Half cells: difference between E(cell) = E(cathode}-E(anode) & E(cell) = E(oxidation) + E(reduction)
Ah, I see what you did wrong.

The first equation is correct. 

$$ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} $$

The second equation should be:

$$ E^\circ_\text{cell} = E^\circ_\text{reduction} - E^\circ_\text{oxidation} $$

Alternatively, you can write it as:

$$ E^\circ_\text{cell} = - E^\circ_\text{oxidation} + E^\circ_\text{reduction} $$

The cathode is actually the reduction step, and the anode is the oxidation step.

A useful mnemonic I learned from my chemistry professor is a red cat and an ox. Interestingly, cathodes are red and are positive. So I have found many uses for this along the way.