Calculate emf of the half-cell given below:

Pt(s)|H₂(g,2 atm)|HCl(aq,0.02M)

E◦_H+/H2 = 0V, 2.303RT/F=0.059,log2 =0.3010

The half‑reaction for the hydrogen electrode is

$2H++2e-\to H{2}_g$

For this reduction the number of electrons transferred is $\mathrm{n}=2$. The Nernst equation for the electrode potential relative to the standard hydrogen electrode (E° = 0 V) is

$E=E_{°}-\frac{0.059}{n}·\log Q_{}$

Because the reaction is written as a reduction, the reaction quotient is

$Q_{}=\frac{p_H}{}$₂ a_H⁺ 2

Given the experimental conditions: $p_H$₂ = 2 atm and a strong acid solution $a_{\mathrm{H⁺}}$. Thus

$Q_{}=\frac{2}{{0.02}^2}=\frac{2}{0.0004}=\mathrm{5 \times 10³}$

Taking the common logarithm:

$\log \left(Q_{}\right)=\log \left(5·103^{}\right)=\log 5+3=0.6990+3=3.6990$

Now insert the values into the Nernst equation (E° = 0 V, n = 2):

$E=0-\frac{0.059}{2}·\log \left(Q_{}\right)=\; -0.0295·3.6990=\; -0.109 V$

Therefore the emf of the given half‑cell is −0.109 V.