[migueldario]

suppose that I have the following electrode potentials

(Cu2+)  + 2e-    --------> Cu           E = 0337

(CU1+) + 1e-    ------------> Cu       E = 0521

and I want to find
 
(Cu2+) + e-   ----------> (CU1+)  E =?

I can multiply the first potential by two and subtract the second and get the correct value

2x 0.337 to 0.521 = 0.153

now my question is because this is so?

I've seen several similar load and I understand the procedure, but i don`t know the reason

I am told that has to do with "the theory of electron volt" but have not found anything about it

appreciate any help

excuse my English

[Ligander]

I am told that has to do with "the theory of electron volt" but have not found anything about it

 Maybe you should express all these processes in terms of energy and use conservation law  .

[freezard7734]

suppose that I have the following electrode potentials

(Cu2+)  + 2e-    --------> Cu           E = 0337

(CU1+) + 1e-    ------------> Cu       E = 0521

and I want to find
 
(Cu2+) + e-   ----------> (CU1+)  E =?

I can multiply the first potential by two and subtract the second and get the correct value

2x 0.337 to 0.521 = 0.153

now my question is because this is so?

I've seen several similar load and I understand the procedure, but i don`t know the reason

I am told that has to do with "the theory of electron volt" but have not found anything about it

appreciate any help

excuse my English

Correction: You shouldn't multiply by two because the number of electrons transferred doesn't affect the electromotive force, E. E is simply the Gibbs energy in disguise:

G = -vEF, where v is the coefficient of the electrons in the half reaction. Therefore, when you multiply a half reaction by a numerical factor, k, G is increased by a factor of k and the number of electrons is increased by a factor of k, and so E is unaffected.