In the activity, click on the e∘cell and keq quantities to observe how they are related. use this relation to calculate keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. the two half-reactions that occur in the cell are

cu2+(aq)+2e−→cu(s) and co(s)→co2+(aq)+2e−

the net reaction is

cu2+(aq)+co(s)→cu(s)+co2+(aq)

use the given standard reduction potentials in your calculation as appropriate. ( keq=5.88*10^20)

in the activity, click on the keq and δg∘ quantities to observe how they are related.

calculate δg∘ using this relationship and the equilibrium constant (keq) obtained in part a at t=298k: keq=5.88*10^20

express the gibbs free energy (δg∘) in joules to three significant figures.

ΔG° = -118x10³ J/mol

Explanation:

The two half-reactions in the cell are:

Oxidation half-reaction:

Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V

Reduction half-reaction:

Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V

The E° of the cell is defined as:

Replacing:

0,34V - (-0,28V) = 0,62V

It is possible to obtain the keq from E°cell with Nernst equation thus:

nE°cell/0,0592 = log (keq)

Where:

E°cell is standard electrode potential (0.62 V)

n is number of electrons transferred (2 electrons, from the half-reactions)

Replacing:

0,62V×2/0,0592 = log (keq)

20,946 = log keq

keq = 8,83x10²⁰≈ 5,88x10²⁰

ΔG° is defined as:

ΔG° = -RT ln Keq

Where R is gas constant (8,314472 J/molK) and T is temperature (298K):

ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰

ΔG° = -118x10³ J/mol

I hope it helps!

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