Equilibrium Potential of Half Cell

[garbagefish]

Homework Statement

Given the following half-reactions:
Ce4+ + e− → Ce3+ E° = 1.72 V
Fe3+ + e− → Fe2+ E° = 0.771 V

A solution is prepared by mixing 9.0 mL of 0.30 M Fe2+ with 7.0 mL of 0.11 M Ce4+.

1. What is the potential of a platinum electrode dipped into the resulting, equilibrated, solution (relative to SHE)?

2. Calculate Keq for the reaction

Ce4+ + Fe2+ ⇌ Ce3+ + Fe3+

3. Calculate [Ce4+] in the solution.

Homework Equations

Nernst equation

The Attempt at a Solution

I got question 2 since its K = 10^(nEo/0.05916), K= 10^(1(1.72-0.771))/0.05916)=1.1*10^16

I tried and can't seem to get #1 correct. #3 I used ICE table method but cannot get a answer that makes sense. Any help is appreciated. Thanks.

 

[nate808]

Thank you for your post. I would be happy to assist you with your questions.

1. The potential of a platinum electrode dipped into the resulting solution can be calculated using the Nernst equation: E = E° - (0.05916/n)logQ, where E is the cell potential, E° is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient. In this case, n = 1 because only one electron is transferred in each half-reaction. The reaction quotient can be calculated using the concentrations of the species involved in the reaction. Since the solution is equilibrated, the concentrations of the species will be the same as the initial concentrations given in the problem. Thus, the reaction quotient can be calculated as follows:

Q = [Ce3+][Fe3+]/[Ce4+][Fe2+]
= (0.11 M)(0.30 M)/(0.30 M)(0.11 M)
= 1

Plugging in the values into the Nernst equation, we get:

E = 1.72 V - (0.05916/1)log(1)
= 1.72 V

Therefore, the potential of the platinum electrode dipped into the resulting solution is 1.72 volts relative to the standard hydrogen electrode (SHE).

2. Your calculation for Keq is correct.

3. To calculate the concentration of Ce4+ in the solution, we can use the equilibrium constant expression for the reaction:

Keq = [Ce3+][Fe3+]/[Ce4+][Fe2+]

Rearranging the equation to solve for [Ce4+], we get:

[Ce4+] = [Ce3+][Fe3+]/(Keq[Fe2+])

Substituting the known values, we get:

[Ce4+] = (0.11 M)(0.30 M)/(1.1*10^16)(0.30 M)
= 2.5*10^-16 M

Therefore, the concentration of Ce4+ in the solution is 2.5*10^-16 M.

I hope this helps. Let me know if you have any further questions.

 

[Blueshift5]

Hello,

Let's first address your attempt at question 2. The equation you have used is not the correct one for calculating equilibrium constant (Keq). The correct equation is Keq = [products]/[reactants], where the concentrations are in Molarity (M). Therefore, for the given reaction, the Keq would be Keq = ([Ce3+][Fe3+])/([Ce4+][Fe2+]).

Now, for question 1, we need to first understand the concept of equilibrium potential. Equilibrium potential is the potential difference between the two half-cells at equilibrium. In this case, the two half-cells are Ce4+/Ce3+ and Fe3+/Fe2+. The potential of the platinum electrode will depend on the difference between the potentials of these two half-cells.

To calculate the equilibrium potential, we can use the Nernst equation: Ecell = E°cell - (0.05916/n)logQ, where E°cell is the standard cell potential, n is the number of electrons transferred, and Q is the reaction quotient. In this case, n = 1 for both half-reactions.

To calculate Q, we need to use the concentrations of the reactants and products at equilibrium. Since the solution is equilibrated, the concentrations of all species will be constant. Therefore, Q = [Ce3+]/[Ce4+] and [Fe3+]/[Fe2+].

Substituting the values in the Nernst equation, we get: Ecell = (1.72 V) - (0.05916/1)log([Ce3+]/[Ce4+]) - (0.05916/1)log([Fe3+]/[Fe2+]).

Now, we can use the concentrations given in the problem to calculate the values of [Ce3+] and [Fe3+]. We know that the total volume of the solution is 16 mL, so the final concentration of Ce3+ will be: [Ce3+] = (0.11 M)(7 mL)/(16 mL) = 0.048 M. Similarly, the final concentration of Fe3+ will be: [Fe3+] = (0.30 M)(9 mL)/(16 mL) = 0.169 M.

Substituting these values in the Nernst equation, we get: Ecell =