For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [Co3+]= 0.765 M , [Co2+]= 0.369 M , and [Cl−]= 0.486 M , and the pressure of Cl2 is PCl2= 9.30 atm ? Express your answer with the appropriate units.

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Answer:

Ecell = 0.500 V

Explanation:

For a chemical cell, the cell potential (Ecell) is:

Ecell = E° - (0.0592/n)*logQ (Nernst equation)

Where n is the number of electrons involved in the redox reaction, and Q is the reaction coefficient.

Let's see a half-reaction:

2Co³⁺(aq) → 2Co²⁺(aq)

The charge goes from +6 (2*3) to +4 (2*2), so n = 6 - 4 = 2 electrons.

The reaction coefficient is the multiplication of the concentration of the products elevated by their coefficients, divided by the multiplication of the concentration of the reactants elevated by their coefficients:

Q = ([Cl₂]*[Co⁺²]²)/([Co⁺³]²*[Cl⁻]²)

The concentration of Cl₂ is the number of moles (n) divided by the volume(V), and can be calculated by the ideal gas law:

PV = nRT

n/V = P/RT  (P is the pressure, R is the gas constant, and T is the temperature)

P = 9.30 atm, R = 0.082 atm.L/mol.K, T = 25°C + 273 = 298 K

n/V = [Cl₂] = 9.30/(0.082*298)

[Cl₂] = 0.381 M

Q = [0.381*(0.369)²]/[(0.765)²*(0.486)²]

Q = 0.051877/0.138228

Q = 0.3753

Ecell = 0.483 - (0.0592/2)*log(0.3753)

Ecell = 0.483 - 0.0296*(-0.4256)

Ecell = 0.500 V

The cell potential of the electrochemical cell with the cobalt and chlorine electrodes has been 0.5 V.

What is an electrochemical cell?

The electrochemical cell has been the cell in which the chemical and electrical energy is converted by the charge transfer.

The cell potential ([tex]E_{cell}[/tex]) can be given as:

[tex]E^\cell=E^\circ-\dfrac{0.059}{n}\;\times\;log\; Q[/tex]

The standard cell potential  ([tex]E^\circ[/tex]) of the cell has been:

[tex]E^\circ=0.483\;\text V[/tex]

The charge transferred in the reaction ([tex]n[/tex]) can be given as 2.

The concentration of chlorine  has been:

[tex]\rm M=\dfrac{P}{RT} \\\\Cl_2=\dfrac{9.30}{0.0821\;\times\;298}\\\\ Cl_2=0.981\;M[/tex]

The value of Q is given as:

[tex]\rm Q=\dfrac{[Cl_2][CO_2]^2}{[CO_3]^2[Cl]^2}\\\\ Q=\dfrac{(0.981)(0.369)^2}{(0.765)^2(0.486)^2} \\\\Q=0.3753[/tex]

Substituting the values for the cell potential:

[tex]E_{cell}=0.483-\dfrac{0.059}{2}\;\times\;\text {log}\;0.3753 \\E_{cell}=0.5\;\rm V[/tex]

The cell potential is 0.5 V.

Learn more about cell potential, here:

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