I don't really study chemistry so while my question may be very obvious, its not obvious to me. If we take an electrochemical reaction like

$$\ce{2Fe^2+ + Au^3+ -> 2Fe^3+ + Au+}$$

we can find its standard potential and standard Gibb's energy by summing half reactions whose are known, and can be consulted in tables.

It's intuitive for me that the standard Gibb's energy will be given by a sum of the corresponding reactions, and that I can multiply a reaction by minus one, or whatever, and use that to construct my complete reaction.

The reason I see Gibbs energy as intuitive is because you're essentially just taking

$$\sum_{\mathrm{products}} \mu_{\mathrm{products}} - \sum_{\mathrm{reactants}} \mu_{\mathrm{reactants}}$$

and this will prove the additive property of "summing up reactions".

But I don't see any reason for additivity being true when considering standard potentials. I don't even see why inverting the reaction should invert my potential if my half cell is in equilibrium and is a kind of "whole" that is compared to the standard hydrogen electrode and should therefore have the same potential no matter how I write it.

How am I screwing up here? Why can we add up potentials?

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    $\begingroup$ Actually standard potentials are not always additive. They turn out to be additive in the case when the final reaction is a complete redox reaction not a half cell reaction. $\endgroup$ – Binary Geek May 17 '15 at 13:08

It helps to view this in terms of the equation

Δ = -nFE°

Reversing the reaction reverses the sign on Δ , and therefore the sign on . The sign on the potential determines the direction in which electricity flows. By convention, a galvanic cell (spontaneous reaction) has a positive , while an electrolytic cell (non-spontaneous reaction) has a -. The magnitude of the potential stays the same, but a sign is added to signify the direction of current flow.

As far as adding half cells to determine the overall potential, this can be done because of the relationship in the above equation. You can see that depends on Δ and n. But for instance, if n is multiplied by 2, Δ will be multiplied by 2 as well (there will be twice as much Gibbs' energy because the reaction involves twice the electron transfer). This means that you can add half-cell potentials to get the overall cell potential. Be careful to note though that if you multiply a reaction by a coefficient that you don't multiply the potentials, as you would for Δ ; the potential does not depend on the stoichiometric coefficients, only on what the cathode and anode are.

  • $\begingroup$ nice! I get it now. Thanks. By the way whats the exact difference between $Eº$ and $E$ ? is $Eº$ the potential you get at the beginning or something, since things are in their standard states? $\endgroup$ – DLV May 17 '15 at 15:10
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    $\begingroup$ $\mathrm{E^\circ}$ is the standard electrode potential; i.e everything is measured under standard conditions. $\mathrm{E}$ is any electrode potential under consideration and should be specified with the conditions under which it applies. $\endgroup$ – bon May 17 '15 at 17:21

Following on Derek's great answer, it is very important to remind that the conventional way we use to add half-cell potentials is a consequence of the conservation of energy. Therefore, we should look at this from the perspective of Hess's Law.

How so?

Well, if we add two reactions, no matter which ones, there's one certainty we keep in our souls: the new free energy will equal the sum of the individual ones.

This is a critical issue when trying to figure out the potential of a new reaction that is not the sum of an oxidation and a reduction.

For instance, take a look at these reactions: $$ \ce{Fe^3+ + e- -> Fe^2+} \qquad {E^\circ}_1 = 0.771\, \text{V}_{SHE} $$ $$ \ce{Fe^2+ + 2e- -> Fe_{(s)}} \qquad {E^\circ}_2 = -0.447 \, \text{V}_{SHE} $$ $$ \ce{Fe^3+ + 3e- -> Fe_{(s)}} \qquad {E^\circ}_3 =\, ? \, \text{V}_{SHE} $$

It is tempting to simply add these potentials and get the third value, but we would be incurring on a major error. The proper way to proceed would be to calculate each reaction free energy and then figure out the standard potential.

$$ \ce{Fe^3+ + e- -> Fe^2+} \qquad {\Delta G^\circ}_1 = -(1)\mathrm{F}{E^\circ}_1 $$ $$ \ce{Fe^2+ + 2e- -> Fe_{(s)}} \qquad {\Delta G^\circ}_2 = -(2)\mathrm{F}{E^\circ}_2 $$ $$ \ce{Fe^3+ + 3e- -> Fe_{(s)}} \qquad {\Delta G^\circ}_3 = -(3)\mathrm{F}{E^\circ}_3 $$

Since we know that ${\Delta G^\circ}_3 = {\Delta G^\circ}_1 + {\Delta G^\circ}_2$, we have: $${E^\circ}_3=\frac{{E^\circ}_1 +2{E^\circ}_2}{3} = -0.041\, \text{V}_{SHE} $$

Why don't we go through all this way when working with cell potentials? Because you always work your equations so that the number of electrons transferred on each of them has the same value. That way, the sum of the potentials is an indirect way of summing the free energies, which are now normalized by a common value.

This trick is evidently very useful, but sometimes it may be very misleading.

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    $\begingroup$ Nice answer! By the way, you can (and should!) use $\ce{...}$ for chemical formulae/equations: so, $\ce{Fe^2+ + 2e- -> Fe(s)}$ renders as $\ce{Fe^2+ + 2e- -> Fe(s)}$. $\endgroup$ – orthocresol Nov 10 '16 at 21:46
  • $\begingroup$ Thanks, orthocresol! I knew this package from LaTeX but I didn't know I could use it here (sort of new around here). $\endgroup$ – truffaut Nov 10 '16 at 23:42

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