For the given cell
$P t | C l_{2} (g, 0.5 bar) | C l^{-} (aq, 0.5 M) | | C l^{-} (aq, 0.05 M) | | C l_{2} (g, 0.05 bar) | P t$
The emf of the above concentration cell $298$ K will be:

- 0.3

volt

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0.03

volt

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- 0.03

volt

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0.3

volt

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Solution

The correct option is B $0.03$ volt
Reactions
At Anode : $2 C l^{-} \to C l_{2} + 2 e^{-}$
At Cathode: $C l_{2} + 2 e^{-} \to 2 C l^{-}$
SO the overall reaction is:
$2 C l_{A}^{-} + C l_{2. C} \to C l_{2. A} + 2 C l_{C}^{-}$
According to Nernst equation,
$E_{c e l l} = 0 - \frac{0.06}{2} log \frac{{({[C l^{-}]}^{2})}_{C} . {(P_{C l_{2}})}_{A}}{{({[C l^{-}]}^{2})}_{A} . {(P_{C l_{2}})}_{C}}$
$= 0 - \frac{0.06}{2} l o g \frac{(.05)^{2} \times (.5)}{(.5)^{2} \times (.05)}$
$= 0 - \frac{0.06}{2} l o g (10^{- 1})$
$= 0.03$ volt

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