Equilibrium constant value

Question

A galvanic cell is based on the following half- reactions:

Cu⁺² (aq) + 2e^- à Cu (s)                  ɛ° = 0.34 V

V⁺² (aq) + 2e^- à V (s)                   ɛ° = -1.20 V

In this cell, the copper compartment contains a copper electrode and [Cu⁺²] = 1.00 M, and the vanadium compartment contains a vanadium electrode and V²⁺ at an unknown concentration. The compartment containing the vanadium (1.00 L of solution) was titrated with 0.0800 M H₂EDTA^2-, resulting in the reaction:

H₂EDTA^2- (aq) + V²⁺ (aq) ßà VEDTA^2- (aq) + 2H⁺ (aq)     K=?

The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of 500.0 mL H₂EDTA^2- solution added. At the stoichiometric point, the cell potential was observed to be 1.98 V. The solution was buffered at a pH of 10.00.

a. Calculate the value of the equilibrium constant, K, for the titration reaction.

Answer 1

For this problem, you must use the Nernst equation to calculate Q using the reported value for E. From Q, you can figure out the concentration of V²⁺.

Plug that into the equation for the equilibrium constant of the titration reaction, and that will be your K.

Okay so Nernst equation first :

E = E⁰ – (RT/nF) * lnQ

Variables:

E = 1.98 V (given in problem)

E⁰ = 1.54 V (see explanation immediately below)

R = 8.3145 J/Kmol

T = I am going to use 25C / 298 K, since you didn't provide a temperature in the problem. If you have different temperature information, use that!

n = 2 moles electron (see explanation immediately below)

F = 96485 Coulomb/mol electrons

Figuring out E⁰:

To get E⁰ you have to write and balance the overall reaction for the Galvanic cell.

Comparing the reduction potentials of the 2 reactions, you see that Cu2+ has the higher reduction potential. So the copper reaction is the reduction (at the Cathode).

That means that the Vanadium is the oxidation reaction (the anode). You have to “flip” that reaction and change the sign of the reduction potential from negative to positive:

Cu²⁺ (aq) + 2 e^- -> Cu (s)                        0.34

V(s) -> V²⁺ (aq) + 2 e^-                    1.20

Overall reaction: Cu²⁺ (aq) + V (s) + 2 e^- ---> Cu (s) + V²⁺ (aq) + 2 e-

Since the balanced equation has 2 e^-, then n=2 in Nernst Equation above.

2 electrons on both sides of the equation cancel so that the overall reaction is written as:

Cu²⁺ (aq) + V (s) ---> Cu (s) + V²⁺ (aq)

And E⁰ = 0.34 + 1.20 = 1.54 Volts.

Plugging values into Nernst Equation:

1.98 = 1.54 - [(8.3145*298)/(2)(96485)]*lnQ

Solve for Q.

0.44 = (-0.01284) * lnQ

-34.27 = lnQ

Q = 1.311 x10^-15

Using the value of Q to find the concentration of [V²⁺]

Q = [V²⁺]/[Cu²⁺]

[Cu²⁺] stays constant at 1.00 M , so [V²⁺] = 1.311 x 10^-15 M

plug the value of [V²⁺] into the expression of K for the titration reaction:

We use this information to determine K for the titration reaction:

H₂EDTA^2- (aq) + V²⁺ (aq) --> VEDTA^2- (aq) + 2H⁺ (aq)

K = [VEDTA^2-][H⁺]² / [H₂EDTA^2-][V²⁺]

So remember to get to this point in the titration, we reacted 0.04 mol H₂EDTA^2- with 0.04 mol V2+, and now (1.3108x10^-15 M * 1.5 L ) 1.966 x 10^-15 mole V²⁺ remain in a total volume of 1.5 L

[VEDTA^2-] = 0.04 mol/ 1.5 L = 2.67 M

[H⁺] = is determined from the pH:

pH = -log [H+]

10 = -log[H+]

[H+] = 1 x 10^-10

[V²⁺] = 1.311 x 10^-15 (determined above by Voltage measurement and Nernst Equation)

[H₂EDTA^2-] = 1.311 x 10^-15 (this is by stoichiometry)

Notice, I don't subtract the 1.966 x 10^-15 moles that remain on the reactant side from the 0.04 moles that are on the product side. This is due to sig figs. Let me know if you want further clarification on that.

Plugging values into the formula for the equilibrium constant:

K = [.0267][1.0 x 10^-10]² / [1.311x10^-15][1.311 x 10 ^-15]

K = (2.67 x 10^-22) / (1.719 x 10^-30)

K = 1.55 x 10⁸

^

Let me know if I can clarify anything!