The spontaneity or feasibility of a reaction can be described by both a positive Ecell⦵ value or a negative value for the change in Gibbs energy, ΔrG⦵, so is it is no surprise that the two are mathematically related.
ΔrG⦵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 …
ΔrG⦵ = – zF Ecell⦵
Example:
Calculate ΔrG⦵ 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
Ecell⦵ = +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 ΔrG⦵ .
(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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