Question

$Pt|C{l}_2$ ($P_1$ atm) |$HCl$($0.1M$)| $C{l}_2$($P_2$ atm)|$Pt$, cell reaction will be spontaneous if

• A. $P_1=P_2$
• B. $P_1>P_2$
• C. $P_2>P_1$
• D. $P_1=P_2=1$ atm

Answer 1

Answer: (C)

$E_{\text{cell}}=E_{\text{cell}}^{\circ }-\frac{0.0591}{2}$ log $\frac{P_{C{l}_2\left(\text{anode}\right)}}{P_{C{l}_2\left(\text{cathode}\right)}}$
$=0-\frac{0.0591}{2}$ log $\frac{P_1}{P_2}$
$E_{\text{cell}}=+ve$(spontaneous), $P_1<P_2$