# The photoelectric current from

Question:

The photoelectric current from $\mathrm{Na}$ (work function, $\mathrm{w}_{0}=2.3 \mathrm{eV}$ ) is stopped by the output voltage of the cell $\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_{2}(\mathrm{~g}, 1$ bar $)|\mathrm{HCl}(\mathrm{aq} ., \mathrm{pH}=1)| \mathrm{AgCl}(\mathrm{s}) \mid \mathrm{Ag}(\mathrm{s})$.

The $\mathrm{pH}$ of aq. $\mathrm{HCl}$ required to stop the photoelectric current from $\mathrm{K}\left(\mathrm{w}_{0}=2.25 \mathrm{eV}\right)$, all other conditions remaining the same, is _____________. $\times 10^{-2}$ (to the nearest integer).

Given, $2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V} ; \mathrm{E}_{\mathrm{AgCl}|\mathrm{Ag}| \mathrm{Cl}^{-}}^{0}=0.22 \mathrm{~V}$

Solution:

(142)

Sodium metal :

$\mathrm{E}=\mathrm{E}_{0}+(\mathrm{KE})_{\max } ; \mathrm{E}_{\text {cell }}^{0}=0.22 \mathrm{~V}$

Cell reaction

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

Anode $: \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-}$

Overall: $\mathrm{AgCl}(\mathrm{s})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})$

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

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

$=0.22+0.12=0.34 \mathrm{~V}$

$(\mathrm{KE})_{\max }=\mathrm{E}_{\text {clll }}=0.34 \mathrm{eV}$

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

For potassium metal :

$\mathrm{E}=\mathrm{E}_{0}+(\mathrm{KE})_{\max }$

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

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

Cell reaction

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

Anode $: \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-}$

Overall: $\mathrm{AgCl}(\mathrm{s})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})$

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

$0.39=0.22-0.06 \log \left[\mathrm{H}^{+}\right]^{2}$

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

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

$\mathrm{pH}=17 / 12=1.4166 \simeq 1.42=142 \times 10^{-2}$