Calculate emf of the half cell given below :
\(Pt ( s )\left| H _2(g, 2 atm)\right| HCl ( aq , 0.02 M )\)
\(E_{ H _2 / H ^{+}}^{\circ}=0 V\)
\(\left(\right.\) Given : \(\left.\frac{2.303 R T}{F}=0.059, \log 2=0.3010\right)\)

(D) The given half-cell is represented as an oxidation electrode: \(Pt ( s )\left| H _2(g)\right| H ^{+}( aq )\).
The corresponding oxidation half-cell reaction is:
\(H_2(g) \rightarrow 2 H^{+}(aq)+2 e^{-}\)
Using the Nernst equation for the oxidation potential:
\(E=E_{H_2 / H^{+}}^{\circ}-\frac{0.059}{n} \log\frac{\left[H^{+}\right]^2}{P_{H_2}}\)
Given values:
\(E_{H_2 / H^{+}}^{\circ}=0 V\) 
n=2 
\({\left[H^{+}\right]=0.02 M \text { (since } HCl \text { is a strong monoprotic acid) }}\)  
\(P_{H_2}=2 atm\) 
Substituting the values into the Nernst equation:
\(E=0-\frac{0.059}{2} \log \frac{(0.02)^2}{2}\) 
\(E=-0.0295 \log \frac{4 \times 10^{-4}}{2}\)  
\(E=-0.0295 \log \left(2 \times 10^{-4}\right) \) 
\(E=-0.0295\left(\log 2+\log 10^{-4}\right)\) 
\(E=-0.0295(0.3010-4) \) 
\(E=-0.0295 \times(-(C)699) \) 
\(E=0.10912 V \approx 0.109 V\)