Thursday

September 29, 2016
Posted by **Tom** on Tuesday, May 13, 2014 at 11:43pm.

electrolyte to a 1.30 M Ag+

(aq) solution containing a silver electrode. (Note that the solution

concentrations are not standard). Determine the initial voltage of the cell at 298 K. Show your

work. The Nernst equation will help you answer this question.

- Chemistry -
**DrBob222**, Wednesday, May 14, 2014 at 12:01amIt 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. - Chemistry -
**CHEM HELP ASAP**, Wednesday, May 14, 2014 at 12:11amwhat is n?

and isnt it ln not log?

so Ecell=1.56-(.0592/n)ln(.76/.80)? - Chemistry -
**DrBob222**, Wednesday, May 14, 2014 at 12:23amn = 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. - Chemistry -
**CHEM HELP ASAP**, Wednesday, May 14, 2014 at 12:25amnever mind you are right with the log part so i just dont understand what n is

- Chemistry -
**CHEM HELP ASAP**, Wednesday, May 14, 2014 at 12:26amOh duh ok i get that thanks a million

- Chemistry -
**CHEM HELP ASAP**, Wednesday, May 14, 2014 at 12:29amand i leave Zn's positive right? or do i change it back to negative now?

- Chemistry -
**DrBob222**, Wednesday, May 14, 2014 at 12:37amZn^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