## Chapter 14, Problem 14.19

Raul Hernandez
Posts: 28
Joined: Fri Jun 17, 2016 11:28 am

### Chapter 14, Problem 14.19

Hello, I have a question regarding the answer to the problem.
"A student was given a standard Cu(s)|Cu2(aq) halfcell
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?"

In this case the Cu(s)|Cu2 is the anode (because it's being oxidized). The value for the Reduction half reaction for Cu2 + 2e- --> Cu is +0.34 V, but this is for the reduction not the oxidation. This means that our value for the copper oxidation should be "-0.34V." However, when looking at the answer and the process in the solution manual, they use the POSITIVE value (which should be for the reduction) which gives a totally different answer.

The answer says that the metal's potential should be "-0.349" but I think this is wrong. Anyone care to tell me why they used the positive value for the copper instead of the negative one? Any help will be highly appreciated.

Maggie Bui 1H
Posts: 35
Joined: Fri Jul 22, 2016 3:00 am
Been upvoted: 1 time

### Re: Chapter 14, Problem 14.19

The solutions manual uses the positive reduction value for the anode, but notice that this positive reduction potential is being subtracted from the reduction potential of the cathode.

We learned to reverse the sign of the reduction potential of the anode and then add the two half-reaction potentials together, but it is also valid to use the positive reduction potentials and subtract the reduction potential of the anode from that of the cathode. These two methods of calculating total standard cell potential are equivalent.

1) $-E^{o}_{anode} + E^{o}_{cathode} = E^{o}$
2) $E^{o}= E^{o}_{cathode} - (E^{o}_{anode})$

Alexandria_Leaf_2F
Posts: 30
Joined: Fri Jul 15, 2016 3:00 am

### 14.13

The reaction in 14.13 was as follows:

Ce^4+(aq) + I^- (aq) ---> I2 (s) + Ce^3+(aq)

I understand that in the final cell Diagram, there needs to be Pt(s) by the cerium because both the oxidized and reduced species are in the same solution (aq). But, in the solutions manual, there was a Pt(s) by the Iodine too as follows:

Pt(s)| I^-(aq) | I2(s) || Ce^4+(aq), Ce^3+(aq) | Pt(s)

Why is that? Iodine is going from (aq) to (s) so the oxidized and reduced versions aren't in the same solution. Shouldn't just it be:

I^-(aq) | I2(s) || Ce^4+(aq), Ce^3+(aq) | Pt(s)

Plus, in the course reader, there was an example given between iron and copper as follows:

2Fe^3+(aq) + Cu(s) ---> Cu^2+(aq) + 2Fe^2+(aq)

and when the cell diagram was drawn out:

Cu(s) | Cu^2+(aq) || Fe^3+(aq), Fe^2+(aq)| Pt(s)

So, here, Pt was only on the side of the cell diagram where the oxidized and reduced species are in the same solution.

Raul Hernandez
Posts: 28
Joined: Fri Jun 17, 2016 11:28 am

### Re: Chapter 14, Problem 14.19

Maggie Bui 1H wrote:The solutions manual uses the positive reduction value for the anode, but notice that this positive reduction potential is being subtracted from the reduction potential of the cathode.

We learned to reverse the sign of the reduction potential of the anode and then add the two half-reaction potentials together, but it is also valid to use the positive reduction potentials and subtract the reduction potential of the anode from that of the cathode. These two methods of calculating total standard cell potential are equivalent.

1) $-E^{o}_{anode} + E^{o}_{cathode} = E^{o}$
2) $E^{o}= E^{o}_{cathode} - (E^{o}_{anode})$

Oh, ok! That makes sense. Sorry, the way that it was done in the solution manual was very confusing without an explanation as to why it was done that way. Thank you!