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Suppose we have a redox reaction and we want to see in what direction the reaction is spontaneous. We do this by comparing the standard reduction potentials of the sub-reactions available in the main reaction and say if $$E^\circ = E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$ then the reaction is spontaneous in the direction mentioned otherwise in the reverse direction. The same result can be obtained by comparing the standard oxidation potentials. However, as every spontaneity has its root in the second law of thermodynamics I am eager to see how the spontaneity of a redox equation in a given direction can be checked based on thermodynamics.

So basically how can we derive the condition $$E^\circ = E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$ for spontaneity of a reaction from the fundamental thermodynamics laws?


To clarify the question better let me add that I already know that the second law for a common chemical reaction yields into the condition $$dG \le 0\,,$$ my question is about to find the condition $$E^\circ =E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$ from that inequality. Also let me add further that I even already know that the electric work equals $-n F E^\circ_{\text{cell}}$ but I cannot justify for myself why $\Delta G$ is equal to this work. (Note. the electric work is an internal work not crossing the boundaries of the system, it is not the work of the system, and if it was then at best we had $\Delta G=W_{\text{max}}$ but the inequality $\Delta G\le 0$ no longer hold.)

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The cell potential is related to the change in Gibbs free energy by this equation:

$$ \Delta G^\circ= -nFE^{\circ}_\text{cell} $$

where $n$ is the number of moles of electrons transferred in the balanced reaction equation and $F$ is Faraday's constant.

The UC Davis chem wiki has a brief explanation of the connection between cell potential, the reaction quotient (via the Nernst Equation), and the equilibrium constant. You should be able to find a more thorough explanation in a general chemistry textbook. My copy of Tro's Chemistry has a derivation of all of these relations, but to summarize the first relation:

Work against any potential is given by force times distance, and in the case of moving a charge, that works out to:

$$ w = -qE^{\circ}_\text{cell} $$

Faraday's constant $F$ gives the charge in coulombs on one mole of electrons, which means the total charge is given by:

$$ q = nF $$

Where $n$ is the number of moles of electrons.

We can substitute this back into the work equation to get:

$$ w = -nFE^{\circ}_\text{cell} $$

Since $\Delta G^{\circ}$ is equal to the maximum amount of work that can be done by a chemical reaction,

$$ \Delta G^\circ = w = -nFE^{\circ}_\text{cell} $$

In other words, saying that change in Gibbs free energy is negative is the same as saying that the cell potential is positive – in both cases we are simply looking at the total usable energy change and saying that "things move downhill" – the reaction proceeds in the direction that minimizes the thermodynamic potential energy of the system.

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  • $\begingroup$ thank you for your answer, but let me clarify my question a little more. I already knew that second law leads to $\Delta G\le0$ and also knew that the electric work equals $-n F E^\circ_{\text{cell}}$, the problem I have is why we are allowed to set equal $\Delta G$ to this work, so to have the relation between the second law and $E^\circ_{\text{cell}}$? Note that here the electric work is not a work of the system on hand, it is internal not crossing the boundaries of the system. Even if it was an external work then the 2nd law had led to $W_{max}=\Delta G$ and not $\Delta G\le 0$. Am I wrong? $\endgroup$ – topology Aug 4 '14 at 7:01

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