Posted by Tom on Tuesday, May 13, 2014 at 11:43pm.
It really helps us help you if you use the same screen name. Tom, Tyler, help ASAP etc is not helpful at all.
You know from your previous post how to calculate the standard Zn/Ag^+ cell potential. That is Eocell and that is for standard concentrations of 1M Zn^2+ and Ag^+. When they are not at standard conditions, then
Ecell = Eocell - (0.0592/n)log(Zn^2+/(Ag^+). Plug in Zn^2+ and Ag^+ from the problem and calculate Ecell. Post your work if you get stuck.
what is n?
and isnt it ln not log?
so Ecell=1.56-(.0592/n)ln(.76/.80)?
n = 2 and you know that from the equation I wrote earlier of
Zn + 2Ag^+ ==> Zn^2+ + 2Ag
The term you are talking about is (RT/nF)ln[(Zn^2+)/(Ag^+)] and if you want to substitute R, T, n, and F in there you may. However, if you do all of that AND multiply by 2.303 to convert ln to log, all of those constants turn out to be 0.05916/n if you use 298 for T (which the problem asks for). Just a nice round number to remember. The history of that for you young snappers (I'm retired and well up there in age) is that when I was in graduate school we used a slide rule (I know, what's a slide rule?) and it was more convenient to read log base 10 than ln base e so obviously we never used ln for anything but always converted. Now with these pocket calculators you can get almost anything out of them AND you don't even need to know what a slide rule is. :-)
Now extra charge for the history lesson.
never mind you are right with the log part so i just don't understand what n is
Oh duh ok i get that thanks a million
and i leave Zn's positive right? or do i change it back to negative now?
Zn^2+ + 2e ==> Zn from the table is -0.76v. Your half cell reaction is
Zn ==> Zn^2+ + 2e E = +0.76v
Ag^+ + e ==> Ag E = 0.80 v.
---------------------------
Add rxnx; add half cell oxidation half to reduction half.
Cell rxn is
Zn + 2Ag^+ ==> Zn^2+ + 2Ag
Ecell = 0.76oxdn + 0.80 rdn = 1.56v for Zn^+2(aq) = 1M and Ag^+(aq) = 1M