6M.1

[Ellis Song 4I]

A student was given a standard Cu(s)uCu21(aq) half-cell and another half-cell containing an unknown metal M in 1.00 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)?

The solutions manual for this problem flips the cell diagram given and says that Copper is the anode and the Metal is the cathode but how do they know this?

[Charlyn Ghoubrial 2I]

An easy way to do this problem is to use the equation Ecell=Er-El
Ecell= E(Cu^(2+)/Cu) - E(M^(2+)/M)
E(M^(2+)/M)= (-0.689+0.34)V = -0.349V

[Prasanna Padmanabham 4I]

Isn't Cu2+ being reduced to Cu? So shouldn't the textbook Standard Cell Potential be +0.34 as the textbook says? Isn't the left side of the cell diagram always the cathode (reduction half-reaction)?
I'm really confused now...

Thank you?

[Julie Park 1G]

In most of the example problems we've encountered, Ecell is positive. However, in this question, Ecell is negative!
When Ecell is positive, the cathode is on the right side and the anode is on the left side
When Ecell is negative, the anode is on the right side and the cathode is on the left side

[Althea Zhao 1B]

I'm confused as well. Why isn't the answer a positive value?

[Elizabeth Bowen 1J]

I'm confused as well. Why isn't the answer a positive value?

It's because they flipped the cell diagram and put the cathode on the left when we learned that conventionally it should be on the right, my TA told me that we shouldn't worry too much about this kind of situation because they would just write it normally on the test (we would be able to safely assume that the cathode is always on the right and the anode is on the left). But yes if I'm correct I think that the way we would know that this cell diagram is flipped is because Ecell is negative, which indicates that the cathode is on the left, I'm pretty sure.