Which of the following will not produce hydrogen gas?

Electrolysis of mildly acidified water

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Electrolysis of NaCl in Nelson's cell

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Reaction of Zn with concentrated $H_{2} S O_{4}$

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Reaction of Zn with NaOH

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Solution

The correct option is C
Reaction of Zn with concentrated $H_{2} S O_{4}$

Water on electrolysis decomposes into $H_{2}$ and $O_{2}$ gases.

In Nelson's cell, aqueous $N a C l$ is electrolyzed where:

The reaction at the cathode is:

$H_{2} O (l) + 2 e^{-} \to H_{2} + 2 O H^{-}$
The reaction at the anode is:

$C l^{-} \to \frac{1}{2} C l_{2} (g) + 1 e^{-}$

The overall reaction is as follows:

$N a C l (a q) + H_{2} O (l) \to N a^{+} (a q) + O H^{-} (a q) + H_{2} (g) + 1 / 2 C l_{2} (g)$

$Z n$ on reacting separately with excess dil $H_{2} S O_{4}$ and excess NaOH liberates $H_{2} (g)$

$Z n + 2 N a O H \to N a_{2} Z n O_{2} + H_{2}$

$Z n + H_{2} S O_{4} (d i l .) \to Z n S O_{4} + H_{2}$

But when Zn reacts with concentrated $H_{2} S O_{4}$ ,

$4 Z n + 5 H_{2} S O_{4} (m o d e r a t e l y s t r o n g a n d h o t) \to 4 Z n S O_{4} + H_{2} S + 4 H_{2} O$

$3 Z n + 4 H_{2} S O_{4} (m o d e r a t e l y s t r o n g a n d h o t) \to 3 Z n S O_{4} + S + 4 H_{2} O$

$Z n + 2 H_{2} S O_{4} (c o n c .) \to Z n S O_{4} + S O_{2} + 2 H_{2} O$

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