[plu]

An excess of finely divided iron is stirred up with a solution that contains Cu2+ ions. The system is
allowed to come to equilibrium. The solid materials are filtered off and electrodes of solid copper and
solid iron are inserted into the remaining solution. What is the ratio [Fe2+]/[Cu2+] at 25°C?
(a) 2 (b) 1 (c) 2.2 x 1026 (d) 4.2 x 10-27 (e) none of these

I am at a complete loss with this problem.  Any help would be appreciated

[Borek]

Two half reactions, two Nernst equations, two potentials - they must be equal. Follow this line and its a piece of cake

[plu]

I'm sorry but I don't quite follow.  How did you arrive at those steps?  

[Borek]

I just know  

Seriously - that's the general idea behind all electrochemistry equilibriums - you have redox systems in the solution, at equilibrium all potentials of all system must be identical. So you have to write down Nernst equation for every equilibrium (Nernst equation) and solve set of equations assuming all potentials are identical.

Full set of equations must contain also all mass balances and charge balances, but often you don;t have to worry about these.

[plu]

Hm.  What would be the significance of the solid copper and solid iron electrodes then?  And would the removal of the solid precipitates have any effect on the calculations?

[Borek]

Hm.  What would be the significance of the solid copper and solid iron electrodes then?  And would the removal of the solid precipitates have any effect on the calculations?

Removing precipitates and then putting iron and copper solid electrodes into the solution doesn't change anything. Before and after you have the same solution in equilibrium with the same solids.

Note, that both halfreactions contain ion and solid - and theat activity of oslid is assumed to equal 1, thus is removed from Nernst equation.

[plu]

Ah, I see.  So at equilibrium, the potentials of the two half cells in a reduction-oxidation reaction are always equal.  As a result, the potential of the entire cell (which equals the difference between the potentials of the two half cells) is 0, meaning deltaG is also 0.  Thanks, mate!