# chem- electrochem-HELP

Given a cell based on the spontaneous reaction
2AgCl(s) + Zn(s) ® 2Ag(s) + 2Cl– + Zn2+
If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should
increase by 0.06 V
increase by 0.18 V.
decrease by 0.06 V.
decrease by 0.18 V.
increase by 0.35 V.

i keep getting the wrong answer- none that are listed anyways.
i have put into half-rxn, then used
G=-nFE and the dG=-RT InK, but cannot get the right answer pls help. its for an assignment due tonight- aust time

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1. got my questions meixed up!
I have tried this one by getting the 3 half-rxn and adding them up. then using E=E*-RT/nF InQ. i get close to an answer but not exact. i get 0.088V

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2. Look up E* for Zn (as written). I find 0.763.
Look up AgCl(s) + e ==> Ag(s) + Cl^- (as written). I find 0.222
Add to find E*cell. I find 0.985

Then
Ecell = E*cell -(0.06/n)log(products/reactants).

Ecell = E*-(0.06/n)log [(Zn^+2)(Ag)^2(Cl^-)^2/(Zn)(AgCl)]
Now substitute for
n = 2
(Zn^+2) = 1 M
(Ag) = standard state = 1
(Cl^-) = 1 M
(Zn) = standard state = 1
(AgCl) = standard state = 1
Plugging all that in gives you the log term of 1, log 1 = 0 and Ecell is just E* = 0.985 v

Now to the other scenario.
(Zn^+2) = 1M
(Ag)(s) = standard state = 1
(Cl^-) = 0.001 (remember to square it).
(Zn) = standard state = 1
(AgCl) = standard state = 1.
So that part becomes
Ecell = E*cell - (0.06/2)log(1 x 10^-6)
Ecell = 0.985 - (0.03)(-6)
So the change is + 0.18 and that is one of the choices.
You may have been using 0.0592 which would give you 0.1776 which still rounds to 0.18v.
(

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3. thank you so much. I stuffed up at the start. i had 3 half-rxn. i didn't see that there was one for AgCl- it all makes thank. thankyou for your patience, It appriciated. I had to cram this subject in to get this assignment done. we are not supposed to start this topic til monday which is when the assignment is due- not fun so again thankyou!

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