[Sophia7X]

Calculate the cell potential for a silver-silver chloride electrode immersed in 0.800 M KCl at 25° C. [Ag⁺ + e^- --> Ag, E° = 0.799 V; Ksp = 1.8 x 10^-10]

First, I calculated the concentration of Ag⁺ at equilibrium...
AgCl --> Ag⁺ + Cl^-
1.8 x 10^-10 = (x)(0.8+x)
x = [Ag⁺] = 2.25 x 10^-10 M.
I tried using the Nernst equation, but I did not get the correct answer.
By the way, the answer choices are a) 1.37, b.) 0.80, c.) 0.57, and d.) 0.23.

another question:
The radius of which ion is closest to that of the Li⁺ ion?
a.) Na⁺ b.) Be⁺² c.) Mg⁺² or d.) Al⁺³
The answer is supposed to be C... why isn't it B?

Sorry, one more!
What is the major product when nitrobenzene is treated with a mixture of HNO3 and H2SO4?
A.) 1,1-dinitrobenzene  B.) 1,2-dinitrobenzene. C.) 1,3-dinitrobenzene D.) 1,4-dinitrobenzene
I'm unfamiliar with the reaction... I just chose D because I thought it had the least steric hindrance. The answer is C— can someone elaborate?

Thanks if you could explain any one of these questions!

[Olympiad_Tutor]

.
Sorry, one more!
What is the major product when nitrobenzene is treated with a mixture of HNO3 and H2SO4?
A.) 1,1-dinitrobenzene  B.) 1,2-dinitrobenzene. C.) 1,3-dinitrobenzene D.) 1,4-dinitrobenzene
I'm unfamiliar with the reaction... I just chose D because I thought it had the least steric hindrance. The answer is C— can someone elaborate?

Thanks if you could explain any one of these questions!

Nitro group directs the electrophile (the second nitro group) to the meta (3 and 5) positions.  Electrophilic Aromatic Substitution is a large topic in organic chemistry. http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/benzrx1.htm

[Borek]

I tried using the Nernst equation, but I did not get the correct answer.

Show how.

[fledarmus]

another question:
The radius of which ion is closest to that of the Li⁺ ion?
a.) Na⁺ b.) Be⁺² c.) Mg⁺² or d.) Al⁺³
The answer is supposed to be C... why isn't it B?

What effect does an increased positive charge in the nucleus have in the radius of the electron orbitals?

You might want to look here for some actual data... http://en.wikipedia.org/wiki/Ionic_radius

[Sophia7X]

I tried using the Nernst equation, but I did not get the correct answer.

Show how.

E_cell = 0.799 - (0.0591)(log[2.25x10^-10*0.800) = 1.37 V

The answer is supposed to be D, 0.23 V.

another question:
The radius of which ion is closest to that of the Li⁺ ion?
a.) Na⁺ b.) Be⁺² c.) Mg⁺² or d.) Al⁺³
The answer is supposed to be C... why isn't it B?

What effect does an increased positive charge in the nucleus have in the radius of the electron orbitals?

You might want to look here for some actual data... http://en.wikipedia.org/wiki/Ionic_radius

Smaller radius.

And I think I just figured this one out. Be is in the second family so it is smaller than Li so a Be⁺² ion would be even smaller? So Mg⁺² would be the better answer since Mg is bigger than Li and becomes smaller as a cation?

[Borek]

E_cell = 0.799 - (0.0591)(log[2.25x10^-10*0.800) = 1.37 V

Why 2.25x10^-10*0.800?

The answer is supposed to be D, 0.23 V.

I got 0.215.

[Sophia7X]

Q = [Ag⁺][Cl^-]?

Also, I tried -0.799 V and I got -0.23 V.

[Borek]

No. If you use 0.799 this is standard potential for Ag/Ag⁺, so Nernst equation takes form

$$E = E_0 + \frac{RT}{F} \ln [Ag^+]$$

Activity of pure metal equals 1, so is not included, although technically it is there in the denominator, as the reaction is

Ag(s) -> Ag⁺(aq) + e^-


Use K_sp to calculate [Ag⁺] and you are ready.

[XGen]

Why is the equation for silver not Ag⁺ + e^- -> Ag? Isn't 0.799 the standard reduction potential, not the oxidation potential?

[Borek]

Doesn't matter - reaction goes both ways at the same potential. I am just used to this convention. If you want to work with the reduction you have to do two things - flip the Q and change the sign in the Nernst equation. That means using

$$E = E_0 - \frac{RT}{F} \ln \frac {1}{[Ag^+]}$$

It is mathematically equivalent and yields exactly the same final result.

[XGen]

Your error seems to be in using (2.25 x 10^-10)(0.800). It's probably a little late, but it's a good clarification for me haha

[Sophia7X]

Doesn't matter - reaction goes both ways at the same potential. I am just used to this convention. If you want to work with the reduction you have to do two things - flip the Q and change the sign in the Nernst equation. That means using

$$E = E_0 - \frac{RT}{F} \ln \frac {1}{[Ag^+]}$$

It is mathematically equivalent and yields exactly the same final result.

Ohh, it's supposed to be 1/[Ag⁺]! I was using the AgCl dissociation equation for Q. Thanks!