Redox Reactions and standard emf calculations

Click For Summary
SUMMARY

The discussion focuses on calculating the value of n for a redox reaction given a standard emf of +0.17V and an equilibrium constant of 5.5 X 10^5 at 298K. The relevant equations discussed include E°cell = E°cathode - E°anode, E°cell = (RT/nF)(lnk), and E = E° - (0.0592/n)(logk). The correct approach involves setting E to zero when Q equals K, leading to the calculation n = (0.0592)(log(5.5*10^5)) / 0.17, resulting in n = 1.9989.

PREREQUISITES
  • Understanding of redox reactions and standard electrode potentials
  • Familiarity with the Nernst equation and its components
  • Basic logarithmic calculations
  • Knowledge of equilibrium constants in electrochemistry
NEXT STEPS
  • Study the Nernst equation in detail, focusing on its applications in electrochemistry
  • Learn about the relationship between standard emf and equilibrium constants
  • Explore advanced redox reaction mechanisms and their calculations
  • Investigate the significance of the number of electrons transferred (n) in electrochemical cells
USEFUL FOR

Chemistry students, electrochemists, and anyone involved in studying or applying redox reactions and electrochemical calculations.

kumarium
Messages
4
Reaction score
0
What is the equation I should be using for the following question:

At 298K a cell reaction has a standard emf of +0.17V. The equilibrium constant for the cell reaction is 5.5 X 10^5. What is the value of n for the cell reaction.

Grateful for anyone's help!
 
Physics news on Phys.org
What equations have you learned while studying redox potentials?
 
E°cell = E°cathode - E°anode
E°cell = (RT/nF)(lnk)
E = E° - (0.0592/n)(logk)
 
Why don't you just select the equation containing all known values and n? This is almost a simple plug and chug.
 
The equation that should be most likely used is E = E°-(0.0592/n)(logk).
I have the value for the standard emf = E°, and value of k.

Do I assume E to be zero then? If I do so, then the solution should look like the following:
0= E° - (0.0592/n)(logk)
0= 0.17 - (0.0592/n)(log(5.5*10^5))
0.17 = (0.0592/n)(log(5.5*10^5))
n= (0.0592)(log(5.5*10^5)) / 0.17
n= 1.9989
 
Yes.

When Q=K cell potential is zero.
 
Thank you for the help!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K