September 26, 2021, 08:32:49 PM
Forum Rules: Read This Before Posting


Topic: Determine [Ag+] in an electrochemical cell  (Read 2952 times)

0 Members and 1 Guest are viewing this topic.

Offline NYM

  • Regular Member
  • ***
  • Posts: 53
  • Mole Snacks: +2/-0
Determine [Ag+] in an electrochemical cell
« on: April 05, 2008, 01:14:10 PM »
An electrochemical cell consists of a Zn electrode in 0.200 M Zn2+ (aq) and a Ag electrode in 0.100 M Ag+ (aq).
KCl (s) is added to the Ag electrode. E is measured to 1.04 V and Eo is 1.56 V. Determine [Ag+].


Nernst equation:
E = Eo - 0.059 V/2 * log ([Zn2+]/[Ag+]) =>
1.04 V = 1.56 V - 0.059V/2 * log (0.200/[Ag+]2 =>
[Ag+] = 6.87*10-10 M.

The right answer is 7.36*10-10 M. What am I doing wrong here?
« Last Edit: April 05, 2008, 02:37:04 PM by NYM »

Offline LQ43

  • Chemist
  • Full Member
  • *
  • Posts: 250
  • Mole Snacks: +32/-9
  • Gender: Female
Re: Determine [Ag+] in an electrochemical cell
« Reply #1 on: April 05, 2008, 02:49:51 PM »
Nernst equation:
E = Eo - 0.059 V/2 * log ([Zn2+]/[Ag+]) =>
1.04 V = 1.56 V - 0.059V/2 * log (0.200/[Ag+]2 =>
[Ag+] = 6.87*10-10 M.

The right answer is 7.36*10-10 M. What am I doing wrong here?


recalculate using 0.0592

Offline NYM

  • Regular Member
  • ***
  • Posts: 53
  • Mole Snacks: +2/-0
Re: Determine [Ag+] in an electrochemical cell
« Reply #2 on: April 05, 2008, 03:43:23 PM »
Nernst equation:
E = Eo - 0.059 V/2 * log ([Zn2+]/[Ag+]) =>
1.04 V = 1.56 V - 0.059V/2 * log (0.200/[Ag+]2 =>
[Ag+] = 6.87*10-10 M.

The right answer is 7.36*10-10 M. What am I doing wrong here?


recalculate using 0.0592

D'oh! Thank you for pointing that out :D

Sponsored Links