Calculating the potential of an electrochemical cell (Nernst equation) This is the problem:

Calculate the ${E_\mathrm{cell}}$ (not ${E°}$) at 25°C.
I found that the reaction is $$\ce{Cu(s) + Fe^2+(aq) -&gt; Cu^2+(aq) + Fe(s)}$$ $\ce{Cu}$ gets oxidized and $\ce{Fe^2+}$ gets reduced.
I added the ${E°_\mathrm {Cu^{2+}/Cu}}$ and ${E°_\mathrm{Fe^{2+}/Fe}}$ to get ${E°_\mathrm{cell} = -.101}$ . But I'm not even sure if I'm supposed to add them or subtract them because I've seen both done, which is really confusing me. In what situations do you add, and which situations do you subtract half reaction potential values?
I then plugged this calculated value into the Nernst equation. 
I said Z = 2 and plugged in the correct constants. However, I am not sure if I calculated Q correctly. I did [anode]/[cathode], but is it supposed to be [products]/[reactants]? It is unclear to me because my instructor did the latter, but I keep seeing the former everywhere else because that is how you would calculate it in an equilibrium problem. I know it doesn't make a difference in this specific problem, but I want to know for future problems when concentration of all components are given (such as if they are all aqueous).
From these steps, I calculated that ${E_\mathrm{cell} = -.164 V}$
How to approach this question?
If you are not sure about the use of the parameter Q, go back to Nernst's law for each electrode, namely $$\ce{E_{M^{2+}/M} = \pu{ E°_{M^{2+}/M}} + \pu{0.0296 V} \times  \log[M^{2+}]}$$ Here the redox potentials of the two half-cells are, according to Nernst's law :$$\ce{E_{Cu} = \pu{+ 0.339 V} + \pu{0.0296 V} \times  \log0.10} = \pu{0.309 V}$$  $$\ce{E_{Fe} = \pu{-0.44 V} + \pu{0.0296 V} \times  \log0.0030} = \pu{- 0.515 V}$$
So the overall performance of the cell is $$\ce{E_{cell} = E_{Cu} - E_{Fe}}=  \pu{0.309 V} - \pu{(-0.515 V)  =  0.824 V}$$