Question

Calculate the reduction potential of the following given electrode:-Pt,H2(2atm)|HCl(0.2M)

Answer 1

Answer:

The two electrodes collect H and O separately . We know that hydrogen composition in water is double than oxygen , so we find volume of gas collected in one electrode is double than volume of gas collected in another electrade .

Answer 2

Answer:

The reduction potential of the given electrode is equal to 0.0501V.

Explanation:

We have given the cell electrode: Pt, H₂ (2atm)| HCl (0.2M)

Firstly, write the cell reaction:

H₂ (g) $\longrightarrow$  2H⁺ (aq) + 2e⁻

We have given the pressure of hydrogen gas = 2atm

The concentration of hydrogen ions = 0.2M

From the Nernst equation, we can write the electrode potential:

$E_{cell}=E^{o}_{cell}-\frac{0.059}{n} log\frac{[H^{+}]^2}{p_{H_{2}}}$

For standard hydrogen electrode, the standard reduction potential is equal to zero volt. $E^{o}_{cell}=0$ and n = 2.

Putting the all values in equation (1);

$E_{cell}=0-\frac{0.059}{2}log\frac{(0.2)^2}{2}$

$E_{cell}=(-0.0295)(-1.699)$

$E_{cell}=0.0501V$

Therefore, the reduction potential of given hydrogen electrode is 0.0501 volt.