Chemistry - Electrochemistry Question with Solution | TestHub

The given reaction is $Pb(s)+2H^{+}(aq)\to P{b}^{2+}(aq)+H_2(g)$.

At equilibrium, $E_{cell}=0$.

The Nernst equation for this reaction is:

$$E_{cell}=E_{cell}^{\circ }-\frac{0.065}{2}\log \frac{[P{b}^{2+}]P_{H_2}}{[H^{+}{]}^2}$$

We are given $E_{Pb/P{b}^{2+}}^{\circ }=0.13V$. For the reverse reaction, $E_{cell}^{\circ }=-E_{Pb/P{b}^{2+}}^{\circ }=-0.13V$.

However, the reaction as written is $Pb\to P{b}^{2+}$ (oxidation) and $2H^{+}\to H_2$ (reduction).

So, $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=E_{H^{+}/H_2}^{\circ }-E_{P{b}^{2+}/Pb}^{\circ }=0-(-0.13V)=0.13V$.

Given $\mathrm{pH}=3$, so $[H^{+}]=10^{-3}M$.

Given $[P{b}^{2+}]=0.001=10^{-3}M$.

Substitute these values into the Nernst equation at equilibrium ($E_{cell}=0$):

$$0=0.13-\frac{0.065}{2}\log \frac{10^{-3}\times P_{H_2}}{(10^{-3}{)}^2}$$

$$0.13=\frac{0.065}{2}\log \frac{10^{-3}\times P_{H_2}}{10^{-6}}$$

$$0.13=\frac{0.065}{2}\log (10^3\times P_{H_2})$$

$$\frac{0.13\times 2}{0.065}=\log (10^3\times P_{H_2})$$

$$4=\log (10^3\times P_{H_2})$$

$$10^4=10^3\times P_{H_2}$$

$$P_{H_2}=\frac{10^4}{10^3}=10bar$$

The final answer is $\boxed{10}$.