[BaO]

 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: E⁰_cell= E⁰_red - E⁰_ox, the answer is always negative.

thank you

[Dan]

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, Cr₂O₇^2- is dichromate ion.

[sdekivit]

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:

E⁰_cell = E⁰_Ox - E⁰_Red

[BaO]

The equation for the E0 should actually be:

E0cell = E0Ox - E0Red

really? but my book shows different

[BaO]

--> 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 ?

[sdekivit]

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