[kimi85]

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]

E = RT/nF ln Ksp

Where is E₀?

[kimi85]

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]

What I am hinting at is that the formula you mention doesn't use standard potential - don't you think it should?

[kimi85]

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]

Write equation for E for Ag/Ag⁺ electrode, then put in Ag⁺ concentration calculated assuming [Cl^-] = 1. That's all.

[kimi85]

Thank you very much. I'll try it

[kimi85]

Hi. I still don't understand how to do it.
Can you show me how?

[Borek]

Show what you did.

[kimi85]

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]

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 K_so comes into play.

K_so = [Ag⁺][Cl^-]

So

[Ag⁺] = K_so/[Cl^-]

Insert it into Nernst equation, see what it gives.

[kimi85]

Your solution is correct.Thank you very much!!!!

God bless you.

[kimi85]

Why is the formula a plus?

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

[Borek]

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.

[kimi85]

Now, I got it. Thank you very much.