My advice to you, sailcrazy, is to not use either version of the formula, as it will only confuse you. (Guess I didn't need to tell you that, right?
😛 ) Instead, what you should do is attack any electrochemistry problem by first considering the following:
What is the form in which the two half-cell reactions are written? Generally, but not always, you will see both half-reactions written as reduction potentials. You can determine this because you will see that the electrons are being added on the left side of the equation. Remember, when you add electrons, it's a reduction. Conversely, if you lose electrons, it's an oxidation, and you will see the electrons on the right side of the equation.
If both half-reactions are written as reductions, you must turn one of the them backward into an oxidation half-reaction. (If something is being reduced, something else better be getting oxidized because those electrons have to come from somewhere.) The tough part is deciding which half-reaction to turn backward. If you have a galvanic cell (most common case), you will want to turn the half-reaction with the smaller reaction potential backward. (Don't forget that a number like -1.2 would be smaller than one like -0.3). You do this because galvanic cells will always have a positive Ecell (they have to, because they are spontaneous, and the equation that relates G with Ecell has a minus sign in it: G=-nFEcell). If the cell is electrolytic, then it is not spontaneous, Ecell should be negative (making G positive) and you will turn the half-reaction with the larger reaction potential backward instead. Remember, in either case, when you turn the half-reaction backward, you change the sign of that half-reaction's reaction potential.
Ok, so now let's consider your specific example. The first half-reaction (the Cu one) is a reduction half-reaction. Again, I know that because the electrons are being added on the left side; Cu is gaining electrons. The second one, in contrast, is an oxidation half-reaction. The electrons are on the right side; O is losing electrons. Good, so in this case, nothing needs to be turned around, because we already have one reduction and one oxidation. You can see that this cell is an electrolytic one as written. (Add the two reaction potentials together, and you are going to get a negative number for Ecell) Looking at the third equation that combines both half-reactions, you can see that it is written in the same fashion as the two half-reactions above it: Cu is getting reduced, and O is getting oxidized. Thus, again, you have an electrolytic cell, nothing needs to be turned backward because one partner is oxidizing and one is reducing, and you merely need to add the two reaction potentials together to get your -0.89.
Really, there isn't any need to do any math on this problem. As soon as you recognize that you have an electrolytic cell, there is only one possible right answer for this question, because the others are all positive for Ecell.
Medikit is right that you do not have to multiply the values of the reaction potentials by the number of moles of electrons. The reason why is that reaction potential is an intrinsic property; it is independent of the amount of material you have. That is, one gram of Cu has the same reaction potential as 1 kg of Cu does. (Temperature and density are other examples of intrinsic properties.) The properties where you do have to multiply by the number of moles are extrinsic properties; those do depend on the amount of material. For example, when you use Hess's law, it matters greatly how much material you have because the more compound you have, the more heat it will give off. (Energy and volume are some other examples of extrinsic properties.)
Hope this helps.
