Chemistry - Electrochemistry Question with Solution | TestHub

Oxidation is loss of electron and in a galvanic cell it occurs at anode. Reduction is gain of electron and in a galvanic cell it occurs at cathode.

At anode :$\frac{1}{2}{\mathrm{H}}_2( \mathrm{g})⟶\underset{\left(\mathrm{aq}\right)}{{\mathrm{H}}^{+}}+{\mathrm{e}}^{-}$

At cathode $\underset{\left(\mathrm{S}\right)}{\mathrm{AgCl}}+\overline{\mathrm{e}}⟶\mathrm{Ag}+\underset{\text{ (aq) }}{{\mathrm{Cl}}^{-}}$

Overall reaction

$\frac{1}{2}{\mathrm{H}}_2( \mathrm{g})+\mathrm{AgCl}\left(\mathrm{s}\right)⟶{\mathrm{H}}^{*}+{\mathrm{Cl}}^{-}+\mathrm{Ag}\left(\mathrm{s}\right)$

Cell representation :

Anode / Anode electrolyte || Cathodic electrolyte / Cathode.

The cell representation for the above cell will be

$\mathrm{Pt}\left|{\mathrm{H}}_2( \mathrm{g})\right|\mathrm{HCl}\left({\text{ soln }}^{\mathrm{n}}\right)\left|\mathrm{AgCl}\right(\mathrm{s}\left)\right|\mathrm{Ag}$