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?
The equation you mentioned, Ecell = Cathode - Anode, is used to calculate the cell potential of a fuel cell or any other electrochemical cell. However, it's important to understand the sign conventions involved.
In the case of a fuel cell, the half-reaction at the cathode involves reduction (gain of electrons), while the half-reaction at the anode involves oxidation (loss of electrons). The convention is to assign a positive sign to the potential of the reduction half-reaction (cathode) and a negative sign to the potential of the oxidation half-reaction (anode).
In the given reaction, h2 + o2 -> 2h2o, the reduction half-reaction is the reduction of oxygen (O2) to water (H2O) at the cathode:
O2 + 4e- + 4H+ -> 2H2O
The standard reduction potential for this half-reaction is 1.23 V, which is positive.
On the other hand, the oxidation half-reaction is the oxidation of hydrogen (H2) to protons (H+) at the anode:
H2 -> 2H+ + 2e-
The standard reduction potential for this half-reaction is 0 V, as it is used as the reference electrode.
Substituting the values into the equation Ecell = Cathode - Anode:
Ecell = 1.23 V - 0 V
Ecell = 1.23 V
Therefore, the cell potential for the given fuel cell reaction is 1.23 V, not -0.43 V. The negative sign indicates oxidation at the anode, but it does not affect the overall sign of the cell potential.