JEE Main 2020 — Electrochemistry Question with Solution

Sodium metal :

$\mathrm{E}={\mathrm{E}}_0+{\left(\mathrm{KE}\right)}_{\max }$   $; {\mathrm{E}}_{\mathrm{call}}^0=0.22 \mathrm{V}$

Cell reaction

Cathode : $\mathrm{AgCl}\left(\mathrm{s}\right)+{\mathrm{e}}^{-} \to \mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

Anode : $\frac{1}{2} {\mathrm{H}}_2\left(\mathrm{g}\right) \to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}$

Overall ; $\mathrm{AgCl}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right) \to \mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.06}{1}\log \left[{\mathrm{H}}^{+}\right] \left[{\mathrm{Cl}}^{-}\right]$

${\mathrm{E}}_{\mathrm{cell}}=0.22-\frac{0.06}{1}\log \left[{10}^{-1}\right]\left[{10}^{-1}\right]=0.22+0.12=0.34 \mathrm{V}$

${\left(\mathrm{KE}\right)}_{\max }={\mathrm{E}}_{\mathrm{cell}}=0.34 \mathrm{eV}$

So $\mathrm{E}=2.3+0.34=2.64 \mathrm{eV}=$ Energy of photon incident

For potassium metal :

$\mathrm{E}={\mathrm{E}}_0+{\left(\mathrm{KE}\right)}_{\max }$

$2.64=2.25+{\left(\mathrm{KE}\right)}_{\max }$

${\left(\mathrm{KE}\right)}_{\max }=0.39={\mathrm{E}}_{\mathrm{cell}}$

Cell reaction

Cathode : $\mathrm{AgCl}\left(\mathrm{s}\right)+{\mathrm{e}}^{-} \to \mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

Anode : $\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right) \to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}$

Overall : $\mathrm{AgCl}\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right) \to \mathrm{Ag}\left(\mathrm{s}\right) + {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.06}{1}\log \left[{\mathrm{H}}^{+}\right] \left[{\mathrm{Cl}}^{-}\right]$

$0.39=0.22-0.12 \log \left[{\mathrm{H}}^{+}\right]$

$0.17=0.12\times \mathrm{pH}$

$\mathrm{pH}=17/12=1.4166=1.42$