What are the products of the overall reaction in the elec-trolysis of an aqueous solution of sodium hydroxide? (Refer to Table 19.1 for standard reduction potentials.) (a) Na(s) and O2(g) (b) H2(g) and O2(g) (c) Na(s) and H2(g) (d) Na(s) and H2O2(aq)

Video transcript

Hello. Everyone in this video, we're trying to see which of the following are going to be the end products of the overall process. That takes place when an acquis barium iodide solution is electoral ized. Alright, so we're gonna have to determine the cathode half reaction and the reaction for the electrolysis of this solution. So the possible cathode half reaction. Let's go ahead and write this out. So that is two moles of H 20 in its liquid state, reacting with two electrons to yield one mole of H two gas, two moles of hydroxide. The standard reduction potential for this half reaction is negative 0. volts. My second possible cathode half reaction is when we have one mole of the barium two plus cat ion reacting with two electrons to yield one mole of my barium in its solid state. The standard reduction reduction potential is equal to negative 2. volts. Since the reduction potential of water has a less negative value than the reduction potential of bay two plus ion. Then the water is more likely to be reduced than our B A two plus ion. So the cattle have reaction. Then it's going to be the one above which I'll go ahead and just circle. So this one right over here and then for my possible a node half reaction, The choices that we have here is going to be two moles of my H20. And its liquid state reacting or yielding one mole of 02 gas as well as four moles of my H plus in its agree a state As well as for electrons, the standard reduction potential here is negative 1.23V. My second possible and would half reaction is going to be two moles of our iodine or I minus, which is a quiz in This yields one mole of R I two and a solid state as well as two electrons. The standard reduction potential here is equal to negative 0.54 volts. So we can see here that I minus will be oxidized because although H D O is easier to be oxidized, the formation of gasses, 02 needs an over voltage to form. So the answer would have reaction that we're gonna go ahead and choose from. This possible list is going to be the bottom one. So this one I'm gonna go ahead and circle so now that we have these two half reactions, we can find the overall net reaction. But just to make it more clear here, I'm gonna go ahead and first right out my possible or not possible but my actual a node and cathode half reactions again, starting off with my cathode that we have written in red, That is the two moles of H20 in its liquid state reacting with two electrons that yields are H2 gas as well as two moles of our hydroxide. And then for our anodes, which we had written in blue earlier, this is equal to two moles of R I minus which is a quiz. And this yields one mole of I two in a solid state as well as two electrons. So how we get the overall net reaction is seeing what cancels out. And the only thing that I see canceling out is going to be our two electrons here. Since in separate size of our reaction arrow, this means that my overall net reaction For my starting material side is going to be two moles of my H20 in its liquid state as well as two moles of my I minus and ion. This goes ahead to yield one mole of H two guests, as well as two moles of hydroxide. And of course as well as our eye to in its solid state. So my answer choice then from out of a through B or a through D rather is going to be answer choice A where we have a church two and I two solid. So this right here is going to be my final answer for this problem.