Midterm 2013 #7

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904564128
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Joined: Fri Sep 25, 2015 3:00 am

Midterm 2013 #7

Postby 904564128 » Tue Feb 09, 2016 5:26 pm

The first part of this question that requires you to calculate the cell potential does not make sense to me. The cell potential of the cathode (the half equation involving Br2) is positive, which makes sense, however the cell potential for the anode (Mn2+ equation) should have the sign of its given cell potential reversed (from +1.49V to -1.49V) since the equation is reversed. However, when the final cell potential is calculated, the equation is not E = (+1.07V) - (-1.49V) = +2.56V. Is the solution incorrect or have I misunderstood how to solve for E?

navessor
Posts: 16
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm 2013 #7

Postby navessor » Tue Feb 09, 2016 9:29 pm

Because the reaction is already reversed in the "given," it is unnecessary to use the equation E = Cathode - Anode. You can instead just add the two equations together to get Ecell.

904564128
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm 2013 #7

Postby 904564128 » Tue Feb 09, 2016 9:56 pm

Then when is the equation ever used?

rebeccawaggoner 1H
Posts: 29
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm 2013 #7

Postby rebeccawaggoner 1H » Tue Feb 09, 2016 10:27 pm

If you use the equation Ecathode - Eanode, you dont flip the sign of the anode, you leave it positive. If you flip the sign of the anode, then you should add the two cell potentials.


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