Chemistry - Electrochemistry Question with Solution | TestHub

Relation between${\mathrm{K}}_{\mathrm{e}\mathrm{q}}$and${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}$is
${\text{E}}_{\text{cell}}^{\text{o}}=\frac{2.303\text{RT}}{\text{nF}}{\text{logK}}_{\text{eq}}$
${\text{E}}_{\text{cell}}^{\text{o}}=\frac{0.0591}{\text{n}}{\text{logK}}_{\text{eq}}$
$0.591=\frac{0.0591}{1}{\text{logK}}_{\text{eq}}$
${\text{logK}}_{\text{eq}}=10$
${\text{K}}_{\text{eq}}=1\times {10}^{10}$