Question

Which option is incorrect for the working cell?
$P t\left| Cl _{2( g )}\right| Cl _{\left(C_{1}\right)}^{-}|| Cl _{\left(C_{2}\right)}^{-}\left|\underset{1 \text { bar }}{ Cl _{2( g )}}\right| P t$

• A. $\Delta G=- ve$
• B. $C_2 > C_1$
• C. $E_{\text {cell }}^{\circ}=0$
• D. $C_1 > C_2$

Answer 1

Answer: (B)

$\underset{c_1}{Cl^{\ominus}} \rightarrow \underset{c_2}{Cl^{\ominus}}$
$E_{\text {cell }} =-0.059 \log \frac{c_{1}}{c_{2}} $
$=0.059 \log \left(\frac{c_{1}}{c_{2}}\right) $
$c_{1} > c_{2}$
For spontaneous process so incorrect is $B$