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### Topic: Calculating the standard half-cell reduction potential  (Read 30152 times)

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#### kimi85

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• Mole Snacks: +0/-12 ##### Calculating the standard half-cell reduction potential
« on: June 23, 2007, 10:54:14 AM »
Hi everyone.

The problem is:

The standard half-cell reduction potential for Ag+/Ag is 0.7996 V at 25 degree Celsius.  Given experimental value Ksp = 1.56 x 10raised to -10 for AgCl, calculate the standard half-cell reduction potential for the Ag/AgCl electrode.

I used this equation: E = RT/nF ln Ksp, and my answer is wrong. The correct answer is  0.2198 V. I don't know how to come up with the answer.

Thank you very much.

#### Borek ##### Re: Calculating the standard half-cell reduction potential
« Reply #1 on: June 23, 2007, 11:12:04 AM »
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#### kimi85

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• Mole Snacks: +0/-12 ##### Re: Calculating the standard half-cell reduction potential
« Reply #2 on: June 23, 2007, 09:08:52 PM »
E is the En.  I think that's the formula to use when you are using standard cell potentials to find the equilibrium constants

#### Borek ##### Re: Calculating the standard half-cell reduction potential
« Reply #3 on: June 24, 2007, 03:35:20 AM »
What I am hinting at is that the formula you mention doesn't use standard potential - don't you think it should?
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#### kimi85

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• Mole Snacks: +0/-12 ##### Re: Calculating the standard half-cell reduction potential
« Reply #4 on: June 24, 2007, 08:56:43 AM »
What I did is 0.7996 - x  = 0.0257/1 ln 1.56 x 10-10.. x is the half-cell reduction potential for the Ag/AgCl electrode.  But it's wrong. Do you know what should be done?

#### Borek ##### Re: Calculating the standard half-cell reduction potential
« Reply #5 on: June 24, 2007, 09:03:49 AM »
Write equation for E for Ag/Ag+ electrode, then put in Ag+ concentration calculated assuming [Cl-] = 1. That's all.
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#### kimi85

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• Mole Snacks: +0/-12 ##### Re: Calculating the standard half-cell reduction potential
« Reply #6 on: June 24, 2007, 09:41:14 AM »
Thank you very much. I'll try it

#### kimi85

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• Mole Snacks: +0/-12 ##### Re: Calculating the standard half-cell reduction potential
« Reply #7 on: June 27, 2007, 09:23:11 AM »
Hi. I still don't understand how to do it.
Can you show me how?

#### Borek ##### Re: Calculating the standard half-cell reduction potential
« Reply #8 on: June 27, 2007, 09:58:26 AM »
Show what you did.
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#### kimi85

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• Mole Snacks: +0/-12 ##### Re: Calculating the standard half-cell reduction potential
« Reply #9 on: June 27, 2007, 11:40:13 AM »
This is what I did:

I get the square root of the Ksp to get the concentration of Ag+ which is equal to 1.25 x 10^-5 , then

0.7996 - x = 0.0257/ n ln 1.25 x 10^-5

I still didn't get the correct answer.

#### Borek ##### Re: Calculating the standard half-cell reduction potential
« Reply #10 on: June 27, 2007, 01:54:10 PM »
Nope.

E = E0 + RT/nF ln Q

where Q is reaction quotient.

Now, in the simplest case (metal and its ions) it translates to

E = E0 + RT/nF ln [Men+]

Now, if the silver electrode is covered with AgCl, [Ag+] on the electrode surface depends on the [Cl-]. That's where the Kso comes into play.

Kso = [Ag+][Cl-]

So

[Ag+] = Kso/[Cl-]

Insert it into Nernst equation, see what it gives.
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#### kimi85

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• Mole Snacks: +0/-12 ##### Re: Calculating the standard half-cell reduction potential
« Reply #11 on: June 27, 2007, 08:43:13 PM »
Your solution is correct.Thank you very much!!!! God bless you. #### kimi85

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• Mole Snacks: +0/-12 ##### Re: Calculating the standard half-cell reduction potential
« Reply #12 on: June 27, 2007, 10:48:14 PM »
Why is the formula a plus?

Shouldn't it be E = Eo - RT/nF In Q?

#### Borek ##### Re: Calculating the standard half-cell reduction potential
« Reply #13 on: June 28, 2007, 04:01:11 AM »
Depends on the way you construct Q. For half reaction

Q = [Oxidised]/[Reduced]

and Nernst equation gets form

E = E0 + RT/nF ln Q

Same goes for more complicated reactions, like

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

Just put oxidized form in numerator:

Q = [MnO4-][H+]8/[Mn2+]

(note that water concentration is taken from the Q, as it is assumed to be constant and not changing, thus it is moved to E0)

It is just a matter of convention used, you may as well use minus and reverse Q exchanging numerator and denominator.

Looks like I was not precise in my previous post, hopefully it is cleared now.
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#### kimi85

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• Mole Snacks: +0/-12 ##### Re: Calculating the standard half-cell reduction potential
« Reply #14 on: June 28, 2007, 04:34:19 AM »
Now, I got it. Thank you very much. 