Chemistry - Electrochemistry Question with Solution | TestHub

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.059}}{\text{n}}\text{log}\frac{{{\text{[H}}^{\text{+}}\text{]}}^{\text{2}}}{{{\text{[Ag}}^{\text{+}}\text{]}}^{\text{2}}}$
$\text{=}{\text{E}}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.059}\text{×}\text{2}}{\text{2}}\text{log}\frac{\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{Ag}}^{\text{+}}\right]}$
${\text{E}}_{\text{cell}}^{\text{o}}-\text{0}\text{.059}\text{×}\text{log}\frac{\left[{\text{H}}^{\text{+}}\right]}{\text{1}}$
$\text{=}{\text{E}}_{\text{cell}}^{\text{o}}\text{+}\text{0}\text{.059}{\text{(}-{\text{log[H}}^{\text{+}}\text{])}}^{\text{1}}$
$\text{=}{\text{E}}_{\text{cell}}^{\text{o}}\text{+}\text{0}\text{.059}\text{pH}$
So$\text{pH}\text{=}\frac{{\text{E}}_{\text{cell}}-{\text{E}}_{\text{cell}}^{\text{o}}}{\text{0}\text{.059}}\text{=}\frac{\text{1}\text{.02}-\text{0}\text{.80}}{\text{0}\text{.059}}\text{=}\text{3}\text{.73}$