Posted by Mima on Sunday, April 29, 2012 at 1:20pm.
The set up looks ok but I would use 8.314 for R and 96,485 for F. If I run through that I obtain 0.438.
E= 0.438?
I got the same answer (1.987) when I used the numbers you mentioned. Is the E standard potential = -0.26 -(-0.76)?
Am I doing that right?(its cathode-Anode) right?
Ecell = 0.5v. You are correct. I plugged in your numbers and obtained 0.438 with them, also. This is like looking for a needle in haystack. The BEST think to have done was type in your work and let someone look at it. As it is both are floundering. Here is your work.
E= 0.5 - (8.315)(352)/(2)(96480) ln 0.819/0.014.
In pieces:
8.314*352/(2*96480) = 0.01517
0.819/0.014 = 58.5
ln 58.5 = 4.060
0.01517*4.069 = 0.0617 and
0.5-0.0617 = 0.438
Check your work. One of us is punching the wrong keys.
Got it, you are right, it's 0.439. Thank you so much, I was doing an easy math mistake (substracting before multiplying). Thanks again.
Related Questions
chemistry - Calculate the cell potential for the following reaction as written ...
chemistry - Calculate the cell potential for the following reaction as written ...
Chemistry - I'm having trouble with the following question, any help would ...
Chemistry - Please check a Nernst equation for me? Calculate the cell potential ...
Chemistry/Electrochem - A concentration cell is built based on the following ...
Chemistry - Consider the standard cell at 25C based on the following half-...
chemistry - The reaction described by the equation Zn + Hg2Cl2 <-->...
Electrochemistry - In lab, we did an experiment with electrochemical cells with ...
college chemistry - A zinc-copper battery is constructed as follows: Zn | Zn+2(0...
Chemistry - A zinc-copper battery is constructed as follows at 25°C. Zn|Zn^2...
For Further Reading
