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A galvanic cell represented as
$P t, H_{2 (1 a t m)} / H_{(0.1 M)}^{+} / / F_{(0.1 M)}^{-} / F_{2 (1 a t m),} P t$ . The emf of the cell will be [ $E_{c e l l}^{o}$ is 2.87V]

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2.99 V

2.75 V

287 V

3.78 V

answer is A.

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Detailed Solution

$\frac{1}{2} H_{2} \to H^{+} + e^{-}$

$\frac{1}{2} F_{2} + e^{-} \to F^{-}$

$\begin{array}{l} \frac{1}{2} H_{2} + \frac{1}{2} F_{2} \to H^{+} + F^{-} \\ 1 a t m 1 a t m 0.1 m 0.1 m \end{array}$

$E M F = 2.87 - \frac{0.06}{1} \log [10^{- 1}] \times [10^{- 1}]$

$= 2.87 - \frac{0.06}{1} \times - 2$

= 2.99V

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