This statement deals with the relationship between standard electrode potentials and Gibbs free energy. You won't be able to understand this unless you know about Gibbs free energy from statements in section 23.4. The statement is new to the syllabus for exams in 2022 onwards, and there isn't a question on it in the specimen papers. There is no way at the time of writing to judge what sort of questions CIE will ask. The key equation is:
You will often find this in books or on the web rearranged slightly as:
The terms: **ΔG°**is the standard Gibbs free energy change.**n**is the number of electrons transferred in the reaction.**F**is the Faraday constant, 96500 coulombs mol^{-1}.**E°**is the standard cell potential in volts._{cell}
We will look at a straightforward example of this, using what you should already know about calculating cell potentials from previous statements. Consider the reaction: If we split this into two half-reactions and find the standard electrode potentials for each of them, these are the E° values you would find from a data book: | |

Note: E° value equations are always written with the electrons on the left-hand side. Don't worry about the fact that in the overall equation we are looking at, we would need the nickel one the other way around.
That sorts itself out in the equation E° | |

Here is the overall equation again: The Pb Now we use the equation:
E°
You have a positive value for E° Now let's fit this into the equation this statement is talking about:
n is the number of electrons involved - in this case, 2. F is the Faraday constant, 96500 coulombs mol
| |

Note: The electrode potentials are only quoted to 2 significant figures, and so you can't quote your answer to more than that.
You might well wonder where the unit joules comes from, and what happens to the unit v in the E° | |

Notice that the calculation has produced a negative value for ΔG°. That means that the reaction is feasible. You will only get that result if E° The minus sign in the equation ΔG° = - nE° You would end up with a positive ΔG° value, and that would make the reaction not feasible. This equation links together the two conditions for feasibility you have met: E° _{cell}has to be positive.ΔG° has to be negative.
To return to the list of learning outcomes in Section 24 To return to the list of all the CIE sections This will take you to the main part of Chemguide.
© Jim Clark 2020 |