A galvanic cell is constructed using the redox reaction, it is represented as:
$\frac{1}{2} H_{2} (g) + A g C l (s) ⇌ H^{+} (a q) + C l^{-} (a q) + A g (s)$

P t | H_{2} (g) | H C l (s o l n .) | | A g N O_{3} (s o l n .) | A g

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A g | A g C l (s) | K C l (s o l n .) | H C l (s o l n .), H_{2} (g) | P t

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P t | H_{2} (g) | K C l (s o l n .) | | A g C l (s) | A g

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P t | H_{2} (g), H C l (s o l n .) | | A g C l (s) | A g

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Solution

The correct option is D $P t | H_{2} (g), H C l (s o l n .) | | A g C l (s) | A g$
Reduction occurs at the silver electrode and oxidation occurs at the hydrogen electrode.
The cell notation is $P t | H_{2} (g), H C l (s o l n .) | | A g C l (s) | A g$

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Δ G^{o}

A g C l (s) + 1 / 2 H_{2} (g) \to A g (s) + H^{+} (a q .) + {C l}^{-} (a q)

- 21.52 k J

2 A g C l (c) + H_{2} (g) \to A g (s) + 2 H^{+} (a q .) + 2 {C l}^{-} (a q .)

P t | H_{2} (g) 1 a t m | H C l (1.0 M) | A g C l | A g

\frac{1}{2} H_{2} (g) + A g C l (s)

\to

H^{+} (a q .) + C l^{-} (a q .) + A g (s)

1 / 2 H_{2} (g) + A g C l (s) \to H^{⨁} (a q) + C l^{⊝} (a q) + A g (s)

Δ G^{\circ}

A g C l (s) + \frac{1}{2} H_{2} (g) \to A g (s) + H^{+} + C l^{-} is - 21.52 K J . Δ G^{\circ} of 2 A g C l (s) + H_{2} (g) \to 2 A g (s) + 2 H^{+} 2 C l^{-}

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