The cell reaction for the given cell is spontaneous if:
$P t | C l_{2} (P_{1} a t m) | C l^{-} (1 M) | C l_{2} (P_{2} a t m) | P t$

P_{1} > P_{2}

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P_{1} < P_{2}

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P_{1} = P_{2}

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P_{1} = 1 a t m

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Solution

The correct option is A $P_{1} < P_{2}$
Cathode:- $(P_{2}) C l_{2} + 2 e^{-} ⟶ 2 C l^{-}$
Anode:- $2 C l^{-} ⟶ 2 e^{-} + C l_{2} (P_{1})$
$(P_{2}) C l_{2} ⟶ C l_{2} (P_{1})$
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{2} log \frac{P_{1}}{P_{2}}$
As for concentration cell $E_{c e l l}^{o}$ is zero .
$E_{c e l l} = \frac{- 0.0591}{2} log \frac{P_{1}}{P_{2}}$
$Δ G = - n F E_{c e l l}$
For a reaction to be spontaneous, $Δ G$ is to be negative. Thus $E_{c e l l}$ must be positive.
$\Rightarrow E_{c e l l} = - l n \frac{P_{1}}{P_{2}}$
$\Rightarrow E_{c e l l} = \frac{0.0591}{2} log \frac{P_{2}}{P_{1}}$
For $E_{c e l l}$ to be positive;
$P_{2} > P_{1}$ .

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