The overall reaction is
: 4 Fe – cytochrome c + 4 H + O → 4 Fe – cytochrome c + 2 HO                ΔG = - 218 kJ/mol,  E = +565 mV
Two electrons are passed from two cytochrome cs, through the Cu and cytochrome a sites to the cytochrome a–Cu binuclear center, reducing the metals to the Fe form and Cu. The hydroxide ligand is protonated and lost as water, creating a void between the metals that is filled by O. The oxygen is rapidly reduced, with two electrons coming from the Fe-cytochrome a, which is converted to the ferryl oxo form (Fe=O). The oxygen atom close to Cu picks up one electron from Cu, and a second electron and a proton from the hydroxyl of Tyr(244), which becomes a tyrosyl radical. The second oxygen is converted to a hydroxide ion by picking up two electrons and a proton. A third electron from another cytochrome c is passed through the first two electron carriers to the cytochrome a–Cu binuclear center, and this electron and two protons convert the tyrosyl radical back to Tyr, and the hydroxide bound to Cu to a water molecule. The fourth electron from another cytochrome c flows through Cu and cytochrome a to the cytochrome a–Cu binuclear center, reducing the Fe=O to Fe, with the oxygen atom picking up a proton simultaneously, regenerating this oxygen as a hydroxide ion coordinated in the middle of the cytochrome a–Cu center as it was at the start of this cycle. Overall, four reduced cytochrome cs are oxidized while O and four protons are reduced to two water molecules.