Gibbs Free Energy

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Tony Ong 3K
Posts: 63
Joined: Fri Sep 28, 2018 12:23 am

Gibbs Free Energy

Postby Tony Ong 3K » Tue Feb 26, 2019 10:39 pm

The reaction taking place in a nicad (nickel–cadmium) cell is Cd(s) +
2 Ni(OH)3(s) S Cd(OH)2(s) + 2 Ni(OH)2(s), and the cell potential when fully
charged is +1.25 V. What is the reaction Gibbs free energy?

Kobe_Wright
Posts: 83
Joined: Fri Sep 28, 2018 12:16 am

Re: Gibbs Free Energy

Postby Kobe_Wright » Tue Feb 26, 2019 11:09 pm

You'd use dG=-nFE. Where you need to find the electrons that flow between the two cells and that is your n while E is given and F is a constant, this gives you your gibbs free energy of the cell.

Xingzheng Sun 2K
Posts: 62
Joined: Fri Sep 28, 2018 12:29 am

Re: Gibbs Free Energy

Postby Xingzheng Sun 2K » Wed Feb 27, 2019 3:56 am

First, you need to write down the half-oxid and half-red equations for the reaction. When you balance it out, I find the electron flow of the reaction which is the number of e- being canceled on each side. Use this number as n to find the △G with △G= -nFE.


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