Pt | Fe2+ (6.4010-5 M), Fe3+ (0.0460 M) || Sn4+ (2.7010-4 M),Sn2+ (0.0750 M) | Pt
i need to calculate the voltage of the cell.
I know to use E=emf- (RT/nF)lnQ .... I think my problem is with setting up Q. I'm confused on what goes on top and what goes on bottom for the ratio in this type of problem. Up until now I have only seen prolems in which only one concentration for each side has been given.
Also, I found n to be 2 (2 electrons transfer). If someone could verify that this is correct that would be great :)
2 answers
Nevermind, I made a silly mistake in my calculations...I figured it out :)
n is 2.
To answer your question, you write the cell reaction and write the equilibrium constant expression for that reaction. That is what goes in for Q (or K). Same thing for half cells.
If I write
Fe^+2 -->Fe^+3 + e
Then E = Eo-(0.059/n)[Fe^+3/Fe^+2].
If I choose to write it as
Fe^+3 + e ==> Fe^+2, then
E = Eo - (0.059/n)log [Fe^+2/Fe^+3]
But, from experience, I think you are going about it the long way.
It is MUCH easier to convert each half cell (which you know how to do and how to write the log part) from standard conditions (1 M et al.) to current conditions, turn the one around that needs to be reversed, then add the two half cells to obtain the cell voltage. In my opinion, it is far too easy to lose a sign or a term even, when trying to do the entire reaction.
To answer your question, you write the cell reaction and write the equilibrium constant expression for that reaction. That is what goes in for Q (or K). Same thing for half cells.
If I write
Fe^+2 -->Fe^+3 + e
Then E = Eo-(0.059/n)[Fe^+3/Fe^+2].
If I choose to write it as
Fe^+3 + e ==> Fe^+2, then
E = Eo - (0.059/n)log [Fe^+2/Fe^+3]
But, from experience, I think you are going about it the long way.
It is MUCH easier to convert each half cell (which you know how to do and how to write the log part) from standard conditions (1 M et al.) to current conditions, turn the one around that needs to be reversed, then add the two half cells to obtain the cell voltage. In my opinion, it is far too easy to lose a sign or a term even, when trying to do the entire reaction.
0 answers