E cell


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melissa_dis4K
Posts: 106
Joined: Fri Sep 28, 2018 12:28 am

E cell

Postby melissa_dis4K » Wed Feb 27, 2019 11:14 pm

For part b, n is equal to 6 but I do not understand why. I thought n was how many electrons were transferred and here we have a charge of 24+ and 24+ on both sides so why isn't n=0? Also, am I doing the charge counts wrong? Thank you!
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Jessica Tsui 1H
Posts: 30
Joined: Fri Sep 28, 2018 12:21 am

Re: E cell

Postby Jessica Tsui 1H » Wed Feb 27, 2019 11:20 pm

n is the transfer of electrons for the chemicals being reduced and oxidized. In this case, n is equal to 6 because Fe loses 6 electrons when oxidized and Cr gains 6 electrons when reduced after the reaction occurs.

melissa_dis4K
Posts: 106
Joined: Fri Sep 28, 2018 12:28 am

Re: E cell

Postby melissa_dis4K » Wed Feb 27, 2019 11:21 pm

Thank you!!! :) That makes sense now I was looking at it wrong.

melissa_dis4K
Posts: 106
Joined: Fri Sep 28, 2018 12:28 am

Re: E cell

Postby melissa_dis4K » Wed Feb 27, 2019 11:26 pm

I can see clearly from the given equation that Fe loses 6 e- but I do not see how Cr gains the 6 e-. Is this seen after the reaction occurs?

Christopher Tran 1J
Posts: 77
Joined: Fri Sep 28, 2018 12:15 am

Re: E cell

Postby Christopher Tran 1J » Wed Feb 27, 2019 11:42 pm

To find n, you can write out the two half-reactions for the cell reaction:

Reduction: 6 Fe3+ (aq) + 6 e- -> 6 Fe2+ (aq)
To balance the charges on both sides, 6 e- is added to the left side.

Oxidation: 2 Cr3+ (aq) + 7 H2O (l) -> Cr2O72- (aq) + 14 H+
The left side has a total charge of 6+ and the right side has a total charge of 12+.
Therefore, to balance the charges on both sides, 6 e- is added to the right side.

2 Cr3+ (aq) + 7 H2O (l) -> Cr2O72- (aq) + 14 H+ + 6 e-

Since the number of electrons for both reduction and oxidation are the same, the number of electrons transferred is 6.
(If they were not the same, you would balance out the half reactions first so that both reactions have the same number of electrons)


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