What is the approximate pH of the H⁺/H₂electrode compartment through which H₂ gas is bubbled at a pressure of 1.00 atm if this electrode, when coupled with a Ni/(0.10 M) Ni²⁺ electrode, forms a galvanic cell having an initial potential of 0.10 V? [The discharge reaction involves oxidation of nickel to Ni²⁺.]

Question

Quizplus · Accepted Answer

The given electrode can be represented as:

Ni/(0.10 M) NiS1U1P12+S1S1P0 // HS1U1P1+S1S1P0/HS1U1B12 S1U1B0

The given initial potential of the cell is 0.10 V, and since the Ni/NiS1U1P12+S1S1P0 half-cell is the reduction half-cell, the HS1U1P1+S1S1P0/HS1U1B12 S1U1B0 half-cell must be the oxidation half-cell.

The Nernst equation for this half-cell is given by:

E = E° - (RT/nF) * ln(Q)

where E° is the standard potential, R is the gas constant, T is the temperature in Kelvin, F is the Faraday constant, n is the number of electrons transferred in the cell reaction, and Q is the reaction quotient.

Since the given electrode is being bubbled with HS1U1B12S1U1B0 gas at 1.00 atm pressure, we can assume that the reaction quotient Q is equal to the partial pressure of HS1U1B12S1U1B0 gas, which is 1.00 atm.

The electrode reaction at the HS1U1P1+S1S1P0/HS1U1B12 S1U1B0 half-cell is:

HS1U1B12S1U1B0(g) + e- -> HS1U1P1+S1S1P0(aq)

The number of electrons transferred in this reaction is 1.

The standard potential for this half-cell is not given, so we cannot calculate the exact value of E using the Nernst equation. However, we can make some assumptions to estimate the approximate pH.

At low pH values, the HS1U1P1+S1S1P0 species will be protonated to form H2S1U1P1+S1S1P0, and at high pH values, the HS1U1P1+S1S1P0 species will be deprotonated to form S1U1P01S1S1P0- ions.

If we assume that the standard potential for this half-cell is approximately 0.00 V (which is reasonable for a half-cell involving H2S1U1P1+S1S1P0/S1U1P01S1S1P0 species), then the Nernst equation simplifies to:

E = - (RT/F) * ln(1/atm)

where atm is the activity of HS1U1B12S1U1B0 gas.

Solving for pH:

E = 0.10 V = - (RT/F) * ln(1/atm)
pH = (-E * F / (2.303 * R * T)) - log10(atm)
pH ≈ 3.15

Therefore, the approximate pH of the HS1U1P1+S1S1P0/HS1U1B12 S1U1B0 electrode compartment is 3.15, and the correct answer is (D).

What Is the Approximate PH of the H+/H2 electrode Compartment

What Is the Approximate PH of the H⁺/H₂electrode Compartment