Solve Galvanic Cell Problem: Iron Oxidation

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Discussion Overview

The discussion revolves around a galvanic cell problem involving the oxidation of iron and the calculation of cell potentials. Participants are examining the correct approach to determine the overall cell potential based on the reduction and oxidation potentials provided in a textbook example.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions whether the negative cell potential for iron indicates that the equation should be flipped to designate it as the anode, suggesting that this would affect the overall calculation of cell potential.
  • Another participant expresses confusion about why the iron cell potential is subtracted, proposing that since the reduction potential is negative, it should become positive when converted to oxidation potential, leading to a different addition with the cathode's cell potential.
  • A later reply asserts that the textbook's explanation is unclear and presents two methods for calculating the overall cell potential: either subtracting the lower reduction potential from the higher or flipping the half-equation for the lower reduction potential, reversing the sign, and then adding the two values.

Areas of Agreement / Disagreement

Participants do not appear to reach consensus on the correct method for calculating the overall cell potential, with multiple competing views on how to handle the potentials and the steps involved in the calculation.

Contextual Notes

There are limitations in the clarity of the textbook's explanation and the assumptions made regarding the treatment of reduction and oxidation potentials, which remain unresolved in the discussion.

Chemistry314
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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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