The cell reaction of the following cell will | vedclass

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(C) At anode: $2Cl^{-}_{(1 \, M)} \rightarrow Cl_{2(P_1)} + 2e^-$At cathode: $Cl_{2(P_2)} + 2e^- \rightarrow 2Cl^{-}_{(1 \, M)}$Net reaction: $Cl_{2(P

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$P_1 < P_2$

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(C) At anode: $2Cl^{-}_{(1 \, M)} \rightarrow Cl_{2(P_1)} + 2e^-$At cathode: $Cl_{2(P_2)} + 2e^- \rightarrow 2Cl^{-}_{(1 \, M)}$Net reaction: $Cl_{2(P_2)} \rightarrow Cl_{2(P_1)}$$E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \frac{P_1}{P_2}$Since $E^0_{cell} = 0$ for a concentration cell,$E_{cell} = - \frac{0.0591}{2} \log \frac{P_1}{P_2} = \frac{0.0591}{2} \log \frac{P_2}{P_1}$For a spontaneous process,$E_{cell} > 0$,which implies $\log \frac{P_2}{P_1} > 0$,so $\frac{P_2}{P_1} > 1$,which means $P_2 > P_1$ or $P_1 < P_2$.

The cell reaction of the following cell will be spontaneous if $Pt|Cl_{2(P_1 \, atm)}|Cl^{-}_{(1 \, M)}||Cl^{-}_{(1 \, M)}|Cl_{2(P_2 \, atm)}|Pt$:

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