Calculation of Electrode Potential of Hydrogen Electrode

The electrode potential of hydrogen electrode is given by the Nernst equation:

E = E° + (RT/nF)ln(Q)

Where,
E = Electrode potential
E° = Standard electrode potential
R = Gas constant
T = Temperature
n = Number of electrons transferred
F = Faraday constant
Q = Reaction quotient

For the hydrogen electrode, the half-reaction is:

H+ + e- → 1/2 H2

The standard electrode potential for this half-reaction is 0 V. Therefore, the Nernst equation simplifies to:

E = (RT/nF)ln(Q)

At standard conditions (pH2 = 1 atm, [H+] = 1 M), Q = 1 and E = 0 V.

However, if the concentration of H+ ions is different from 1 M, then we need to calculate the reaction quotient.

The reaction quotient, Q, is given by:

Q = ([H+]^1/2/[H2])

Where,
[H+] = Concentration of H+ ions
[H2] = Partial pressure of H2 gas

In this question, the H3O+ ion concentration is given as 10^-5. Since H3O+ and H+ ions are in equilibrium, we can assume that [H+] = 10^-5 M.

The partial pressure of H2 is given as pH2 = 1 atm. The partial pressure of H2 is always written in the denominator of the reaction quotient. However, in this question, the partial pressure is given in the numerator. This is because pH2 is the pressure of H2 gas in atm, which is equivalent to the mole fraction of H2 gas in the mixture. Therefore, we can write pH2 as [H2]/(Total pressure).

Total pressure = pH2 + pN2 + pO2 + ... (partial pressures of other gases)

Since the partial pressures of other gases are negligible compared to pH2, we can assume that the total pressure is equal to pH2. Therefore, we can write pH2 as [H2]/pH2.

Substituting the values in the Nernst equation, we get:

E = (RT/nF)ln([H+]^1/2/[H2]/pH2)
E = (8.314 × 298.15/2 × 96485)ln(10^-5/1/1)
E = 0.000198 V

Therefore, the electrode potential of hydrogen electrode when [H3O+] = 10^-5 (pH2 = 1 atm) is 0.000198 V.