Chemistry - Electrochemistry Question with Solution | TestHub

$\frac{{\displaystyle \stackrel{Cl₂\left(P_2\right) + 2e⁻ \to 2Cl⁻ }{2Cl⁻ \to Cl₂ + 2e⁻}}}{\left(P_2\right) \left(P_1\right) }$

$C{l}_2\to C{l}_2$

(P₂) (P₁)

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

For the cell to be spontaneous

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