Question

JEE Mains chemistry question is in the image

Answer 1

we have to find the pH of HCl solution ( if the condition remains same).

solution : it's unique question. let's try to solve step by step.

for sodium metal :

from photoelectric effect,

we know, E = Φ + K_(max)

where Φ is work function.

given potential of the cell = 0.22V

so, maximum kinetic = workdone by electron

= potential × charge on electron

= 0.22 eV

now, E = 2.5 eV + 0.22 eV = 2.72 eV

for potassium metal :

again, E = Φ' + K'_(max)

here Φ' = 2.3 eV and E = 2.72 eV

so, K'_(max) = 2.72 eV - 2.3 eV = 0.42 eV

so, potential, E_(cell) = 0.42 V

now see the chemical reaction,

at cathode : AgCl + e¯ ⇔Ag + Cl¯

at anode : 1/2 H₂ - e¯ ⇔ H⁺

overall reaction, AgCl + 1/2 H₂ ⇔Ag + Cl¯ + H⁺

so, n = 1,

now from Nernst equation,

E_(cell) = E⁰_(cell) - 0.06/n log[H⁺][Cl¯]

but [Cl¯] = 1 M, E_(cell) = 0.42V

and E⁰_(cell) = 0.22 V

now 0.42 = 0.22 - 0.06/1 log[H⁺]

⇒0.20/0.06 = -log[H⁺]

⇒3.33 = -log[H⁺]

we know from Arrhenius formula, pH = -log[H⁺]

so, pH = 3.33

Therefore the pH of HCl solution must be 3.33