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### Topic: Redox question- urgent  (Read 5945 times)

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#### 21385

• Regular Member
•   • Posts: 55
• Mole Snacks: +2/-1 ##### Redox question- urgent
« 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: 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.

Thanks a million

#### Borek ##### Re: Redox question- urgent
« Reply #1 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Hunt ##### Re: Redox question- urgent
« Reply #2 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 )