Timeline for How can we directly add half cell potentials to measure the EMF of a galvanic cell?
Current License: CC BY-SA 3.0
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Mar 4, 2018 at 13:52 | comment | added | porphyrin | no, much, much simpler, I wanted only to give a familiar example where addition does not work. We can measure the difference in temperature between inside (say 20 C) and outside (say -5 C) which is 25 degrees difference but we do not add the temperatures as this makes no sense (15 degrees). Similarly with potentials. | |
Mar 4, 2018 at 13:25 | comment | added | Gaurang Tandon | Oh, I see where you're going. Can you elaborate on that relation with temperature? Do you mean to say that a solid at 300K when touched with a solid at 400K, through conduction their final temperature will be...? | |
Mar 4, 2018 at 13:22 | comment | added | porphyrin | The value you need is always the difference in potential which is ok just as we can have a difference in temperature but not a sum. The values themselves are ultimately referenced to the hydrogen electrode at $0$V. Your confusion seems to be that you want to add potentials. (The addition bit I mentioned is just a way of remembering how to do the calculation, so I hope I did'n't confuse ) | |
Mar 4, 2018 at 11:45 | comment | added | Gaurang Tandon | Actually, my question was exactly why the addition of redox potentials works the way it does, because according to me their addition should be invalid as they are intensive properties. | |
Mar 4, 2018 at 11:30 | history | edited | porphyrin | CC BY-SA 3.0 |
typo
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Mar 4, 2018 at 11:28 | comment | added | porphyrin | Its the same if you make $E^\text{o}$ the emf of the cell which is the convention. I used $\Delta E^\text{o} $ only because you measure a difference in $E$ but I will change it, :) | |
Mar 4, 2018 at 11:25 | comment | added | Gaurang Tandon | Actually, I was taught $\Delta G=-nFE^\circ$ and not $\Delta G=-nF\Delta E^\circ$. The latter is your first equation... | |
Mar 4, 2018 at 11:18 | history | answered | porphyrin | CC BY-SA 3.0 |