CHEMISTRY COMMUNITY

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?

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.

Has anyone confirmed if this is a typo? I was confused when looking at the answer key.

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.

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.