Question

The photoelectric current from $Na$ (work function, $w _{0}=2.3 eV$ ) is stopped by the output
voltage of the cell
$Pt(s)\left| H _{2}( g , 1 bar )\right| HCl ( aq ., pH =1)| AgCl (s)| Ag (s)$
The $pH$ of aq. $HCl$ required to stop the photoelectric current from $K \left( w _{0}=2.25 eV \right),$ all other conditions remaining the same, is_____.$\times 10^{-2}$ (to the nearest integer).
Given, $2.303 \frac{ RT }{ F }=0.06 V ; E _{\text {AgCl|Ag|CI }^{-}}^{0}=0.22 V$

Answer 1

$\frac{1}{2} H _{2} \rightarrow H ^{+}+ e ^{\Theta}$

$e ^{\Theta}+ AgCl _{( s )} \rightarrow Ag _{( s )}+ Cl ^{\Theta}$

______________________________

$\frac{1}{2} H _{2}+ AgCl _{( s )} \rightarrow H _{( aq )}^{+}+ Ag _{( s )}+ Cl _{( aq )}^{\Theta}$

$E =\varepsilon^{0}-\frac{.06}{1}$log $\frac{\left[ H ^{+}\right]\left[ Cl ^{\Theta}\right]}{ P _{ H _{2}}^{\frac{1}{2}}}$

$E =0.22-.06 \log \frac{\left(10^{-1}\right)\left(10^{-1}\right)}{1^{\frac{1}{2}}}$

$E=0.22+.12=.34$ volt

$\Rightarrow $ total energy of photon will be (for Na)

$=2.3+0.34=2.64 eV$

$\Rightarrow $ stopping potential required for $K$

$=2.64-2.25=0.39$ volt

$E =\varepsilon^{0}-\frac{.06}{1} \log \frac{\left[ H ^{+}\right]\left[ Cl ^{\Theta}\right]}{ P _{ H _{2}}^{\frac{1}{2}}}$

as $\left[ H ^{+}\right]=\left[ Cl ^{\ominus}\right]$ so

$0.39=0.22-.06 \log \frac{\left[ H ^{+}\right]^{2}}{1^{\frac{1}{2}}}$

$0.17=+.12 pH$

$pH =1.4166$

$\Rightarrow 1.42$