In which of the following electrolysis reactions $H_{2}$ is given off only at anode?

Acidulated

H_{2} O

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{N a}_{2} {S O}_{4} (a q)

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Molten

C a H_{2}

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H C O O N a (a q)

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Solution

The correct option is C Molten $C a H_{2}$
The fused ${C a H}_{2}$ on electrolysis produces $H_{2}$ gas on anode instead of cathode because ${C a H}_{2}$ yields $H^{-}$ ion on fusion and as it is anion gets discharged at anode that is positive electrode.
${C a H}_{2} ⟶ C a^{+} + 2 H^{-}$
At anode : $2 H^{-} ⟶ H_{2} + 2 e$
At cathode : $C a^{+} + 2 e ⟶ C a$
${C a H}_{2}$ is an ionic hydride, which conducts electricity in the molten state.

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{S O}_{2} (g) + H {N O}_{2} (a q) ⟶ H_{2} {S O}_{4} (a q) + N O (g)

B + N a O H \to 2 N a_{3} B O_{3} + H_{2}

P_{4} + N a O H + H_{2} O \to N a H_{2} P O_{2} + P H_{3}

S + N a O H \to N a_{2} S_{2} O_{3} + N a_{2} S + H_{2} O

H_{2} {S O}_{4} (a q) + {B a (O H)}_{2} \to

H_{2} {S O}_{4}

(a) H_{2} {S O}_{4} + N a O H \to N a H {S O}_{4} + H_{2} O

(b) H_{2} {S O}_{4} + 2 N a O H \to {N a}_{2} {S O}_{4} + 2 H_{2} O

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