Question

The standard half cell reduction potential for ${Ag}^{+}|Ag$ is $0.7991V$ at ${25}^{o}C$. Given that the experimental value of $K_{sp}=1.56\times {10}^{-10}$ for $AgCl$, calculate the standard half cell reduction potential for the $Ag|AgCl$ electrode.

• A. $0.2192V$
• B. $-0.2192V$
• C. $-1.2192V$
• D. $1.219V$

Answer 1

The correct option is A $0.2192V$

For desired reaction,

$AgCl+1e^{-}\to Ag+{Cl}^{-};E^o=\frac{-{\Delta }_r{G}^o}{\left(1\right)F}$

The needed ${\Delta }_r{G}^o$ for the reactions:

${Ag}^{+}+1e^{-}\to Ag;\Delta G^o=-n{FE}^o$

$AgCl\to {Ag}^{+}+{cl}^{-};\Delta G^o=-RT\ln K_{sp}$,

${\Delta }_r{G}^o=-\left(1mol\right)(9.468\times {10}^{4}J{mol}^{-1}{V}^{-1})\left(0.7991V\right)$

$-\left(8.314J{K}^{-1}{mol}^{-1}\right)\left(298K\right)\ln (1.56\times {10}^{-10})$

$=-77.10kJ+55.95kJ=-21.15kJ$

The potential is $E^o=\frac{-(-21.15)}{\left(1\right)\left(96.485\right)}=+0.2192V$