Cell Potentials

  • Thread starter τheory
  • Start date
  • #1
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Homework Statement


Calculate the cell potential at 25°C for the cell:

Fe (s) | (Fe^2+ (0.100 M) || Pd^2+ (1.0*10^-5 M) | Pd (s)

given that the standard potential for Fe^2+/Fe is -0.45 V and for Pd^2+/Pd is +0.95 V.

A. +1.28 V
B. +1.16 V
C. +1.52 V
D. +1.68 V

Homework Equations


E Cell potential = E Cathode - E Anode


The Attempt at a Solution


By looking at the two potentials for Fe and Pd, I deducted that the species at the cathode was Pd because 0.95 > -0.45. Then, I took the difference:

E Cell potential = E Cathode - E Anode
E Cell Potential = 0.95 - (-0.45)
E Cell potential = 1.40 V (which isn't even an answer choice)

However upon looking at answer choices and the answer key, the answer was in fact A. +1.28 V; can someone explain to me what went wrong in my calculations? Or if there is a typo?
 

Answers and Replies

  • #2
Borek
Mentor
28,808
3,308
Solutions are not in a standard state - you have to correct for concentrations using Nernst equation.
 

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