14.19

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Lucia H 2L
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Joined: Mon Nov 14, 2016 3:00 am
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14.19

Postby Lucia H 2L » Wed Feb 21, 2018 10:58 pm

A student was given a standard Cu(s) 0 Cu (aq) half-cell and another half-cell containing an unknown metal M immersed in 1.00 m M(NO3)2(aq). When the copper was connected as the anode at 25 C, the cell potential was found to be -0.689 V. What is the reduction potential for the unknown M2 /M couple?

Since copper is the anode, then Cu(s) --> Cu2+(aq) + 2e-
And the reduction potential is .34
So shouldn't the cell potential be -.689(cathode) - .34 (anode) = -1.029?
The answer is -.698+ .34 = -.349

why is this?

Thu Uyen Tran 1B
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

Re: 14.19

Postby Thu Uyen Tran 1B » Wed Feb 21, 2018 11:46 pm

E°= E°(cathode) - E°(anode). The cell potential given (-0.689 V) is the E°, and you have to look for the E°(anode),0.34 V, which is the standard potential of the copper couplet. Solve for the E°(cathode). It seems that you were attempting to plug in the numbers to solve for the cell potential which rather than the reduction potential.

Minie 1G
Posts: 63
Joined: Fri Sep 29, 2017 7:04 am

Re: 14.19

Postby Minie 1G » Thu Feb 22, 2018 12:33 am

Yeah, what's given to us is the cell potential. The question is asking for the M2/M couple, which is the cathode. So rearranging the potential = cathode-anode will give us the correct answer, -0.689+0.34=-0.349 V.


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