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 :)

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 calculate the voltage of the cell, you can use the Nernst equation:

E = E° - (RT/nF) * ln(Q)

where:
E is the cell voltage
E° is the standard cell voltage
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of electrons transferred
F is Faraday's constant (96,485 C/mol)
ln is the natural logarithm
Q is the reaction quotient

In this case, you mentioned that n is equal to 2, which means that 2 electrons are being transferred in the overall cell reaction. That is correct.

To set up the Q ratio for this type of problem, you need to consider the concentrations of the species involved on each side of the cell.

In this case, the cell diagram Pt | Fe2+ (6.40x10^-5 M), Fe3+ (0.046 M) || Sn4+ (2.70x10^-4 M), Sn2+ (0.075 M) | Pt suggests that the following reaction is occurring in the cell:

Fe2+ + Sn4+ ⇌ Fe3+ + Sn2+

To set up Q, you need to write the ratio of product concentrations over reactant concentrations, with each concentration raised to the power of its stoichiometric coefficient.

Q = ([Fe3+]/[Sn2+]) * ([Fe2+]/[Sn4+])

In this case, since the reaction is happening in a galvanic cell, the concentrations of the species are at equilibrium. Therefore, you can directly use the given concentrations to calculate Q:

Q = (0.046 M / 0.075 M) * (6.40x10^-5 M / 2.70x10^-4 M)

Now that you have the Q value, you can substitute it into the Nernst equation along with the other known values (E°, R, T, n, and F) to calculate the cell voltage (E).