So, this is the products of lekylisys part 2. First question is, elaclysis of accus msCl using platinum electrode. Again, you know platinum electrode in our electrode. Now, accus msCl give mn plus sign CL-1 and accus means water that give H3 plus I's and OH-1. Again, compare auto nA plus and H3O plus, you know H3O plus will gain electron and we get half of H2 gas plus H2 liquid. And what about CL-1 OH-1? You know CL-1 has lesser discharge potential. So, CL-1 lose electron to give half CL2 gas plus electron. It is at cathode and this is at a note. Now, come to the next part. If we are using mercury as cathode, then what would happen? So, see the change. Here in this case, when we use platinum electrode, then accus msCl has lesser discharge potential. But when we use mercury as cathode, then sodium positive has lower discharge potential than H3O plus and sodium will gain electron to give sodium. And in the presence of mercury, it forms sodium and welcome. Sodium welcome in aqueous solution, it gives sodium hydroxide half of H2 gas and mercury. In case of dilute and acL, it is just like only H3O plus sign will gain electron to give half H2 plus H2 liquid. And at anode OH-1 loses electron. It gives plus 4. And the electrolysis of H2 is so 4, first case at low current density. You know H2 is so 4 give, H plus sign and as so 4 2 negative sign, water will give H3O plus and OH negative. These are ions from H2 so 4 and these are the ions from water. Again at low current density, out of H3O plus and H plus cathode H3O plus put gain electron to give half H2 gas H2 liquid. And at anode OH negative ion will lose electron to give to H2O plus 4 electron. Now when current density is high, high current density at 50% H2O4. In this case at cathode reaction remains same but at anode. At a so 4 negative ion loses electron to give H2 H2O8 plus 2 electron. And this H2 H2O8. And the still with water give H2O2 like first it is hydrolyzed to give H2O4 plus H2O5. And then H2O5 give H2O4 and H2O2. This structure is like this. This is peroxodislifuric acid also known as Marshall acid. And when it is treated with water, it give H2O4 and H2O5. This H2O5 is peroxod monosylifuric acid also known as carose acid. Electrolysis of RCONA using platinum electrode. You know this is always electrolysis. In this case we have RCONA plus and from water we have OHNA as Q plus. So at cathode reaction simple, electrode is platinum. So obviously H3O plus Gaye electron. Now at anode tell me out of RCONA, OHNA keep in mind RCONA has lower discharge of radiation. RCONA gives RCO. which is called free radical and 2L3 radical combined together to give an LK. So overall reaction is RCONA2. So this is 2L2 to give LK plus 2 CO2 plus 2 electrons.