Mikaela B.

asked • 05/26/19# Galvanic cell based on half-reactions

**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 V**^{2+}** at an unknown concentration. The compartment containing the vanadium (1.00 L of solution) was titrated with 0.0800 M H**_{2}**EDTA**^{2-}**, resulting in the reaction:**

**H**_{2}**EDTA**^{2-}** (aq) + V**^{2+}** (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 H**_{2}**EDTA**^{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.**

## 1 Expert Answer

Ellen E. answered • 05/26/19

Chemistry Tutor - High School, AP, College

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 = E^{0} – (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:

-E^{0}

-Q

-And in order to get Q, we need the concentration of V^{2+}, which has been determined by a titration.

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 Cu^{2+} 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:

Cu^{2+} (aq) + 2 e^{-} --> Cu (s) E^{0} =0.34

V(s) --> V^{2+} (aq) + 2 e^{-} E^{0} = 1.20

Overall reaction:

Cu^{2+} (aq) + V (s) --> Cu (s) + V^{2+} (aq) E^{0} = 0.34 + 1.20 = 1.54 Volts.

Now to use the information from the titration to figure out the concentration of V^{2+} (which will be used to calculate Q in the Nernst Equation).

You’re given the balanced equation:

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

And you’re told how much H_{2}EDTA^{2-} was used to get to the stoichiometric point:

500.0 ml of 0.08 M H_{2}EDTA2- was added to get to the stoichiometric point.

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

The stoichiometric ratio of H_{2}EDTA^{2-} and V^{2+} is 1:1.

So that means 0.0400 mol V^{2+} 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 = [V^{2+}]/ [Cu^{2+}] = 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 = E^{0} – (RT/nf)*lnQ

Variables:

E = ?

E^{0} = 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 25^{0}C, 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 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

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

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

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, 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 by Voltage measurement)

[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.

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}

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Ellen E.

Did they specify a temperature?05/26/19