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

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Redox Reactions and Electrochemistry

The photoelectric current from Na (work function,

w_{0} = 2.3 eV

) is stopped by the output voltage of the cell

Pt (s) / H_{2} (g, 1 bar) ∣ HCl (aq ., pH = 1) ∣ AgCl (s) ∣ Ag (s) .

The pH of aq. HCl required to stop the photoelectric current from

K (w_{0} = 2.25

eV ), all other conditions remaining the same, is ____

\times 1 0^{- 2}

(to the nearest integer).

Given,

2.303 \frac{RT}{F} = 0.06 V; E_{AgCl ∣ Ag ∣ Cl^{-}}^{0} = 0.22 V

Answer: 142

The stopping voltage equals the cell emf, and the photon energy is the work function plus that stopping energy. For the

AgCl∣Ag

cell,

E = E^{\circ} - 0.06 lo g ([H^{+}] [C l^{-}])

; at

pH = 1

, with

[H^{+}] = [C l^{-}] = 1 0^{- 1}

E = 0.22 - 0.06 lo g (1 0^{- 2}) = 0.22 + 0.12 = 0.34

V, so the photon carries

2.3 + 0.34 = 2.64

eV. The same photon on potassium leaves

2.64 - 2.25 = 0.39

eV as the stopping energy, so the new cell emf is

0.39

V. Then

0.39 = 0.22 + 0.12 pH

, using

- lo g ([H^{+}] [C l^{-}]) = 2 pH

, which gives

pH = \frac{0.17}{0.12} = 1.42

, that is

142 \times 1 0^{- 2}

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Integer answer