Verified Solution • 2025–26 Session moderate

QJEEM20Ch3db9: JEE Main 2020 Chemistry - Class 12 JEEM20Ch3db9 Solution: The photoelectric current from Na (Work function, w 0 = 2.3 eV) …

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 _______ ×\times 10-2 (to the nearest integer).

Given, 2.303 RTF{{RT} \over F} = 0.06 V;
EAgCl∣Ag∣Cl−0E_{AgCl|Ag|C{l^ - }}^0 = 0.22 V

Step-by-Step Verified Solution

12{1 \over 2} H2 + AgCl →\to H+ + Ag + Cl-

From Nernst Equation,

Ecell = Ecell0E_{cell}^0 - 0.061log⁡[H+][Cl−]{{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]

= 0.22 – 0.06 log 10–2 = 0.34 V

Work function of Na metal = 2.3 eV

KE of photoelectron = 0.34 eV

Energy of incident radiation = 2.3 + 0.34 = 2.64 eV

For K atom,

Energy of incident radiation for K metal = 2.64 eV

Work function of K metal = 2.25 eV

KE of photoelectrons = 2.64 – 2.25 = 0.39 eV

∴\therefore Ecell = 0.39 V

Using Nernst Equation,

0.39 = 0.22 - 0.061log⁡[H+][Cl−]{{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]

As [H+]=[Cl−]\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]

0.39 = 0.22 - 0.061log⁡[H+]2{{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}

⇒\Rightarrow 0.39 = 0.22 - 0.12×log⁡[H+]0.12 \times \log \left[ {{H^ + }} \right]

⇒\Rightarrow 0.39 = 0.22 + 0.12 ×\times pH

⇒\Rightarrow pH = 1.42 = 142 ×\times 10-2

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