Chemistry - Electrochemistry Question with Solution | TestHub

$-0.625=E_{cathode}-E_{anode}$

$=E_{C{N}^{-}/Ag(CN{)}_2^{-}/Ag}-0.245$

$\therefore E_{C{N}^{-}/Ag(CN{)}_2^{-}/Ag}=-0.625+0.245$

$=-0.380\mathrm{V}$

Now, $[\mathrm{Ag}(\mathrm{CN}{)}_2{]}^{-}+e^{-}⟶\mathrm{Ag}+2{\mathrm{CN}}^{-}$

$E_{C{N}^{-}/Ag(CN{)}_2^{-}/Ag}=E_{A{g}^{+}/Ag}$

$-0.380=0.8-\frac{0.059}{1}\log \frac{1}{[{\mathrm{Ag}}^{+}]}$

$$\log [{\mathrm{Ag}}^{+}]=\frac{-0.380-0.8}{0.059}=\frac{-1.180}{0.059}=-20$$

$$[{\mathrm{Ag}}^{+}]=10^{-20}\mathrm{M}$$

$$\therefore K_f=\frac{[\mathrm{Ag}(\mathrm{CN}{)}_2^{-}]}{[{\mathrm{Ag}}^{+}][{\mathrm{CN}}^{-}{]}^2}=\frac{7.50\times 10^{-3}}{10^{-20}\times (2.5\times 10^{-2}{)}^2}$$

$$=\frac{7.50\times 10^{17}}{6.25\times 10^{-4}}=1.2\times 10^{21}$$

$$\therefore x=21$$