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Posted: Wed Feb 21, 2018 10:58 pm
by Lucia H 2L
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?

Re: 14.19

Posted: Wed Feb 21, 2018 11:46 pm
by Thu Uyen Tran 1B
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.

Re: 14.19

Posted: Thu Feb 22, 2018 12:33 am
by Minie 1G
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.