The fuel cell of h2 + o2 --> 2h2o has a cell potential of 1.23 V... If The equation for Ecell = Cathode - Anode, why wouldn't the cell be -.43 because anode = oxidation and cathode = reduction, right?

In electrochemical cells, the cell potential is determined by the difference between the reduction potentials of the cathode (where reduction occurs) and the anode (where oxidation occurs). The equation for calculating the cell potential (Ecell) is indeed Ecell = Cathode - Anode.

However, in your example of the fuel cell reaction 2H2 + O2 → 2H2O, you need to consider the reduction potentials of the half-reactions involved.

For the cathode, the reduction half-reaction is:
2H+ + 2e- → H2

For the anode, the oxidation half-reaction is:
O2 + 4H+ + 4e- → 2H2O

To calculate the cell potential, you need to know the reduction potentials of these half-reactions. The standard reduction potentials for these reactions are known. The standard reduction potential for the cathode half-reaction is 0 V (by convention), and for the anode half-reaction, it is 1.23 V.

Since the cell potential is calculated as Ecell = Cathode - Anode, with the anode potential being subtracted, the calculation would be:
Ecell = 0 V - 1.23 V = -1.23 V

So, the cell potential for the given fuel cell reaction is -1.23 V, not -0.43 V as you suggested.

It is important to note that the negative sign indicates that the reaction is not spontaneous under standard conditions. In this case, the fuel cell reaction requires an external energy source to proceed, such as the input of electrical energy.