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# Redox equilibria Watch

1. Can someone help me with 2biii

https://9493747455b802768f762a6eb92f...Equilibria.pdf

I don't really get the question. I tried learning some method.

Where you write out the overall equation

Which I got as like Fe2+ + 2bro3- -> Fe3+ + Br2.
haven't bothered balancing it.
Then I just read somewhere if you increase products E cell decreases.
But I don't get why, idk if I am doing it right tbh.
Le chatelier's principle causes the equilibrium to shift to the left if you add Fe3+? How come the ms to the right. I

If someone can explain it would be appreciated
2. To get a larger emf, you want to obtain a smaller/ more neg electrode potential for the Fe2+ reaction as that will be taken away from the more positive electrode potential.

The reaction that occurs is Fe2+ ---> Fe3+ + e
More neg electrode potential = favouring oxidation I believe so you would need to decrease Fe3+ conc so position of eq shifts to replace it.
3. (Original post by Manexopi)
To get a larger emf, you want to obtain a smaller/ more neg electrode potential for the Fe2+ reaction as that will be taken away from the more positive electrode potential.

The reaction that occurs is Fe2+ ---> Fe3+ + e
More neg electrode potential = favouring oxidation I believe so you would need to decrease Fe3+ conc so position of eq shifts to replace it.
I dont get it
4. (Original post by Super199)
I dont get it
Okay so you know you work out emf by doing more positive E value minus the more negative one? If you want a larger emf, the number you're taking away needs to be smaller. To get a more negative/smaller electrode potential, the reaction needs to favour oxidation so you need to change the concentration of Fe3+ so that the reaction that happens is the oxidation of Fe2+ to Fe3+. In order to do that, you decrease the conc of Fe3+ so the position of equilibrium shifts to replace that lost Fe3+ meaning oxidation has occurred. I don't really know how else to explain it unfortunately.

Try thinking of it as how in the electrochemical series, the more negative electrode potentials are the strongest reducing agents i.e those that end up getting oxidised so thats what we're aiming for.

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