14.1

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Palmquist_Sierra_2N
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14.1

Postby Palmquist_Sierra_2N » Thu Feb 02, 2017 9:42 pm

H+(aq) + Cr2o7(2-)(aq) + C2H5OH(aq)-->Cr(3+)(aq) + C2H4O(aq) + H2O(l)
How do you know the C is being oxidized?

Wendy_Liu_3A
Posts: 27
Joined: Wed Sep 21, 2016 2:55 pm

Re: 14.1

Postby Wendy_Liu_3A » Thu Feb 02, 2017 10:05 pm

C only appears in 2 compounds in the reaction: C2H5OH the reactant and C2H4O the product.
First calculate the charge of C in C2H5OH. The charges of H (+1) and O (-2) will be constant, and the overall charge of C2H5OH is 0. C*2 +(1*6)-(2) = 0, C = -2.
Then calculate the charge of C in C2H4O. Again, the charges of H (+1) and O (-2) will be constant, and the overall charge of C2H4O is 0. C*2 +(1*4)-(2) = 0, C = -1.

Since the charge of C increases from reactant (-2) to product(-1), we know that C has lost an electron, so C is oxidized.


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