Which of following cell can produce more electrical work? (given $P_{H_{2}} = 1 a t m$ )

P t, H_{2} (g) | 0.1 M N H_{4} C l | | 0.1 M C H_{3} C O O H | H_{2} (g), P t

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P t, H_{2} (g) | 0.1 M H C l | | 0.1 M N a O H | H_{2} (g), P t

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P t, H_{2} (g) | 0.1 M N H C l | | 0.1 M C H_{3} C O O K | H_{2} (g), P t

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P t, H_{2} (g) | 0.1 M C H_{3} C O O K | | 0.1 M H C l | H_{2} (g), P t

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Solution

The correct option is D $P t, H_{2} (g) | 0.1 M C H_{3} C O O K | | 0.1 M H C l | H_{2} (g), P t$
For max electrical work, $E_{c e l l} = M a x .$
$E_{c e l l} = - \frac{0.0591}{1} l o g \frac{[H^{+}]_{a}}{[H^{+}]_{c}}$
For $E_{c e l l}$ to be the highest [H $^{+}$ ] $_{a}$ should be lower and [H $^{+}$ ] $_{c}$ should be higher and that is why anode compartment should be more basic and cathodic compartment should be more acidic.
Here cell given in option D has more basic and more acidic anodes and cathodes. Thus the correct answer is option D.

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Similar questions

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

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

P t | H_{2} (g) | B O H (a q) | | H A (a q) | H_{2} (g) | P t; (0.1 b a r; 1 M; 0.1 M; 1 b a r

K_{a} (H A) = 10^{- 7}, K_{b} (B O H) = 10^{- 5}

P t, H_{2 (g)} | H^{+} (10^{- 6} M) | | H^{+} (10^{- 4} M) | H_{2 (g)}, P t

298 K

P t, H_{2} (g) | 2 M C H_{3} C O O N H_{4} | | 2 M N a C l | H_{2} (g), P t

p K_{a} (C H_{3} C O O H) = 4.74

p K_{b} (N H_{4} O H) = 4.74

\frac{2.303 R T}{F} = 0.059)

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