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  4. Pt|Cl2 (P1 atm) |HCl(0.1 M)| Cl2(P2 atm)|Pt, cell reaction will be spontaneous if

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Q. $Pt|Cl_{2}$ ($P_{1}$ atm) |$HCl$($0.1 M$)| $Cl_{2}$($P_{2}$ atm)|$Pt$, cell reaction will be spontaneous if

Electrochemistry

Solution:

$E_{\text{cell}} =E^{\circ}_{\text{cell}} -\frac{0.0591}{2}$ log $\frac{P_{Cl_2\left(\text{anode}\right)}}{P_{Cl_2\left(\text{cathode}\right)}}$
$ = 0- \frac{0.0591}{2}$ log $\frac{P_{1}}{P_{2}}$
$E_{\text{cell}} = +ve$(spontaneous), $P_{1}< P_{2}$