deltaH
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Half reaction you want is not listed, but...
Often an electrochemical half reaction you require is not readily available, but it looks like you can construct it by a combination of other half
reactions that are in order to determine the correct half reaction potential.
I will be using wiki's table of standard electrode potentials
I want to determine the standard half reaction potential for the reduction of bromate to bromide ions in basic aqueous media:
BrO3-(aq) + 3H2O(l) + 6e- <=> Br-(aq) + 6OH-(aq) E = ??
Okay, a quick look through the table shows that this reaction is not listed, however, you can find the following that are:
(i) 2 BrO3-(aq) + 12 H+(aq) + 10 e- <=> Br2(l) + 6 H2O(l) E = +1.48V
(ii) Br2(l) + 2 e- <=> 2 Br-(aq) E = +1.066V
(iii) 2H2O(l) + 2 e- <=> H2(g) + 2 OH-(aq) E = −0.8277V
(iv) H2(g) <=> 2 H+(aq) + 2 e- E = 0.0000V
EDIT: Now, as pointed out by DraconicAcid:
dG = nFE which means that you need to multiple by the number of electrons when you add reactions' E and then divide by the equation's new number of
electrons.
So we can add (i) + (ii) and divide the result by 2 to get:
(v) BrO3-(aq) + 6H+(aq) + 6e- <=> Br-(aq) + 3H2O(l)
E = (1.48V*10+1.066*2)/12 = +1.411V
Note, I divided by 12 electrons, not six, because the addition yields a 12 electron system. After this I divide straight through by 2 to simplify the
equation, but this doesn't affect E.
By summing 3*(iii) and (v):
(vi) BrO3-(aq) + 3H2O + 6H+ + 12e- <=> Br-(aq) + 3H2(g) + 6OH-
E = (1.411V*6-0.8277V*6)/12 = 0.29165V
Finally (vi) + 3(iv) yields:
BrO3-(aq) + 3H2O(l) + 6e- <=> 2Br-(aq) + 6OH-(aq)
E = (0.29165V*12 + 0V)/6 = 0.5833V
So final E = +0.583V
/EDIT
Is my thing correct? Thanks
[Edited on 17-10-2013 by deltaH]
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DraconicAcid
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The E values are not additive; deltaG is additive. Since deltaG = -nFE, you have to multiply the potentials by the number of electrons before you
add, and then divide by the number of electrons when you're done.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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deltaH
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| Quote: | | The E values are not additive; deltaG is additive. Since deltaG = -nFE, you have to multiply the potentials by the number of electrons before you add,
and then divide by the number of electrons when you're done. |
Thanks for that DraconicAcid
I have made the changes as you suggested, have a look at the edit on the origional post, is this now correct?
Thanks for your help again, much appreciated!
[Edited on 17-10-2013 by deltaH]
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deltaH
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In my potassium ferrate(VI) synthesis thread, I use the previously determined half reaction for bromate as well as the half potential for oxidising
Fe3O4 (magnetite) to FeO4(2-) in basic media. For completeness and as a second example, I will include the calculation by summation of the latter here
as well:
Half reaction required: Fe3O4(s) + 16OH- <=> 3FeO4(2-) + 8H2O + 10e- E = ??
I can look up the following half reactions from wiki's table of standard electrode potentials:
(i) Fe3O4(s) + 8H+ +8e- <=> 3Fe(s) + 4H2O E = +0.085
(ii) Fe(s) <=> Fe3+ + 3e- E = +0.04V
(iii) 3Fe3+ + 12H2O <=> 3FeO4(2-) + 9e- + 24H+ E = -2.2V
(iv) 2H2O(l) + 2 e- <=> H2(g) + 2 OH-(aq) E = -0.8277V
(v) H2(g) <=> 2 H+(aq) + 2 e- E = 0.0000V
(i) + 3*(ii) = (vi):
(vi) Fe3O4(s) + 8H+ <=> 3Fe3+ + e- + 4H2O
E = (0.085V*8 + 0.04V*9)/1=+1.04V
(vi) + (iii) = (vii):
(vii) Fe3O4(s) + 8H2O <=> 3FeO4(2-) + 16H+ + 10e-
E = (1.04V*1 -2.2V*9)/10 = -2.096V
(vii) + (8*(iv) turned around) = (viii)
(viii) Fe3O4(s) + 8H2 + 16OH- <=> 3FeO4(2-) + 16H+ + 8H2O + 26e-
E = (-2.096V*10 + 0.8277V*16)/26 = -0.297V
(viii) + 8*(v) = ANSWER:
ANSWER: Fe3O4(s) + 16OH- <=> 3FeO4(2-) + 8H2O + 10e-
E = (-0.297V*26 + 0V)/10 = -0.772V
[Edited on 18-10-2013 by deltaH]
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The_Davster
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Also, check the BIG list of half reactions:
http://www.efunda.com/materials/corrosion/electrochem_list.c...
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deltaH
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Thanks The_Davster
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