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Galvanic Cell Problem

  1. Dec 5, 2016 #1

    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?
  2. jcsd
  3. Dec 5, 2016 #2


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    No image. Write out your question.
  4. Dec 5, 2016 #3
    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?
  5. Dec 5, 2016 #4


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    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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