Understanding the relationship between the potential difference for an electrochemical cell and Gibbs energy

The spontaneity or feasibility of a reaction can be described by both a positive E_cell^⦵ value or a negative value for the change in Gibbs energy, Δ_rG^⦵, so is it is no surprise that the two are mathematically related.

Δ_rG^⦵for a reaction not only tells us if that reaction is spontaneous, but its value is also equivalent to the maximum amount of work a system can do. This makes is really useful for determining the electrical work that can be obtained from a chemical reaction in a battery or fuel cell. 

Imagine an electrochemical cell where we replace the high resistance voltmeter with an electric motor, for example.  The electrons flowing from the oxidation (negative) half cell to the reduction (positive) half cell can be used to do useful work such as drive a motor.

The energy required to transfer charge around a circuit represents the electrical work that the electrochemical cell can perform.

charge (Q) = zF
z = no. of mol of electrons produced in the reaction
F = Faraday’s constant (charge on 1 mol of electrons 96500 Cmol^-1)

It takes 1 joule of work to move 1 coulomb of charge through a potential difference of 1 volt.

So if we combine these equations …

 Δ_rG^⦵  =  – zF E_cell^⦵ 

Example:

Calculate Δ_rG^⦵ for the following reaction, under standard conditions, using standard electrode potential data

Fe²⁺_(aq)  +  Ag⁺_(aq)  ⇾  Fe³⁺_(aq)  +  Ag_(s)

Ag⁺  +  e^–  ⇌  Ag     E^⦵=  + 0.80V    (reduction / positive half cell)

Fe³⁺  +  e^–  ⇌  Fe²⁺     E^⦵=  + 0.77V (oxidation / negative half cell)

1 mol of e^– transferred

E_cell^⦵ =  +0.80 – 0.77 = + 0.03V

Practice questions

1. Calculate E^⦵_cell for each of the following electrochemical cells, write an equation for the spontaneous reaction that occurs and calculate a value for Δ_rG^⦵ .

(a) Zn²⁺_(aq) / Zn_(s)  and  Cd²⁺_(aq) / Cd_(s)

(b) Cu²⁺_(aq) Cu_(s)  and  Cd²⁺_(aq) / Cd_(s)

(c) Ag⁺_(aq) / Ag_(s)  and Cd²⁺_(aq) / Cd_(s)

Half cell	E⦵ / V
Zn2+(aq)  +  2e-  ⇌  Zn(s)	-0.76
Cd2+(aq)  +  2e-  ⇌  Cd(s)	-0.40
Cu2+(aq)  +  2e-  ⇌  Cu(s)	+0.34
Ag+(aq)  +  e-  ⇌  Ag(s)	+0.80

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2. The first stage in making a useful cobalt compound requires Co²⁺_(aq) ions to be oxidised to Co³⁺_(aq) ions in the presence of a catalyst. A student wanted to work out whether it was possible to carry out this reaction without a catalyst. 

Co³⁺_(aq) + e^– ⇌ Co²⁺_(aq)     E^⦵  = +1.82V

O_2(g) + 4H⁺_(aq) + 4e^– ⇌ 2H₂O_(l)    E^⦵  = +1.23V

1. Write the overall equation for the oxidation of Co²⁺ by O₂.
2. Use standard electrode potentials to calculate ΔG^⦵ for the reaction, given that Faraday’s constant is 96500 C mol^-1.
3. With reference to your answer above, explain whether this reaction is feasible.

4. How would the feasibility of this reaction change if the solution were to become less acidic?  Explain your answer. 

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3. Iron(II) ions or iron(III) ions can be used to catalyse the reaction between iodide ions and peroxodisulphate ions, S₂O₈^2-

2I^–_(aq) +    S₂O₈^2- _(aq)   ⇾  2SO₄^2-_(aq)  +  I_2(aq)

(a) Use the standard electrode data given below to calculate E^⦵_cell for the reaction above.

Fe3+(aq)  + e-  ⇌  Fe2+(aq)	E⦵ = +0.77 V
S2O82-(aq)  + 2e-  ⇌  2SO42-(aq)	E⦵ = +2.01 V
I2(aq) + 2e-  ⇌  2I–(aq)	E⦵ = +0.54 V

(b)   Use your answer to part (a) to calculate the Gibbs energy for this reaction (Faraday’s constant is 96500 Cmol^-1).

(c)   Explain how your answers prove that the reaction is feasible.

(d)   Suggest, with reasons, why the reaction is not seen to occur without a catalyst present. 

(e)   Explain why, with reference to the standard electrode potential data, both iron(II) ions and iron(III) ions are able to catalyse this reaction. Include balanced full equations in your answer. 

Answers

2. (a)   4Co²⁺_(aq) + O_2(g) + 4H⁺_(aq) ⇾ 4Co³⁺_(aq) + 2H₂O_(l)

(b) E^⦵_cell = +1.23 – 1.82 = -0.59V (O₂ is oxidising Co²⁺ (the opposite of what E^⦵ would predict))

Δ_rG^⦵ = -zF.E_cell  = – 4 x 96500 x (-0.59) = +228 kJ mol^-1

(c) The reaction is not feasible as ΔG is positive.

(d) O_2(g) + 4H⁺_(aq) + 4e^– ⇌ 2H₂O_(l)    E^⦵  = +1.23V
If the solution is less acidic, the position of equilibrium shifts to the left producing more electrons and E^⦵ will be less positive. This means E_cell would be even less positive, ΔG more positive and the reaction even less feasible.

3. (a) E^⦵_cell = +2.01 – (+0.54) = +1.47V
(b) Δ_rG^⦵ = -zF.E_cell  = – 2 x 96500 x (+1.47) = -284 kJ mol^-1
(c) E^⦵_cell is positive and ΔG is negative, both showing that the reaction is feasible. The magnitude of ΔG also indicates that the reaction would go to completion.

(d) The need for a catalyst suggests the reaction has a high activation energy because we need to overcome the repulsion between two negative ions coming together.

(e) S₂O₈^2- would be reduced to SO₄^2- by Fe²⁺ because the Fe³⁺ / Fe²⁺ half cell has a less positive E^⦵. I^– would then reduce Fe³⁺ back to Fe²⁺ regenerating the catalyst and I^– are oxidised to I₂ because the Fe³⁺ / Fe²⁺ half cell has a more positive E^⦵ than the I₂ / 2I^– half cell. Alternatively, the reactions could happen the other way around, with Fe³⁺ oxidising I^– to I₂ first. 

2Fe²⁺_(aq) + S₂O₈^2-_(aq) ⇾  2Fe³⁺_(aq) + 2SO₄^2-_(aq) 

2Fe³⁺_(aq) + 2I^–_(aq) ⇾ 2Fe²⁺_(aq) + I_2(aq)