The half-cell, Tl | Tl+(1 mol L-1), is connected to a Pt | H+ | H2(1 atm) half-cell in which the concentration of H+ is unknown. The measured cell voltage is 0.0841 V and the Pt | H2 | H+ half-cell is the cathode. Standard reduction potentials are given on the next page. (a) Draw a diagram of the electrochemical cell, labeling all electrodes and solutions. (b) What is the pH in the Pt | H+ | H2 half-cell? (c) What is the value of the equilibrium constant for the reaction occurring in this cell?
(a) To draw a diagram of the electrochemical cell, we need to understand the setup described in the question. We have two half-cells connected to each other.
The first half-cell is Tl | Tl+(1 mol L-1), where Tl represents thallium and Tl+ represents thallium ions with a concentration of 1 mol L-1.
The second half-cell is Pt | H+ | H2(1 atm), where Pt represents platinum, H+ represents hydrogen ions, and H2 represents hydrogen gas at a pressure of 1 atm.
To draw the diagram, we can represent the two half-cells as follows:
Tl(s) | Tl+(1 mol L-1) || H+(unknown) | Pt(s) | H2(1 atm)
The double vertical line represents the salt bridge or the porous barrier that allows ion flow between the two half-cells.
(b) The pH in the Pt | H+ | H2 half-cell can be determined using the Nernst equation, which relates the cell voltage to the concentrations of the species involved. The Nernst equation is given as:
Ecell = E°cell - (RT/nF) * ln(Q)
In this case, Ecell is the measured cell voltage (0.0841 V), E°cell is the standard cell potential, R is the gas constant (8.314 J/(K·mol)), T is the temperature in Kelvin, n is the number of electrons transferred in the cell reaction, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.
Since we know the standard reduction potentials (given on the next page), we can use those values to calculate E°cell.
For the Pt | H+ | H2 half-cell, the half-reaction is:
2H+ + 2e- -> H2
The standard reduction potential for this half-reaction (E°red) is 0 V (given).
Now, we can use the Nernst equation to find the pH:
0.0841 V = 0 V - (8.314 J/(K·mol))(298 K)/(2 * 96485 C/mol) * ln(Q)
Simplifying the equation, we get:
0.0841 V = -0.12007 ln(Q)
Taking the exponential of both sides, we get:
Q = e^(-0.0841/0.12007)
Calculating this value, we get Q ≈ 0.322
Now, we can use the relationship between Q and the concentration of hydrogen ions (H+) to determine the pH.
Q = [H+]^2 (since the stoichiometric coefficient of H+ in the half-reaction is 2)
Therefore, [H+] = √(Q) = √(0.322) ≈ 0.568 mol L^-1
To convert this concentration into pH, we can use the logarithmic scale.
pH = -log[H+]
Substituting the value of [H+], we get:
pH = -log(0.568) ≈ 0.246
So, the pH in the Pt | H+ | H2 half-cell is approximately 0.246.
(c) The value of the equilibrium constant (K) for the reaction occurring in this cell can be related to the standard cell potential (E°cell) using the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(K)
Simplifying the equation, we get:
0.0841 V = E°cell - (8.314 J/(K·mol))(298 K)/(2 * 96485 C/mol) * ln(K)
Since we know the measured Ecell and E°cell, we can solve for ln(K):
ln(K) = (E°cell - 0.0841 V) * (2 * 96485 C/mol) / (8.314 J/(K·mol))(298 K)
Calculating this value, we get ln(K) ≈ -33.781
Taking the exponential of both sides to solve for K, we get:
K = e^(-33.781)
Calculating this value, we get K ≈ 4.139 x 10^(-15)
So, the value of the equilibrium constant for the reaction occurring in this cell is approximately 4.139 x 10^(-15).