Solve Galvanic Cell Problem: Iron Oxidation

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The discussion centers on the calculation of cell potential in a galvanic cell involving iron oxidation. Participants clarify that when iron is oxidized, its negative reduction potential should be converted to a positive oxidation potential before combining it with the cathode's potential. The textbook's explanation is deemed unclear, leading to confusion about whether to subtract or flip the equation. The correct approach involves either subtracting the lower reduction potential from the higher or flipping the half-equation and reversing the sign. Ultimately, the calculated standard cell potential is 1.95V.
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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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