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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: 21385 on May 24, 2008, 01:50:58 PM

Title: Redox question- urgent
Post by: 21385 on May 24, 2008, 01:50:58 PM
Calculate the ratio of the cumulative stability constants for the formation of [Fe(CN)6](3-) and [Fe(CN)6](4-) ions.

Fe3+ + 1e- => Fe2+ E3=0.772 V
[Fe(CN)6]3- + 1e- => [Fe(CN)6]2- E4=0.356 V

Here is the solution:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi32.tinypic.com%2Fe9t5l0.jpg&hash=022749ea01fa991ed59b0d44d39798bc7d645e40)

What I don't understand is "E4=E3 + 0.059log(formation(Fe(II))/formation(Fe(III))). I understand the Nernst equation but I don't understand how it can just equate E4 with E3; they're different reactions.

Can somebody please help me? This is urgent.

Thanks a million
Title: Re: Redox question- urgent
Post by: Borek on May 24, 2008, 02:28:37 PM
Not checked, but general guidelines that I will follow doing it will be:

Write Nernst equation for the first reaction, then solve both formation formulas for Fe(X) concentration and put these concentrations into the first Nernst equation. Now compare it with Nernst equation for the second reaction. Some expressions must be identical.

Could be I am wrong.
Title: Re: Redox question- urgent
Post by: Hunt on July 17, 2008, 04:38:29 PM
... or you can use thermochemistry.

Delta Go = - RT Ln K
Delta Go  = - nFEo

Remember that G is a state function.

Fe3+ + 1e- => Fe2+    delta Go = - FE3o

Fe3+ + 6 CN- => Fe(CN)63-    delta Go = -RT Ln Box

Fe2+ + 6 CN- => Fe(CN)64-    delta Go = -RT Ln Bred

The target reaction :

Fe(CN)63- + e- => Fe(CN)64-    delta G = - FE4o

Then it's easy to show that :

- FE4o = - FEo - RT Ln Bred + RT Ln Box

E4o = E3o + RT / F Ln (Bred / Box )