Chemistry - Electrochemistry Question with Solution | TestHub

At anode $\mathrm{Cu}\to {\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}$

At cathode $2{\mathrm{Ag}}^{+}+2{\mathrm{e}}^{-}\to 2\mathrm{Ag}$

Cell reaction $\to \mathrm{Cu}+2{\mathrm{Ag}}^{+}\to {\mathrm{Cu}}^{2+}+2\mathrm{Ag}$

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.06}{2}\log \frac{\left[{\mathrm{Cu}}^{2+}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2}$

$0.43={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.06}{2}\log \frac{\left(0.001\right)}{{\left(0.01\right)}^2}$

${\mathrm{E}}_{\mathrm{cell}}^0=0.46$

${\mathrm{E}}_{\mathrm{cell}}^0={\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^0-{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^0$

$0.46=0.80-{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^0$

${\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^0=0.34$ volt

${\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^0=34\times {10}^{-2}$