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caculate electrode potential of ethanol electrode
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the emf of ethanol oxygen fuel cell is 1.00v, calculate electrode potential for theethanol electrode. the oxygen half equation is given as o2+4H+ +4e- ---> 2h20 the electrode potential is +1.23. half equation for c2h5oh +3h2o----> 2co2 + 12h+ +12e-
emf of a cell= reduction + oxidation
the oxygen electrode is undergoing reduction , its more positive so the ethanol electrode must be oxidising, I thought u would have to do 1.00 (emf) - 1.23 (oxidation) = -0.23v but the actual answer is +0.23?
why is the answer positive, this was the method I was taught from school
emf of a cell= reduction + oxidation
the oxygen electrode is undergoing reduction , its more positive so the ethanol electrode must be oxidising, I thought u would have to do 1.00 (emf) - 1.23 (oxidation) = -0.23v but the actual answer is +0.23?
why is the answer positive, this was the method I was taught from school
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#2
2CO2 + 12H+ +12e -> C2H5OH +3H2O E = ?
O2 +4H+ +4e- ---> 2H2O E = 1.23 V
Stop writing H20 for H2O. Water is 2 hydrogen atoms and 1 oxygen atom, not 20 hydrogen atoms.
Ecell = Ered - Eox
1 = 1.23 - ?
therefore ? = 0.23 V
O2 +4H+ +4e- ---> 2H2O E = 1.23 V
Stop writing H20 for H2O. Water is 2 hydrogen atoms and 1 oxygen atom, not 20 hydrogen atoms.
Ecell = Ered - Eox
1 = 1.23 - ?
therefore ? = 0.23 V
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(Original post by BobbJo)
2CO2 + 12H+ +12e -> C2H5OH +3H2O E = ?
O2 +4H+ +4e- ---> 2H2O E = 1.23 V
Stop writing H20 for H2O. Water is 2 hydrogen atoms and 1 oxygen atom, not 20 hydrogen atoms.
Ecell = Ered - Eox
1 = 1.23 - ?
therefore ? = 0.23 V
2CO2 + 12H+ +12e -> C2H5OH +3H2O E = ?
O2 +4H+ +4e- ---> 2H2O E = 1.23 V
Stop writing H20 for H2O. Water is 2 hydrogen atoms and 1 oxygen atom, not 20 hydrogen atoms.
Ecell = Ered - Eox
1 = 1.23 - ?
therefore ? = 0.23 V
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(Original post by usernamenew)
sorry about that, why is the ethanol half equation flipped?
sorry about that, why is the ethanol half equation flipped?
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#5
(Original post by usernamenew)
oh its because its getting reduced right?
oh its because its getting reduced right?
Last edited by username3249896; 2 years ago
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(Original post by BobbJo)
When using Ered - Eox, this means 'the electrode potential of the electrode at which reduction is occuring' - 'the electrode potential of the electrode at which oxidation is occurring'. Electrode potentials are taken as electrode reduction potentials. Hence the way the half-equations are written is in the direction of a reduction process.
When using Ered - Eox, this means 'the electrode potential of the electrode at which reduction is occuring' - 'the electrode potential of the electrode at which oxidation is occurring'. Electrode potentials are taken as electrode reduction potentials. Hence the way the half-equations are written is in the direction of a reduction process.
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#7
(Original post by usernamenew)
so when using ered-eox, u write both equations as reduction?
so when using ered-eox, u write both equations as reduction?
O2 +4H+ +4e- ---> 2H2O E = 1.23 V
Ecell = Ered - Eox
1 = 1.23 - ?
therefore ? = 0.23 V
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