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Application of Electrolysis

Which option ...

Question

Which option is incorrect for the working cell?
$P t | {C l}_{2 (g)} | {C l}_{(C_{1})}^{-} ∥ {C l}_{(C_{2})}^{-} | {C l}_{2 (g)} | P t$

Δ G = - v e

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C_{2} > C_{1}

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E_{c e l l}^{o} < 0

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C_{1} > C_{2}

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Solution

The correct option is B
$C_{2} > C_{1}$

$C C_{1} l^{-} \to C C_{2} l^{-}$
$E_{c e l l} = - 0.059 log \frac{C_{2}}{C_{1}} = 0.059 log \frac{C_{1}}{C_{2}}$
$c_{1} > c_{2}$
For spontaneous process so incorrect is B

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