Question

the cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: pt(s)|h2(g,1atm)|h (aq,1.0m)|au3 (aq,?m)|au(s) . what is the concentration of au3 in the solution if ecell is 1.25 v ?

Answer 1

The concentration of Au³⁺ in the solution is 2.101 x 10⁻²² M.

Step-by-step explanation:

Determining the Au³⁺ concentration. The cell reaction is:

Pt(s) + 3H⁺(aq) + 3Au³⁺(aq) → Au(s) + 3H₂(g)

We are given the following information:

• E_cell = 1.25 V
• [H⁺] = 1.0 M
• P(H₂) = 1 atm
• [Au³⁺] = Unknown (x M)

We need to find [Au³⁺].

Step 1: Identify the standard electrode potentials

E°_H₂ = 0.00 V

E°_Au = 1.50 V

Step 2: Write the Nernst equation for the cathode half-cell

The Nernst equation for the reduction of Au³⁺ to Au is:

E_cathode = E°_Au - (RT/nF) * ln([Au³⁺])

where:

• R is the gas constant (8.314 J/mol K)
• T is the temperature (298 K)
• n is the number of electrons transferred (3)
• F is the Faraday constant (96485 C/mol)

Step 3: Substitute the known values and solve for [Au³⁺]

E_cathode = E_cell = 1.25 V

E°_Au = 1.50 V

[H⁺] = 1.0 M

P(H₂) = 1 atm

n = 3

F = 96485 C/mol

Substituting these values and solving for [Au³⁺] gives:

ln([Au³⁺]) = - [(1.25 V - 1.50 V) * 3 * F] / (R * T)

[Au³⁺] = exp(- [(1.25 V - 1.50 V) * 3 * 96485 C/mol] / (8.314 J/mol K * 298 K))

[Au³⁺] = 2.101 x 10⁻²² M

Therefore, the concentration of Au³⁺ in the solution is 2.101 x 10⁻²² M.