Which of the following reactions is not an oxidation-reduction reaction?

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)

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N a O H (a q) + H C l (a q) \to N a C l (a q) + H_{2} O (l)

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H_{2} O_{2} (l) \to H_{2} O (l) + O_{2} (g)

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C H_{4} (l) + O_{2} (g) \to H_{2} O (g) + {C O}_{2} (g)

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M g (s) + 2 H C l (a q) \to M g {C l}_{2} (a q) + H_{2} (g)

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Solution

The correct option is A $N a O H (a q) + H C l (a q) \to N a C l (a q) + H_{2} O (l)$
Option $B$ is the correct answer.
The equation in option B is an acid-base reaction, where the base $N a O H$ reacts with an acid $H C l$ to give a salt $N a C l$ and water. The oxidation number of all the elements before and after the reaction remains the same. The oxidation number of $N a$ before and after the reaction is $+ 1$ . The oxidation number of $C l$ before and after the reaction is $- 1$ . The oxidation number of Oxygen before as well as after the reaction remains the same $- 2$ . And the total oxidation number of two $H$ atoms before and after remains $+ 2$ cumulatively. Thus it is not an oxidation-reduction reaction.
All the other reactions show a change in oxidation number before and after the reaction.

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C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) ⟶ 2 {C O}_{2} (g) + H_{2} O (l); △ G^{0} = 1234 k J

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H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); △ G^{0} = - 237 k J

H_{2} (g) + 2 C (s) \to C_{2} H_{2} (g)

Δ H \neq Δ U

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); △ H_{1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △ H_{2}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

Δ H_{1}

Δ H_{2}

H_{2_{(g)}} + \frac{1}{2} O_{2} (g) \to H_{2} O_{(l)}

H^{+} (a q) + O H^{-} (a q) = H_{2} O (l); Δ H = - X_{1} k J . m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l); Δ H = - X_{2} k J . m o l^{- 1}

C O_{2} (g) + H_{2} (g) = C O (g) + H_{2} O (l); Δ H = + X_{3} k J . m o l^{- 1}

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) = 2 C O (g) + H_{2} O (l); Δ H = + X_{4} k J . m o l^{- 1}

H_{2} O (l)

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