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

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

Answers

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