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Standard Reduction potential of Fe

  • #1
Krushnaraj Pandya
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Homework Statement


If E°(Fe2+/Fe) is x1 and E°(Fe3+/Fe) is x2 then find E°(Fe3+/Fe2+)

2. The attempt at a solution
We need E for Fe3+ + e ----> Fe2+.
We are given Fe2+ + 2e----->Fe is x1. Reversing will give Fe--->Fe2+ + 2e (-x1)...(i)
also, Fe3+ + 3e----->Fe is x2, ....(ii)
multiply (ii) by 2 and (i) by 3, then adding we have
Fe + 2Fe3+---->3Fe2+ as 2x2-3x1. What now?
 

Answers and Replies

  • #2
mjc123
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Why do you reverse the second equation? Add equation 1 to equation 2 and what do you get?
PS Are you familiar with oxidation state diagrams (also called Frost diagrams)? Drawing one for Fe may help.
 
  • #3
Krushnaraj Pandya
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Why do you reverse the second equation? Add equation 1 to equation 2 and what do you get?
PS Are you familiar with oxidation state diagrams (also called Frost diagrams)? Drawing one for Fe may help.
I am not acquainted with a frost diagram. If it isn't too complex, I'd like to know about them
 
  • #4
Krushnaraj Pandya
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Why do you reverse the second equation? Add equation 1 to equation 2 and what do you get?
PS Are you familiar with oxidation state diagrams (also called Frost diagrams)? Drawing one for Fe may help.
Alright, so adding both reactions gives us the desired reaction. But how do we obtain the result that the sum of their free energies is equal to the free energy of the reaction obtained by adding them? Also, why can't we simply add the E values as well like we do for a cell by summing cathode and anode reaction E values
 
  • #5
Krushnaraj Pandya
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Note that we added E in the NO alkaline medium question as well
 
  • #6
mjc123
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But how do we obtain the result that the sum of their free energies is equal to the free energy of the reaction obtained by adding them?
Have you heard of Hess's law?
G is a state function, so the difference in G between two states is independent of the path taken between them. So it doesn't matter whether you go from Fe3+ to Fe directly, or via Fe2+ as an intermediate - the free energy change is the same.
Also, why can't we simply add the E values as well like we do for a cell by summing cathode and anode reaction E values
Free energies are additive. Potentials are not. They are related by ΔG = -nFE. So
ΔG3 = ΔG1 + ΔG2
implies 3E3 = E1 + 2E2
A Frost diagram would show this clearly (google "Frost diagram"). It is a graph with oxidation number as the x axis and free energy as the y axis. The difference in free energy (y-coordinate) between two states is nFE; the slope of the line joining them is FE (where E is the reduction potential; note positive progression along the x-axis means oxidation, so a positive slope means positive ΔGox, i.e. negative Eox, i.e. positive Ered.)
 
  • #7
i think
F2(g) + 2e- -> 2F-(aq)
 
  • #8
mjc123
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And how is that relevant to the electrochemistry of Fe?
 
  • #9
Krushnaraj Pandya
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Have you heard of Hess's law?
G is a state function, so the difference in G between two states is independent of the path taken between them. So it doesn't matter whether you go from Fe3+ to Fe directly, or via Fe2+ as an intermediate - the free energy change is the same.

Free energies are additive. Potentials are not. They are related by ΔG = -nFE. So
ΔG3 = ΔG1 + ΔG2
implies 3E3 = E1 + 2E2
A Frost diagram would show this clearly (google "Frost diagram"). It is a graph with oxidation number as the x axis and free energy as the y axis. The difference in free energy (y-coordinate) between two states is nFE; the slope of the line joining them is FE (where E is the reduction potential; note positive progression along the x-axis means oxidation, so a positive slope means positive ΔGox, i.e. negative Eox, i.e. positive Ered.)
I thought Hess's Law was for Enthalpies only
 
  • #10
mjc123
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Strictly speaking, perhaps, but the principle applies for all state functions.
 
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  • #11
Krushnaraj Pandya
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Strictly speaking, perhaps, but the principle applies for all state functions.
Alright, so E is not a state function. But G=-nFE, and G is a state function.
Reversing the equation reverses sign of E and G as well, but addition, subtraction or multiplication of reaction can be done only with G? or both?
 
  • #12
Krushnaraj Pandya
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We did add E in some equations directly. For example cathode+anode reactions for a cell
 
  • #13
mjc123
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Consider what happens in a cell. Electrons consumed at the cathode are balanced by electrons released at the anode. Consider the example
Cathode: A2+ + 2e → A ΔG1 = -2FE1
Anode: 2B → 2B+ + 2e ΔG2 = 2FE2 (where E2 is the reduction potential of B+/B)
Overall: A2+ + 2B → A + 2B+ ΔG = -2FE = ΔG1 + ΔG2 = -2FE1 + 2FE2
So E = E1 - E2. The factor -nF always cancels out when the equation is balanced.
Note that when you combine a 2-electron reduction with a 2-electron oxidation, you get a 2-electron redox reaction - the electrons produced in the oxidation are consumed in the reduction (not the exact same ones, but you know what I mean, they balance). The redox reaction consists of the oxidation and reduction half-reactions happening simultaneously.
This is not the case with sequential reactions such as Fe3+ → Fe2+ → Fe. If the first step consumes 1 electron and the second step 2 electrons, the overall reaction consumes 3 electrons. You can't cancel nF. Even if n was the same in the two steps, it would be 2n for the overall reaction, leading to 2E = E1 + E2.
What would we get if we added (taking sign into account) the two potentials? Well, suppose we have
Cathode: 2Fe3+ + 6e → 2Fe ΔG1 = -6Fx2
Anode: 3Fe → 3Fe2+ + 6e ΔG2 = 6Fx1
Overall: 2Fe3+ + Fe → 3Fe2+ ΔG = -6FE = ΔG1 + ΔG2 = -6Fx2 + 6Fx1 E = x2 - x1
This is the potential of the cell Fe I Fe2+ II Fe3+ I Fe. But note that it doesn't help answer your original question.
 

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