Electrolysis of dilute sulfuric acid

-The electrolyte in this reaction is a (dilute) solution of H₂SO₄ and H₂O. This will become H+, SO₄²- and OH-. -Since there is only one positive ion that can be formed at the cathode, hydrogen is formed. Half equation for the reaction at the cathode: -2H+ +2e- -> H₂ (reduction) -At the anode, OH- is formed as it is less reactive than the sulphate (explained below). OH will undergo oxidation and become water and oxygen. Half equation for the reaction at the anode: -4OH²- -4e- -> 2H₂O + O₂ -Both gases are colourless but can be tested for by using the squeaky pop test and the relighting a glowing splint test for hydrogen and oxygen, respectively.