4Te(OH)₂ + H₂O + O₂ → 4Te(OH)₃

This equation shows the oxidation of tellurium(II) hydroxide to tellurium(III) hydroxide. It involves 4 moles of Te(OH)₂, 1 mole of water, and 1 mole of oxygen reacting to form 4 moles of Te(OH)₃. This indicates an increase in the oxidation state of tellurium from +2 to +3.