Electrochemistry Problem

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TexasFluoride

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I am having trouble with the following electrochemistry problem.

http://img101.imageshack.us/img101/4448/kaplanquestion.jpg


Someone please explain why we don't double the rxn of Co3+-----> Co2+ hence doubling the cell potential to 3.64, so the 2electron will cancel when combined with the rxn involving I2.

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It sounds like you're trying to balance these 2 reactions, which you dont need to do. You just have to find the cell potenial.
Both these half reactions as it is are being reduced. In order for the reactions to occur there has to be one oxidation and one reduction. Thus you reverse the bottom reaction and change the sign of +0.53 to -0.53.
And like ZFT said, +1.82 + (-0.53) = +1.29 V
 
It sounds like you're trying to balance these 2 reactions, which you dont need to do. You just have to find the cell potenial.
Both these half reactions as it is are being reduced. In order for the reactions to occur there has to be one oxidation and one reduction. Thus you reverse the bottom reaction and change the sign of +0.53 to -0.53.
And like ZFT said, +1.82 + (-0.53) = +1.29 V

How did you know you had to reverse the bottom reaction, why couldn't you reverse the top equation? That would give you a different answer though
 
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How did you know you had to reverse the bottom reaction, why couldn't you reverse the top equation? That would give you a different answer though


One reaction has to be the reduction and one has to be the oxidation. The higher reduction potential (more positive) will reduce and the other will oxidize. Therefore in this case, I- will oxidize to I so you flip that one to -.53.
 
How did you know you had to reverse the bottom reaction, why couldn't you reverse the top equation? That would give you a different answer though

i think bc if you reverse the other one you'd get a more negative voltage which favors reduction, a smaller negative (more postive) is a better oxidizer so -.53 is a better oxidizer than -1.82...i hope that's clear enough mb someone else can better clarify but its pretty straightforward
 
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