Posted by fenerbahce on Friday, July 20, 2012 at 10:02am.
How do you calculate the theoretical cell voltage?
1) ZnZn^+2 (1M) CuCu^+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:
CuCu^+2 (1M) CuCu^+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.
Answer This Question
Related Questions
 chemistry  The voltage of an electrochemical cell depends on the redox reaction...
 chemistry  2. The voltage of an electrochemical cell depends on the redox ...
 c  The voltage of an electrochemical cell depends on the redox reaction ...
 Chem  Calculate the standard cell voltage, Eo, for the following reactions. ...
 college chemistry  the Cu(No3)2 solution was diluted to 0.001 M. The voltage ...
 chemistry  1. Given that 50 grams of ice is heated at 20.0 °C to steam at 135....
 Chemistry  Determine the cell potential from the following half cells: Pb^o/Pb2...
 Chemistry  The voltage generated by the zinc concentration cell described by, ...
 chemistry  The voltage generated by the zinc concentration cell described by, ...
 Chemistry 104  The voltage generated by the zinc concentration cell described ...
More Related Questions