• Support PF! Buy your school textbooks, materials and every day products Here!

Electrode potential ?

  • Thread starter Puchinita5
  • Start date
  • #1
183
0
I don't know why, but I just can't figure this out.

"Calculate the electrode potential for a silver filled electrode in a solution of 0.00550 M AgNO3 solution."

I got that

Ag+ + e- --> Ag(s) had an E* = .80V

I got that the answer is E=.80 - .0592 log (1/.00550M)= .6655 V

But why do you put 1/.00550M in the parenthesis?

My first thought was that AgNO3 + e- --> Ag(s) + NO3-
and that the concentreation of AgNO3 is the same as NO3.

so that it would be log(1) , which is wrong.

I just don't get the concept I guess.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Borek
Mentor
28,352
2,738
It all depends on how you write the Nernst equation. There are two conventions:

[tex]E = E_0 + \frac {RT}{nF} ln \frac{[Ox]}{[Red]}[/tex]

and

[tex]E = E_0 - \frac {RT}{nF} ln \frac{[Red]}{[Ox]}[/tex]

(personally I prefer the first one, but that's just because I was taught this way).

Note that both equations are equivalent thanks to the sign change.

Oxidized form of silver is Ag+, reduced form is a metallic Ag - with activity equal to 1. So you can write the equation either as

[tex]E = 0.88 + 0.059\ log([Ag^+])[/tex]

or

[tex]E = 0.88 - 0.059\ log\big(\frac 1 {[Ag^+]}\big)[/tex]

Concentration of Ag+ is 0.00550 M, concentration of NO3- doesn't matter.
 
  • #3
183
0
oh! okay that makes sense. I don't know why but I always confuse myself with these electrode problems.

Thank you!
 

Related Threads for: Electrode potential ?

  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
16
Views
6K
  • Last Post
Replies
8
Views
898
Replies
1
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
0
Views
7K
Top