Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: BaO on July 25, 2006, 09:38:37 PM
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Ethanol, C2H5OH, is oxidized by an acidified dichromate solution according to the equation shown. The E° for the reaction above is 2.98 V. The E° for the oxidation of ethanol is
A 0.52 V
B 2.98 V
C 1.75 V
D 1.23 V
this is what i'm not sure about:
dichromate is Cr+2,so its E0 value is -0.41? but with that value i cant find the right answer, because when i use this fomular: E0cell= E0red - E0ox, the answer is always negative.
thank you
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Ethanol, C2H5OH, is oxidized by an acidified dichromate solution according to the equation shown.
what equation? you will need to use it.
this is what i'm not sure about: dichromate is Cr+2
No, Cr2O72- is dichromate ion.
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it's a strange question since the standard electrode potential of ethanol should be negative otherwise you need electrolysis to let the reaction happening.
--> it should be -1,75 V
The equation for the E0 should actually be:
E0cell = E0Ox - E0Red
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The equation for the E0 should actually be:
E0cell = E0Ox - E0Red
really? but my book shows different
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--> it should be -1,75 V
you got -1.75. then, i just wonder how much E0(C2H5OH) is . in the chart , i only found this: Cr2O7-2 +14 H+ + 6e- ->2Cr3+ +7 H2O.....Eo=+1.23. is that what you got too ?
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yes, but when E0 of ethanol as a reductor (it gets oxidized) is higher (here +/- 0,5 V), the reaction won't take place and a voltage is needed to let take place this reaction (the difference between the two standard electrode potentials).
in a table of standard electrode potentials, for a reaction to occur spontaneously, the standard electrode potential should be higher then that of the reductor. This can be shown with the laws of thermodynamics :)