Chemistry - Electrochemistry Question with Solution | TestHub

$\mathrm{E}=\mathrm{E}°-\frac{0.059}{\mathrm{n}}\mathrm{l}\mathrm{o}{\mathrm{g}}_{10}\mathrm{Q}$
At equilibrium,$\mathrm{E}=0,\mathrm{Q}={\mathrm{K}}_{\mathrm{c}}$
$\mathrm{E}°=\frac{0.059}{\mathrm{n}}\mathrm{l}\mathrm{o}{\mathrm{g}}_{10}{\mathrm{K}}_{\mathrm{c}}$
This equation is of the form y = mx, i.e., straight line, passing through origin.