QJEEM19Ch83c2: JEE Main 2019 Chemistry - Class 12 JEEM19Ch83c2 Solution: In the cell Pt

Anode : H₂(g) $$\to$$ 2H⁺(aq) + 2e^-

Cathode : AgCl(s) + e^- $$\to$$ Ag(s) + Cl^-(aq)
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            H₂(g) + 2AgCl(s) $$\to$$ 2Ag(s) + 2H⁺(aq) + 2Cl^-(aq)

From Nernst equation we know,

E_cell = E⁰_cell - $$\frac{0.06}{n}\log Q$$

Here,

E_cell = E⁰_cell - $$\frac{0.06}{2}\log \frac{{\left[H^{+}\right]}^2{\left[C{l}^{-}\right]}^2}{P_{H_2}}$$

$$\Rightarrow$$ 0.92 = E⁰_{AgCl/AgCl ^-} - $$0.03\log \frac{{\left[{10}^{-6}\right]}^2{\left[{10}^{-6}\right]}^2}{1}$$

$$\Rightarrow$$ E⁰_AgCl/AgCl = 0.92 - 0.72 = 0.2 V