Which reaction does not represent autoredox or disproportionation?

C l_{2} + 2 O H^{-} ⟶ C l^{-} + C l O^{-} + H_{2} O

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2 H_{2} O_{2} ⟶ 2 H_{2} O + O_{2}

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2 C u^{+} ⟶ C u^{+ 2} + C u

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(N H_{4})_{2} C r_{2} O_{7} ⟶ N_{2} + C r_{2} O_{3} + 4 H_{2} O

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Solution

The correct option is C $(N H_{4})_{2} C r_{2} O_{7} ⟶ N_{2} + C r_{2} O_{3} + 4 H_{2} O$
Disproportionation reaction is a specific type of redox reaction in which a species is simultaneously reduced and oxidised to form two different products.
In the option A: $C l_{2}$ (0) becomes $C l^{-}$ (-1) and $C l O^{-}$ (+1).
In the option B: $H_{2} O_{2}$ (oxidation state of O = -1) becomes $H_{2} O$ (-2) and $O_{2}$ (0).
In the option C: $C u$ (+1) becomes $C u$ (+2) and $C u$ (0).
In the option D: $(N H_{4}) C r_{2} O_{7}$ (oxidation state of Cr = +6) becomes only $C r_{2} O_{3}$ (+3). There is no oxidized product of Cr.
Hence, option D is the correct answer.

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Similar questions

(N H_{4})_{2} C r_{2} O_{7} \to N_{2} + C r_{2} O_{3} + 4 H_{2} O

{(N H_{4})}_{2} C r_{2} O_{7} ⟶ N_{2} + C r_{2} O_{3} + 2 H_{2} O

C u^{+}

2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)

{C r}_{2} O_{7}^{2 -} + 2 ⊖ O H ⟶ 2 C r O_{4}^{2 -} + H_{2} O

Z n + C u S O_{4} ⟶ Z n S O_{4} + C u

2 M n O_{4}^{⊖} + 3 {M n}^{2 +} + 4 ⊖ O H ⟶ 5 M n O_{2} + 2 H_{2} O

2 {C u}^{\oplus} ⟶ C u + {C u}^{2 +}

{C r}_{2} O_{7}^{2 -} + 2 ⊖ O H ⟶ 2 C r O_{4}^{2 -} + H_{2} O

Z n + C u S O_{4} ⟶ Z n S O_{4} + C u

2 M n O_{4}^{⊖} + 3 {M n}^{2 +} + 4 ⊖ O H ⟶ 5 M n O_{2} + 2 H_{2} O

2 {C u}^{\oplus} ⟶ C u + {C u}^{2 +}

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