Chemistry - Electrochemistry Question with Solution | TestHub

From the given cell reaction and nernst equation.

${\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{o}}-\frac{0\text{.}0591}{\text{n}}\text{ log }\frac{\left[{\text{Ni}}^{2+}\right]}{{\left[{\text{Ag}}^{+}\right]}^2}$

$=1\text{.}05\text{ V}-\frac{0\text{.}0591}{2}\text{ log }\frac{\left[0\text{.}160\right]}{{\left[0\text{.}002\right]}^2}$

$=1\text{.}05-\frac{0\text{.}0591}{2}\text{ log }\left(4\times 10^4\right)$

$=1\text{.}05-\frac{0\text{.}0591}{2}\left(4\text{.}6021\right)$

$=1\text{.}05-0\text{.}14=0\text{.}91\text{ V}$

${\text{E}}_{\text{cell}}=0\text{.}91\text{ V}$