During the preparation of H2S2O8 (per disulphuric acid) by the electrolysis of 75% aqueous Sulphuric acid solution, O2 gas is also released at anode as by-product. If 15.12 L of H2 gas is released at cathode and 5.04 L O2 gas is released at anode at STP, the weight H2S2O8 produced in gram is

2H2SO4=4H+1+2SO4-2 Upon electrolysis by the application of voltage of about 2.1V, At Anode (Oxidation): 2SO4-2→S2O8-2+2e-2;  Also, water undergoes Oxidation as the required potential is only 1.23V 2H2O→4H+1+O2 +4e-1 Altogether, 6e-1 are released during the process of Oxidation. At cathode (Reduction): 6H+1+6e-1→3H2 The over all redox reaction is: 2H2SO4+2H2O→6e-1O2+ 3H2+H2S2O8 The number of the moles of H2, O2 and H2S2O8 will be in 3:1:1 ratio. Number of mole of H2=volume of H2 at STP 15.12LGram molar volume 22.4L=0.675 mole Number of mole of O2=volume of O2 at STP 5.04LGram molar volume 22.4L=0.225 mole No. of mole of H2S2O8 formed=0.225 mole, same as the number of mole of O2 Gram molecular weight of H2S2O8=194 g mole-1 Mass of H2S2O8 formed=194 x 0.225=43.65g