[sinthreck]

An excess of finely pulverised iron is stirred up with a solution that contains 1 M Cu²⁺ ions, and the system is allowed to come to equilibrium. The solid materials are then filtered off and electrodes of solid copper and solid iron inserted into the remaining solution. What is the value of the ratio of [Fe²⁺]/[Cu²⁺] at 25ºC?

Eº(Fe²⁺/Fe) = -0.44 V
Eº(Cu²⁺/Cu) = +0.34 V

Eº = 0.44 + 0.34 = 0.78 V

Fe_(s) Fe²⁺ + 2e^-
Cu²⁺ + 2e^- Cu_(s)
Hence n = 2

Since system is at equilibirum (removing precipitate has no effect because afterwards they add both electrodes):
Ecell = 0 and Q = k

Eº = (.0592/n) * log K

k = 10^ [(Eº)(n)/.0592]
  = 2.2x10²⁶

The problem with my solution is that I do not make use of the initial concentration of 1M Cu2+ ions. Have I gone wrong?

[Borek]

The problem with my solution is that I do not make use of the initial concentration of 1M Cu2+ ions. Have I gone wrong?

If there was an excess of iron, does the initial concentration of copper matter?

[sinthreck]

I think without excess iron, the system could not reach a state of equilibirum.

The net equation is:

Fe_(s) + Cu²⁺  Fe²⁺ + Cu_(s)

Originally we have only Cu2⁺ ions in solutions. When we add the Fe_(s), the right hand side of the equation becomes favourable. As such, Fe_(s) dissociates into Fe²⁺ ions until the ratio of (Fe²⁺/Cu²⁺) = k (a fixed value at that temperature).
We end up with an excess of Fe_(s) and Cu_(s) in the solution.

So what would happen if we did NOT add excess pulverised iron but only a small amount.
I suppose we could not create sufficient Fe²⁺ and hence equilibirum is not attainable.

[Borek]

I think without excess iron, the system could not reach a state of equilibirum.

System would get to equilibrium - just a different one.

Fe_(s) + Cu²⁺  Fe²⁺ + Cu_(s)

OK, just remember it is Fe(s) (no subscript).

Originally we have only Cu2⁺ ions in solutions. When we add the Fe_(s), the right hand side of the equation becomes favourable. As such, Fe_(s) dissociates into Fe²⁺ ions until the ratio of (Fe²⁺/Cu²⁺) = k (a fixed value at that temperature).
We end up with an excess of Fe_(s) and Cu_(s) in the solution.

Fe(s) doesn't "dissociate" - it gets oxidized to Fe²⁺.

So what would happen if we did NOT add excess pulverised iron but only a small amount.
I suppose we could not create sufficient Fe²⁺ and hence equilibirum is not attainable.

As told earlier you will have equilibrium, but a different one - without solid iron. However, that means you will be not able to calculate copper concentration, as presence of Fe(s) and Cu(s) means you know ratio of metals concentrations.

[sinthreck]

Excellent, thanks for your help.

Dissociate It's pretty obvious what I was studying prior to this...

Next time I visit Gdansk, I'll buy you a beer.

[Borek]

Next time I visit Gdansk, I'll buy you a beer.

It will be easier if you will be going through Warsaw