ to $H C l O_{4}$ (ii) ${N H}_{4}^{+}$ to $N H_{3}$
(iii) $N O_{2}$ to $N_{2} O_{4}$ (iv) $H S O_{3}^{-}$ to $S O_{4}^{2 -}$
(v) $H_{2} O_{2}$ to $H_{2} O$

(i) Oxidation (ii) Reduction (iii) None (iv) None (v) Oxidation

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(i) Oxidation (ii) None (iii) None (iv) Oxidation (v) Reduction

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(i) Reduction (ii) Oxidation (iii) Reduction (iv) None (v) Reduction

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(i) Oxidation (ii) Reduction (iii) None (iv) Reduction (v) Reduction

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Solution

The correct option is B (i) Oxidation (ii) None (iii) None (iv) Oxidation (v) Reduction
Oxidation is loss of electron and reduction is the gain of electron. Thereby assigning the oxidation number on central atom in each molecule for each conversion will determine oxidation or reduction process. Considering oxidation number of $H = + 1, O = - 2$ we get oxidation state as:
(i) Oxidation number of $C l$ in $H C l O_{3} \to H C l O_{4}$
$H + 5 C l O_{3} \to + 7 H C l O_{4}$
Oxidation number of $C l$ changes from +5 to +7, by losing two electron is oxidation process.
(ii) Oxidation number of $N$ in $N H_{4}^{+} \to N H_{3}$
$- 3 N H_{4}^{+} \to - 3 N H_{3}$
Oxidation number of $N$ does not change. Thus its neither oxidation nor reduction.
(iii)Oxidation number of $N$ in $N O_{2} \to N_{2} O_{4}$
$+ 4 N O_{2} \to {+ 4 N}_{2} O_{4}$
Oxidation number of $N$ does not change. Thus its neither oxidation nor reduction.
(iv)Oxidation number of $S$ in $H S O_{3}^{-} \to S O_{4}^{2 -}$
$H {+ 4 S O}_{3}^{-} \to + 6 S O_{4}^{2 -}$
Oxidation number of $S$ changes from +4 to +6, by losing two electron is oxidation process.
(v) Oxidation number of $S$ in $H_{2} O_{2} \to H_{2} O$
$H_{2} {- 1 O}_{2} \to H_{2} - 2 O$
Oxidation number of $O$ changes from -1 to -2, by gaining one electron is reduction process.
Therefore: (i) Oxidation(ii) None(iii) None(iv) Oxidation(v) Reduction

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