Solve Galvanic Cell Problem: Iron Oxidation

  • Thread starter Thread starter Chemistry314
  • Start date Start date
  • Tags Tags
    Cell Galvanic cell
AI Thread Summary
The discussion centers on the calculation of cell potential in a galvanic cell involving iron oxidation. Participants clarify that when iron is oxidized, its negative reduction potential should be converted to a positive oxidation potential before combining it with the cathode's potential. The textbook's explanation is deemed unclear, leading to confusion about whether to subtract or flip the equation. The correct approach involves either subtracting the lower reduction potential from the higher or flipping the half-equation and reversing the sign. Ultimately, the calculated standard cell potential is 1.95V.
Chemistry314
Messages
2
Reaction score
0
DXrMA02.jpg

http://imgur.com/a/wGTEQ

The example is straight from my textbook. Since the iron is being oxidized, and it has a negative cell potential to begin with, wouldn't you flip the equation to make it the anode and in the end add it to the other cell potential?
 
Physics news on Phys.org
No image. Write out your question.
 
mjc123 said:
No image. Write out your question.
It seems to be working again. My question is why they subtract the iron cell potential. The reduction potential is negative, so to change it to oxidation potential would it not become positive, and then you add it to the cathodes cell potential?
 
Yes. The textbook is not very clear here, and the final equation is actually wrong. You should either
Subtract the lower reduction potential from the higher (this is the easiest way), or
Flip round the half-equation with the lower reduction potential, reverse the sign (to make it an oxidation potential) and add the two numbers.
Either way you get Eocell = 1.51 - (-0.44) = 1.95V
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top