JEE Main 2020 — Electrochemistry Question with Solution

$$\frac{1}{2}$$H₂ + AgCl $$\to$$ H⁺ + Ag + Cl^-

From Nernst Equation,

E_cell = $$E_{cell}^0$$ - $$\frac{0.06}{1}\log \left[H^{+}\right]\left[C{l}^{-}\right]$$

= 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

$$\therefore$$ E_cell = 0.39 V

Using Nernst Equation,

0.39 = 0.22 - $$\frac{0.06}{1}\log \left[H^{+}\right]\left[C{l}^{-}\right]$$

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

0.39 = 0.22 - $$\frac{0.06}{1}\log {\left[H^{+}\right]}^2$$

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

$$\Rightarrow$$ 0.39 = 0.22 + 0.12 $$\times$$ pH

$$\Rightarrow$$ pH = 1.42 = 142 $$\times$$ 10^-2