The photoelectric current from Na (Work function, w₀ = 2.3 eV) is stopped by the output voltage of the cell
Pt(s) | H₂ (g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).
The pH of aq. HCl required to stop the photoelectric current form K(w₀ = 2.25 eV), all other conditions remaining the same, is _______ × 10⁻² (to the nearest integer).
Given, 2.303RT/F = 0.06 V;
E_(AgCl|Ag|Cl -)⁰ = 0.22 V

Solution

Chemistry ma’am · 4 steps

The cell voltage is the stopping potential, so the same light fixes the voltage you now need.

1. Cell EMF with pH. The cell reaction is H₂ + 2AgCl → 2H⁺ + 2Cl⁻ + 2Ag with n = 2. In HCl, [H⁺] = [Cl⁻] = 10^(-pH), so both ions bring the pH into the log.

   E = 0.22 - 0.06/2 log([H⁺]²[Cl⁻]²) = 0.22 + 0.12 pH

2. Photon energy from Na. At pH 1, E = 0.22 + 0.12 = 0.34 V. This stops electrons of maximum kinetic energy 0.34 eV, so the photon energy is 2.3 + 0.34 = 2.64 eV.
3. Voltage needed for K. With the same light, electrons from K leave with up to 2.64 - 2.25 = 0.39 eV. The cell must now give 0.39 V, so 0.22 + 0.12 pH = 0.39.
4. So. pH = 0.17/0.12 = 1.417 = 141.7 × 10⁻², which rounds to 142.

The common slip: 183. Putting only [H⁺] in the log gives E = 0.22 + 0.06 pH and a pH of 1.83, that is 183. The Cl⁻ concentration also depends on the pH.

The idea to keep. In HCl, both H⁺ and Cl⁻ come from the same acid, so both enter the Nernst log. The stopping voltage in volts equals the maximum kinetic energy in eV.

Answer from the NTA final answer key. Solution by PrepWiser.