Which option is incorrect for the working cel | vedclass

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Question

(B) The given cell is a concentration cell: $Pt \mid Cl_{2(g, 1 \ bar)} \mid Cl_{(C_1)}^{-} \parallel Cl_{(C_2)}^{-} \mid Cl_{2(g, 1 \ bar)} \mid Pt$.

Vedclass · Accepted Answer

$C_2 > C_1$

Vedclass · Answer

(B) The given cell is a concentration cell: $Pt \mid Cl_{2(g, 1 \ bar)} \mid Cl_{(C_1)}^{-} \parallel Cl_{(C_2)}^{-} \mid Cl_{2(g, 1 \ bar)} \mid Pt$. For a concentration cell,$E_{cell}^{0} = 0$. The cell reaction is $2Cl_{(C_1)}^{-} \longrightarrow 2Cl_{(C_2)}^{-}$. The Nernst equation is $E_{cell} = E_{cell}^{0} - \frac{0.0591}{n} \log \frac{C_2}{C_1}$. Since $E_{cell}^{0} = 0$ and $n = 2$,$E_{cell} = -0.0591 \log \frac{C_2}{C_1} = 0.0591 \log \frac{C_1}{C_2}$. For the cell to be spontaneous,$E_{cell} > 0$,which implies $\log \frac{C_1}{C_2} > 0$,so $C_1 > C_2$. If $C_1 > C_2$,then $\Delta G = -nFE_{cell} Therefore,the option $C_2 > C_1$ is incorrect.

Which option is incorrect for the working cell? $Pt \mid Cl_{2(g, 1 \ bar)} \mid Cl_{(C_1)}^{-} \parallel Cl_{(C_2)}^{-} \mid Cl_{2(g, 1 \ bar)} \mid Pt$

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