Chemistry - Electrochemistry Question with Solution | TestHub

$\frac{{\displaystyle \stackrel{C{l}_2\left(P_2\right)+2e^{-}\to 2C{l}^{-}}{2C{l}^{-}\to C{l}_2+2e^{-}}}}{\left(P_2\right) \left(P_1\right) }$

Cl₂ → Cl₂

(P₂) (P₁ )

$E=-\frac{0.0591}{2}\log \frac{P_1}{P_2}$

For the cell to be spontaneous,

$\Delta G=-nFE=-ve$ and $E=+ve$, which means $P_2>P_1$.