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The reduction of O₂ to H₂O involves 4e⁻, but not 2e⁻, i.e.
O₂ + 4H⁺ + 4e⁻ → 2H₂O …… E° = +1.229 V

The reduction potential is a tendency for the half reaction. When the both sides of a half equation is multiplied by a whole number, the reduction potential is unchanged because the tendency is a constant.

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Method 1 :

Fe³⁺ + e⁻ → Fe²⁺ …… E° = + 0.769 V
O₂ + 4H⁺ + 4e⁻ → 2H₂O …… E° = +1.229 V

For the reaction :
O₂ + 4Fe²⁺ + 4H⁺ → 2H₂O + 4Fe³⁺ …… E°(cell)
The conversion from O₂ (O.N. of O = 0) to H₂O (O.N. of O = -2) is reduction.
The conversion from Fe²⁺ (O.N. of Fe = +2) to Fe³⁺ (O.N. of Fe = 3) is oxidation.

E° of the reaction, E°(cell) = E°(red) - E°(oxid) = (+1.229 V) - (+0.769 V) = +0.460 V

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Method 2 :

4Fe²⁺ → 4Fe³⁺ + 4e⁻ …… E° = -(+ 0.769 V) = -0.769 V
O₂ + 4H⁺ + 4e⁻ → 2H₂O …… E° = +1.229 V

Add the above two half equation, the overall equation is :
O₂ + 4Fe²⁺ + 4H⁺ → 2H₂O + 4Fe³⁺ …… E°(cell) = (-0.769 V) + (+1.229 V) = +0.460 V

Electrochemistry.....

Target reaction below, where Fe2+is oxidized to Fe3+, oxygen gas is reduced.
O2 + 4Fe2+ + 4H+ --> 2H2O + 4Fe3+

(4Fe2+ --> Fe3+ + 1e-) .......... .................. E = -0.77V ............. oxidation half-reaction
O2 + 4H+ + 4e- --> 2H2O ...... .................. E = +1.23V ............ reduction half-reaction
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4Fe2+ + O2 + 4H+ --> 4Fe3+ + 2H2O ..... E = +0.46V ............. balanced net ionic equation

E(total) is the sum of the cell potentials for the oxidation and reduction half-reactions. Pay attention to the way the half-reaction is written in the standard reduction table, and to the sign of E. For oxidation, the reduction reaction is reversed and the sign of E is reversed.

The cell potential is positive, therefore, the reaction is spontaneous

In manipulating potentials, we should only change the signs of the values, not the magnitude.

Thus, one does not multiply the E° by the number of electrons in the half reaction. In the case of the Fe2+ equation, you need not multiply by 4.

Fe2+ ==⇒ Fe3+ + e- E° = -0.769 V

O2 + 4H+ + 2e– → 2H2O E° = +1.229
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O2 + 4Fe2+ + 4H+ → 2H2O + 4Fe3+ E° = +0.46

Look at the following website:

https://www.chem.wisc.edu/deptfiles/genchem/netorial/rottosen/tutorial/modules/electrochemistry/05potential/18_52.htm