Question

The equilibrium constant for the reaction
$Sr\left(s\right)+{Mg}^{+2}\left(aq\right)\rightleftharpoons {Sr}^{+2}\left(aq\right)+Mg(s{)}_{+2}$ is $2.69\times {10}^{12}$ at ${25}^{o}C$.
The $E^O$ for a cell made up of the $Sr/{Sr}^{+2}$ and ${Mg}^{+2}/Mq$ half cells is:

• A. $0.3667V$
• B. $\mathrm{0..7346}V$
• C. $0.1836V$
• D. $0.1349V$

Answer 1

The correct option is A $0.3667V$

${Sr}_{\left(s\right)}+{{Mg}^{+1}}_{\left(aq\right)}\rightleftharpoons {{Sr}^{2+}}_{\left(aq\right)}+{Mg}_{\left(s\right)};K_{eq}=2.69\times {10}^{12}$

As we know that,

$\Delta G=-nF{E^0}_{cell}..........\left(1\right)$

and we also know that,

$\Delta G=-RT\ln \left(K_{eq}\right)..........\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\ln K_{eq}=\frac{nF{E^0}_{cell}}{RT}$

where as,

$K_{eq}=$ Equillibrium constant $=2.69\times {10}^{12}$

$T=$ Temperature $=25℃=(273+25)K=298K\left(\text{Given}\right)$

$n=$ no. of electrons involved $=2$

$F=$ Faraday constant $=96500C$

$R=$ Universal gas constant $=8.314J{K}^{-1}{mol}^{-1}$

${E^0}_{cell}=?$

$\therefore \ln (2.69\times {10}^{12})=\frac{2\times 96500\times {E^0}_{cell}}{8.314\times 298}$

$\Rightarrow \ln \left(2.69\right)+\ln \left({10}^{12}\right)=\frac{193000\times {E^0}_{cell}}{2477.572}(\because \ln ab=\ln a+\ln b)$

$\Rightarrow 0.99+(12\times 2.302)=77.9\times {E^0}_{cell}$

$\Rightarrow {E^0}_{cell}=\frac{28.614}{77.9}=0.367\approx 0.3667V$