Question

The photoelectric current from Na (Work function, \(w_0\) = 2.3 eV) is stopped by the output voltage of the cell
Pt(s) | \(\mathrm{H_2}\) (g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).
The pH of aq. HCl required to stop the photoelectric current form K(\(w_0\) = 2.25 eV), all other conditions remaining the same, is _______ \(\times\) \(10^{-2}\) (to the nearest integer).
Given, \(2.303\frac{RT}{F}\) = 0.06 V;
\(E^0_{\mathrm{AgCl|Ag|Cl^-}}\) = 0.22 V

Hint:

The cell emf equals the stopping potential; use the Nernst equation to link emf with pH and the photoelectric equation to link stopping potential with work function.

Answer 1

Correct Answer: 142

Cell reaction: \(\mathrm{H_2(g) + 2AgCl(s) \to 2Ag(s) + 2H^+ + 2Cl^-}\), n = 2, \(E^0_{cell}\) = 0.22 - 0 = 0.22 V.
\(E = 0.22 - \frac{0.06}{2}\log([H^+]^2[Cl^-]^2) = 0.22 - 0.06\log([H^+][Cl^-])\).
Since \([H^+] = [Cl^-] = 10^{-pH}\): \(E = 0.22 + 0.12\,pH\).
For pH = 1: E = 0.34 V = stopping potential for Na.
Photoelectric equation for Na: \(h\nu = w_0 + eV_0 = 2.3 + 0.34 = 2.64\) eV.
For K, the same light: \(eV_0 = 2.64 - 2.25 = 0.39\) V.
So 0.22 + 0.12 pH = 0.39, giving pH = 0.17/0.12 = 1.4167.
pH = 141.67 \(\times 10^{-2}\), which is 142 to the nearest integer.
Answer: 142