Zn_{(s)} + Cl_2(1 atm) to Zn^{2+} + 2Cl^-. T | vedclass

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(B) According to the Nernst equation:$E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log Q$For the reaction $Zn_{(s)} + Cl_2(1 \ atm) 	o Zn^{2+} + 2Cl^-$,

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$[Zn^{2+}]$ should be decreased

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(B) According to the Nernst equation:$E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log Q$For the reaction $Zn_{(s)} + Cl_2(1 \ atm) 	o Zn^{2+} + 2Cl^-$,the reaction quotient is $Q = \frac{[Zn^{2+}][Cl^-]^2}{P_{Cl_2}}$.Substituting $Q$ into the equation: $E_{cell} = E^0_{cell} - \frac{0.0591}{2} \log \frac{[Zn^{2+}][Cl^-]^2}{P_{Cl_2}}$.To increase $E_{cell}$,the value of the logarithmic term must decrease,which means $Q$ must decrease.$Q$ decreases if $[Zn^{2+}]$ is decreased,$[Cl^-]$ is increased,or $P_{Cl_2}$ is increased.Therefore,option $B$ is correct.

$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$:

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