Redox Reactions and standard emf calculations

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To determine the value of n for the cell reaction with a standard emf of +0.17V and an equilibrium constant of 5.5 X 10^5, the equation E = E° - (0.0592/n)(logk) is utilized. By setting E to zero at equilibrium, the equation simplifies to 0 = 0.17 - (0.0592/n)(log(5.5*10^5)). Solving for n yields n = (0.0592)(log(5.5*10^5)) / 0.17, resulting in n approximately equal to 1.9989. The discussion emphasizes the importance of selecting the appropriate equation based on known values. Understanding the relationship between standard emf and equilibrium constant is crucial in redox reaction calculations.
kumarium
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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!
 
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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!
 
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