Question

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. (Answer: 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.

Answer 1

Answer:

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:

$E_{cell} = E_{red} - E_{ox}$

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!