Solve Galvanic Cell Problem: Iron Oxidation

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SUMMARY

The discussion centers on solving the galvanic cell problem involving iron oxidation. Participants clarify that when calculating cell potential, the lower reduction potential must be subtracted from the higher one. Specifically, for iron, which has a negative reduction potential, the oxidation potential is derived by flipping the half-equation and reversing the sign. The correct final equation yields a cell potential of Eocell = 1.51 - (-0.44) = 1.95V.

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  • Understanding of galvanic cells and electrochemical reactions
  • Familiarity with reduction and oxidation potentials
  • Knowledge of half-equations in electrochemistry
  • Basic skills in calculating cell potentials
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  • Study the Nernst equation for calculating cell potentials under non-standard conditions
  • Learn about standard reduction potentials from the electrochemical series
  • Explore the concept of cell efficiency in galvanic cells
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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?
 
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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
 

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