## 6M.1

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

### 6M.1

A student was given a standard Cu(s)|Cu2+(aq)Cu(s)|Cu2+(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)?

Can someone explain this one to me? I think I am just confused about the signs.

205007651
Posts: 53
Joined: Tue Nov 13, 2018 12:17 am

### Re: 6M.1

The textbook just does it using the positive reduction value for the anode instead of the one we went over in class. The two equations are ultimately the same, but it just gets confusing with the signs.
E= -E(anode) + E(cathode)
E= E(cathode) - (E(anode))

Megan Vu 1J
Posts: 101
Joined: Thu Jul 25, 2019 12:15 am

### Re: 6M.1

Make sure to start with the equation of E cell = E cathode - E anode.
Once you put it in the equation, you can solve for the remaining or missing variable.
-0.689 V = -1.038 V - E anode
E anode = -0.349 V.

The negative signs can be confusing, but as long as you plug it in for specific variables and solve, you should be able to get the answer.

kausalya_1k
Posts: 50
Joined: Wed Nov 14, 2018 12:23 am

### Re: 6M.1

Megan Vu 1J wrote:Make sure to start with the equation of E cell = E cathode - E anode.
Once you put it in the equation, you can solve for the remaining or missing variable.
-0.689 V = -1.038 V - E anode
E anode = -0.349 V.

The negative signs can be confusing, but as long as you plug it in for specific variables and solve, you should be able to get the answer.

Where did you get the -1.038V value for the cathode?

Connie Chen 1E
Posts: 51
Joined: Mon Jun 17, 2019 7:24 am

### Re: 6M.1

As mentioned before, Ecell=Ecathode-Eanode.
But since the voltage is negative, we know that copper is actually supposed to be oxidized; it was placed on the wrong side. Since E standard for copper is +0.34 V and E cell is -0.689 V, -0.689 V= Ecathode - 0.34 V. We solve and get -0.349 V.