Posted by fenerbahce on Friday, July 20, 2012 at 10:02am.
How do you calculate the theoretical cell voltage?
1) Zn|Zn^+2 (1M)| Cu|Cu^+2 (1M)
okay so by looking at example this example
Calculate the emf (voltage) for the following reaction:
Zn(s) + Fe2+ → Zn2+ + Fe(s)
Write the 2 half reactions:
Zn(s) → Zn2+ + 2e
Fe2+ +2e → Fe(s)
look up the standard electrode potentials in the table above
Zn2+ + 2e → Zn Eo = -0.76V
This equation needs to be reversed, so the sign of Eo will also be reversed.
Zn(s) → Zn2+ + 2e Eo = +0.76V
Fe2+ +2e → Fe(s) Eo = -0.41V
Add the two equations together:
Zn(s) → Zn2+ + 2e Eo = +0.76V
Fe2+ + 2e → Fe(s) Eo = -0.41V
Zn(s) + Fe2+ → Zn2+ + Fe(s) Eocell = +0.76 + (-0.41) = +0.35V
I came up with the answer: 1.11
So am I on the right track?
AND ALSO VERY VERY IMPORTANT**
for the first one they more have 1 M
but for this question:
Cu|Cu^+2 (1M) Cu|Cu^+2(0.1 M)
see how one has 1M and the other 0.1M will that make a difference in the calculation thank you.
- chemistry - fenerbahce, Friday, July 20, 2012 at 10:03am
NOTE I calculates by using the above example
- chemistry - fenerbahce, Friday, July 20, 2012 at 11:03am
******HOW DO YOU CALCULATE THE REDUCTION POTENTIAL OF 1 M CuSO4 AND 0.1 M CuSO4?
NOTE: THE REDUCTION POTENTIAL OF 1M Zn(NO3)2 IS -0.76 V. I'M REALLY STUCK.
- chemistry - DrBob222, Friday, July 20, 2012 at 10:39pm
You have done the Zn/Cu cell right. The Ecell = 0.35 v.
For the same element (Zn/Zn^2+) and (Zn/Zn^2+) but with the Zn^2+ different concentrations, it is
Ecell = Eocell -(0.0592/n)log(dil soln/concnd soln). Eocell is always 0 since we have th same element. n is fairly obvious and the dilute and concd solns are obvious. Substitute and solve for Ecell.
If you want to do them separately (calc E reduction potential for the 1M then E reduction potential for the 0.1M) you can use
E = EoCell - (0.0592/n)log(Cu/Cu^2+)
For 1M, you substitute 1 for Cu and 1 for Cu^2+, log 1 = 0 and Ehalf cell for 1 M is just Eo. For 0.1M, substitute 1 for Cu and 0.1 for Cu^2+ and Ehalf cell is E = 0.34 -(0.0592/2)log 0.1
E = 0.34 -(0.0296)(-1) = 0.34+0.0296 = about 0.37.
- chemistry - fenerbahce, Sunday, July 22, 2012 at 1:09pm
Thank you very much..And I posted another question, the most important ones that I don't know how to do.
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