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

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#### NYM

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•   • 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 »

#### LQ43

• Chemist
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• Gender:  ##### 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

#### 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 