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6M.1

Posted: Mon Feb 24, 2020 4:25 pm
by JamieVu_2C
A student was given a standard Cu(s)|Cu2+(aq) half-cell and another half-cell containing an unknown metal M in 1.00M M(NO3)2(aq) and formed the cell M(s)|M+(aq)||Cu2+(aq)|Cu(s). The cell potential was found to be -0.689 V. What is the value of E(M2+/M)?

In this problem, since Cu2+(aq)/Cu(s) is written on the right side of the cell diagram, it should be the cathode. So, Cu2+(aq)/Cu(s) should be written as a reduction half reaction. However, the textbook solutions says that it is the anode and is instead written as an oxidation half reaction. Why is this? If the Cu2+(aq)/Cu(s) electrode is on the right of the cell diagram, then shouldn't it be the cathode?

Re: 6M.1

Posted: Mon Feb 24, 2020 5:11 pm
by Brian Tangsombatvisit 1C
You're right, if the Cu half reaction is written on the right side of the equation, it should be the cathode. This is probably a typo in the textbook as the solutions manual writes it on the left side of the cell diagram, making it the anode.

Re: 6M.1

Posted: Tue Feb 25, 2020 10:57 am
by Jessica Esparza 2H
Has anyone confirmed if this is a typo? I was confused when looking at the answer key.

Re: 6M.1

Posted: Tue Feb 25, 2020 4:12 pm
by Alexis Webb 2B
I am not sure if it is a typo, but I was wondering what answer would you get if the Cu2+/Cu is the cathode so that I can check mine.

Re: 6M.1

Posted: Tue Feb 25, 2020 4:29 pm
by Daniel Honeychurch1C
In 6 L.3 of the textbook, it says that a positive cell potential indicates that the right side of the cell diagram is the cathode. Conversely, a negative cell potential indicates that the left side of the cell diagram in the cathode. In this question, the cell potential is negative (-0.689 V), so the left side of the cell diagram (M 2+/M) is the cathode.