Ellen E. answered • 05/27/19

Chemistry Tutor - High School, AP, College

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 V^{2+}.

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 = E^{0} – (RT/nF) * lnQ

Variables:

E = 1.98 V (given in problem)

E^{0} = 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 E**^{0}**:**

To get E^{0} 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:

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

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

Overall reaction: Cu^{2+ }(aq) + V (s) + 2 e^{-} ---> Cu (s) + V^{2+} (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:

Cu^{2+ }(aq) + V (s) ---> Cu (s) + V^{2+} (aq)

And E^{0} = 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 [V**^{2+}**]**

Q = [V^{2+}]/[Cu^{2+}]

[Cu^{2+}] stays constant at 1.00 M , so [V^{2+}] = 1.311 x 10^{-15 }M

**plug the value of [V**^{2+}**] into the expression of K for the titration reaction:**

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

H_{2}EDTA^{2-} (aq) + V^{2+} (aq) --> VEDTA^{2-} (aq) + 2H^{+} (aq)

K = [VEDTA^{2-}][H^{+}]^{2} / [H_{2}EDTA^{2-}][V^{2+}]

So remember to get to this point in the titration, we reacted 0.04 mol H_{2}EDTA^{2-} with 0.04 mol V2+, and now (1.3108x10^{-15 }M * 1.5 L ) 1.966 x 10^{-15} mole V^{2+} 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}

[V^{2+}] = 1.311 x 10^{-15} (determined above by Voltage measurement and Nernst Equation)

[H_{2}EDTA^{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 10^{8}

^{}

Let me know if I can clarify anything!

Ellen E.

I'm thinking, there are several opportunities to build your chemical intuitions in this problem. 1) Our calculation showed that the residual concentration of the V2+ is very small after titration. This makes sense, because that is how titrations are supposed to work. 2) Our calculated K is very large. This also makes sense because the V2+ remaining after titration is very small AND, again, this is what you expect for good titration reaction. Titrations are, as much as possible, based on reactions that "go to completion", which means they must have a very high K and very little reactants remaining after reaction. 3)Another intuition to build: The shift in E after the titration is QUITE LARGE. The standard potential is 1.54, and after the titration the E shifts to 1.98 is quite large. This makes sense if you look at the overall reaction for the Cell: V2+ is on the PRODUCTS side. Depleting the products, means the reaction must shift left, towards the products, to go back to equilibrium. The standard potential (calculated at 1M solution concentration) is much smaller than in the post-titration scenario, because the post-titration scenario has been shifted much further away from equilibrium. I hope this will help you tie together the concepts of equilibrium and electrochemistry.05/27/19