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Emf of the cell $Al (s) | {Al}^{3 +} (10^{- 3} M) | | {Ag}^{+} (10^{- 1} M) | Ag$ at 298 K is $(E_{cell}^{0} = 2.46 V)$

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2.46 V

2.4 V

2.52 V

2.37 V

answer is A.

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Detailed Solution

$Al ⟶ {Al}^{3 +} + 3 e^{-}$

$[{Ag}^{+} + e^{-} ⟶ Ag] \times 3$

over all reaction is :

Al + 3Ag⁺ $\to$ Al⁺³ + 3Ag

$E_{cell} = 2.46 - \frac{0.0591}{3} \log \frac{10^{- 3}}{{(10^{- 1})}^{3}}$

$E_{cell} = 2.46 V$

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