Question

The photoelectric current from $Na$ (work function, $w_0=2.3eV$ ) is stopped by the output
voltage of the cell
$Pt(s)\left|H_2(g,1bar)\right|HCl(aq.,pH=1)|AgCl(s)|Ag(s)$
The $pH$ of aq. $HCl$ required to stop the photoelectric current from $K\left(w_0=2.25eV\right),$ all other conditions remaining the same, is_____.$\times 10^{-2}$ (to the nearest integer).
Given, $2.303\frac{RT}{F}=0.06V;E_{{\text{AgCl|Ag|CI }}^{-}}^0=0.22V$

Answer 1

Answer: 58.00

$\frac{1}{2}{H}_2\to H^{+}+e^{\Theta }$

$e^{\Theta }+AgC{l}_{(s)}\to A{g}_{(s)}+C{l}^{\Theta }$

______________________________

$\frac{1}{2}{H}_2+AgC{l}_{(s)}\to H_{(aq)}^{+}+A{g}_{(s)}+C{l}_{(aq)}^{\Theta }$

$E={\epsilon }^0-\frac{.06}{1}$log $\frac{\left[H^{+}\right]\left[C{l}^{\Theta }\right]}{P_{H_2}^{\frac{1}{2}}}$

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

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

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

$=2.3+0.34=2.64eV$

$\Rightarrow$ stopping potential required for $K$

$=2.64-2.25=0.39$ volt

$E={\epsilon }^0-\frac{.06}{1}\log \frac{\left[H^{+}\right]\left[C{l}^{\Theta }\right]}{P_{H_2}^{\frac{1}{2}}}$

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

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

$0.17=+.12pH$

$pH=1.4166$

$\Rightarrow 1.42$