6M.1

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Ellis Song 4I
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Joined: Thu Jul 11, 2019 12:17 am

6M.1

Postby Ellis Song 4I » Mon Mar 02, 2020 7:26 pm

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
Posts: 50
Joined: Wed Nov 13, 2019 12:26 am

Re: 6M.1

Postby Charlyn Ghoubrial 2I » Mon Mar 02, 2020 7:35 pm

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
Posts: 111
Joined: Thu Jul 25, 2019 12:17 am

Re: 6M.1

Postby Prasanna Padmanabham 4I » Tue Mar 03, 2020 8:30 pm

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
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

Re: 6M.1

Postby Julie Park 1G » Wed Mar 04, 2020 11:06 pm

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
Posts: 101
Joined: Mon Jun 17, 2019 7:24 am

Re: 6M.1

Postby Althea Zhao 1B » Thu Mar 05, 2020 1:21 am

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

Elizabeth Bowen 1J
Posts: 53
Joined: Wed Nov 14, 2018 12:20 am

Re: 6M.1

Postby Elizabeth Bowen 1J » Thu Mar 05, 2020 2:29 am

Althea Zhao 1B wrote: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.


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