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Two_Fishes science forum beginner
Joined: 17 Jan 2006
Posts: 24

Posted: Mon Apr 17, 2006 3:55 am Post subject:
Differing electrode potentials in tables



Could somebody explain to me why the CRC Handbook shows different
standard electrode potentials for halfreactions done stepwise and all
at once? For example, chlorate to chloride (alkaline) is given as 0.6v,
approximately. Chlorate to chlorite to hypochlorite to chloride adds up
to 1.8 volts. Another example, Fe+3 to Fe is 0.3v direct, but Fe+3 to
Fe+2 is +0.77v and Fe+2 to Fe is 0.44, which totals +0.33v. Surely
both go through the Fe+2 stage. 

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Lasse Murtomäki science forum beginner
Joined: 17 Jun 2005
Posts: 35

Posted: Tue Apr 18, 2006 1:55 pm Post subject:
Re: Differing electrode potentials in tables



Look at the stoichiometry. For the reaction
Fe(3+) + 3 e = Fe
we can write
E_1 = E_1^0 + RT/(3F)*log[Fe(3+)]
Accordingly:
Fe(3+) + e = Fe(2+)
E_2 = E_2^0 + RT/F*log{[Fe(3+)]/[Fe(2+)]}
Fe(2+) + 2 e = Fe
E_3 = E_3^0 + RT/(2F)*log[Fe(2+)]
If you sum two latter reactions:
E_2 + E_3 = E_2^0 + E_3^0 + RT/F*log{[Fe(3+)]/[Fe(2+)]^½} which is not
equal to E1.
But 1/3*(E_2 + 2*E_3) = E_1
1/3*(0.77  2*0.44) = 0.036666666...
You have a misprint: for the first reaction E_1^0 = 0.036 V, I checked CRC.
Thanks for a very interesting question! 

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Oscar Lanzi III science forum Guru Wannabe
Joined: 30 Apr 2005
Posts: 176

Posted: Wed Apr 19, 2006 1:37 am Post subject:
Re: Differing electrode potentials in tables



You don't add up voltages. You add up free energies. The free energy
associated with a halfreaction is the voltage times the charge
transferred, with the sign reversed (thus positive voltage > negative
free energy.) The charge transferred is proportional to the number of
electrons you see in the reaction. Thus Fe(2+) + 2e() = Fe involves
twice as much charge transfer as Fe(3+) + 1e() = Fe(2+). The constant
of proportionality in this relation is the accumulated charge of one
mole of electrons, called Faraday's Constant and usually labeled F and
evaluated as 96,487 Coulombs. Thus our product relation fro free energy
is usually written
deltaG = n*F*E.
So if we are given voltages for the above two halfreactions and want to
compute the overall voltage for Fe(3+) + 3e() = Fe, we multiply the
Fe(2+)/Fe reaction by 2F and the Fe(3+)/Fe(2+) reaction by just F to get
the right free energy terms to add up. Then we see that the overall
free energy corresponds to a threeelectron transfer, so we have to
divide by 3F to get back to a voltage. What we ultimately get is a
weighted average of voltages instead of just a sum.
OL 

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Two_Fishes science forum beginner
Joined: 17 Jan 2006
Posts: 24

Posted: Wed Apr 19, 2006 2:56 pm Post subject:
Re: Differing electrode potentials in tables



Thanks for the explanations! So you take the average, weighted
by the number of electrons, to get the halfcell voltage. Then
you add the two halfcells together to get the cell voltage. Then
you multiply the voltage by Faraday's constant and divide by the
molecular weight per mole of electrons to get the theoretical
energy density of an electromotive cell (which is what I was
trying to compute.)
Boron  Air (acidic)
B + 3H2O = H3BO3 + 3H+ + 3e 2.5 v 3.6 g/e
O2 + 4H+ + 4e = 4H2O 1.229v 8 g/e
3.729v, 8615 Whr/kg
Borohydride  Air (alkaline)
BH4 + 8OH = H2BO3 + 5H2O + 8e 1.24v 4.75 g/e
O2 + 2H2O + 4e = 4OH 0.401v 8 g/e
1.641 v, 3449 Whr/kg
I'll see if I can get these to work. Would be nice, wouldn't it? 

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Two_Fishes science forum beginner
Joined: 17 Jan 2006
Posts: 24

Posted: Thu Apr 20, 2006 6:58 pm Post subject:
Re: Differing electrode potentials in tables



Now it seems I can't even read the Handbook correctly. It is pretty
small type. My apologies to anybody who was offended by the wrong
figures.
Boron  Air (acidic)
B + 3H2O = H3BO3 + 3H+ + 3e 0.73 v 3.6 g/e
O2 + 4H+ + 4e = 4H2O 1.229v 8 g/e
1.959v, 4526 Whr/kg
Boron  Air (alkaline)
B + 4OH = H2BO3 + H2O + 3e 2.5 v 3.6 g/e
O2 + 2H2O + 4e = 4OH 0.401v 8 g/e
2.901v, 6702 Whr/kg
Boron  Sodium Chlorate
B + 4OH = H2BO3 + H2O + 3e 2.5 v 3.6 g/e
ClO3 + 3H2O + 6e = Cl + 6OH 0.62 v 17.74 g/e
3.12v, 3918 Whr/kg
The discharged state is Sodium Borate and Sodium Chloride; if the
cell can be recharged from those materials it would be inexpensive. 

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