Galvanic cell based on half-reactions

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 cell potential of the cell before the titration was carried out.

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

Answer 1

Oh how I love AP chem. Such juicy questions.

Part a)

Here you have a Galvanic cell that is at NON-STANDARD conditions. For such things, we need the Nernst equation:

E = E⁰ – (RT/nF) * lnQ

But of course, since this is AP chemistry, almost none of the values we need for the Nernst equation are simply given in the problem. We will have to calculate:

-E⁰

-Q

-And in order to get Q, we need the concentration of V²⁺, which has been determined by a titration. 

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. This means that the copper reaction is the reduction in this Galvanic cell (making copper the cathode).

And Vanadium must be the oxidation reaction (making it the anode, where the electrons come from). You have to “flip” the Vanadium reaction and change the sign of the reduction potential from negative to positive:

Cu²⁺ (aq) + 2 e^- --> Cu (s)                       E⁰ =0.34

V(s) --> V²⁺ (aq) + 2 e^-                   E⁰ = 1.20

Overall reaction:

Cu²⁺ (aq) + V (s) --> Cu (s) + V²⁺ (aq) E⁰ = 0.34 + 1.20 = 1.54 Volts.

Now to use the information from the titration to figure out the concentration of V²⁺ (which will be used to calculate Q in the Nernst Equation).

You’re given the balanced equation:

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

And you’re told how much H₂EDTA^2- was used to get to the stoichiometric point:

500.0 ml of 0.08 M H₂EDTA2- was added to get to the stoichiometric point.

That is, (0.5000 L * 0.0800 M =) 0.0400 mol H₂EDTA^2- were needed to get to the stoichiometric point.

The stoichiometric ratio of H₂EDTA^2- and V²⁺ is 1:1.

So that means 0.0400 mol V²⁺ was present in the original solution. The volume of the original solution was 1.00 L so the concentration of the original solution was (0.04 mol/ 1.0 L) = 0.0400 M. This is the information we need to calculate Q.

Q is the [Products]/[Reactants] for the balanced reaction of the Galvanic cell (not the titration!)

Q = [V²⁺]/ [Cu²⁺] = 0.0400/1.00 = 0.0400

(copper concentration is given in the problem!)

Now we can plug that into the Nernst equation and get the answer to part a:

E = E⁰ – (RT/nf)*lnQ

Variables:

E = ?

E⁰ = 1.54 Volts

R = 8.3145 J/Kmol

T = There is no temperature specified in the problem, so I’m not sure what to do about that. I am going to go with 25⁰C, which is 298 K

n = 2 mol electron (see balanced chemical equation for the Galvanic cell)

F = 96485 Coulomb/mole electrons.

Q = 0.04

E = 1.54 – [(8.3145*298)/(2)(96485)]*ln0.04

E = 1.54 – [(2478)/(192970)]*(-3.219)

E = 1.54 - [0.01284]*(-3.219)

E = 1.54 – (-0.04133)

E = 1.58133 = 1.58 Volts (3 sig figs)

For Part b)

You use the E measured at the stoichiometric point to calculate Q using the Nernst Equation. 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

E0 = 1.54 V

R = 8.3145 J/Kmol

T = I am guessing 25C, since you didn't provide a temperature in the problem. So 298 K

n = 2 moles electron

F = 96485 Coulomb/mol electrons

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

Q = [V²⁺]/[Cu²⁺]

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

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, 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 by Voltage measurement)

[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.

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