[cliverlong]

I have also posted this question on the "talk" page of the following Wikipedia entry (my emphases)

http://en.wikipedia.org/wiki/Standard_electrode_potential

Any ideas here?
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First the wikipedia entry text says ...

    Since the electrode potentials are conventionally defined as reduction potentials, the sign of the potential for the metal electrode being oxidized must be reversed when calculating the overall cell potential. Note that the electrode potentials are independent of the number of electrons transferred -that is, they are set to one mole of electrons transfered- and so the two electrode potentials can be simply combined to give the overall cell potential even if different numbers of electrons are involved in the two electrode reactions

Then we have the following text ..

    Since the table of standard electrode potentials is defined for a transfer of one mole of electrons, care must be made in determining an electrode potential using two other electrode potentials. Adjustments have to be made for the number of electrons being transfered. For example:

(eq1) Fe3+ + 3e- --> Fe(s) is listed as -0.036V
(eq2) Fe2+ + 2E- --> Fe(s) is listed as -0.44V to get a third equation:
(eq3) Fe3+ + e- --> Fe2+ (listed as +0.77V)

one would need to take eq1 and multiply the voltage by 3, reverse eq2 (changes the sign) and multiply the voltage by 2. Adding those two voltages together gives the standard potential for eq3.

Surely that is saying that the calculated electrode potential of the cell does depend on moles of electrons transferred - which contradicts the first bit I quoted.

Ideas anyone?

[DevaDevil]

in your example you are calculating the potential for another half cell reaction, which has to be based on 1 mole electrons as well as there is an electron still present.

In an overall reaction there are no electrons present as they cancel out, and the voltage is defined on mole electrons transfered. So as both half cells are defined in potential per mole electron transfered, the overall potential does not have to be adjusted, as it will automatically be set to Volts for 1 mole electrons transfered

[cliverlong]

<< snip >>

In an overall reaction there are no electrons present as they cancel out, and the voltage is defined on mole electrons transfered. So as both half cells are defined in potential per mole electron transfered, the overall potential does not have to be adjusted, as it will automatically be set to Volts for 1 mole electrons transfered

That's interesting because the textbook I am referring to, Facer Edexcel A2 Chemistry, page 168 gives the example

1/2 Cl₂(g) + e^- <-> Cl^-(aq)   E=+1.36V
Cl₂(g) + 2e^- <-> 2Cl^-(aq)     E=+1.36V

So number of moles has NO effect on E , since units are Volts NOT Volts per mole

How confusing !      


Clive