Write equations for the half-reactions that occur at the anode and cathode for the electrolysis of each aqueous solution. a. Ni(NO3)2(aq)

Video transcript

everyone in this example, we're given the electrolysis of lead to I date and we need to determine the half reaction equations that occur at the anodes and cathodes. So we're going to write out our a quick solution for lead to I date and this is going to dissolve into the following ions where we form the lead two plus catalon as well as two moles of ri date an ion. So we should recognize that our half reactions will come from these products. So for our first product or sorry for our first half reaction, we have our PB two Plus Catalon which forms solid lead as a product. Now we should recognize that we do have bounced atoms of lead here. However, we have a net charge of plus two on the reactant side and a net charge of zero on the product side. So we want to cancel out this charge. This net charge of plus two on the reactant side by canceling out by adding two electrons. Now because we added two electrons to the react inside here, we should recall that this means that this half reaction occurs as a reduction. Our next step is to write out our second half reaction where we have Just one mole of our I date an ion which produces 1/2 most of our di atomic molecule iodine. Now we should recognize that even though our iodine is balanced because one half times two is just one. We would go ahead and recognize that we're missing oxygen on this product side. We have three moles of oxygen on the reactant side. And so we should recall that to balance oxygen, one mole of oxygen is equivalent to one mole of water. So we're going to use water to balance out oxygen. And because we're missing three moles of oxygen on the product side, we're going to fix that by Adding three moles of water to the product side. But now we've also introduced six moles of hydrogen to the product side, which we don't have on the react inside here. And so we should recall that to balance hydrogen, we're going to recall that one atom of hydrogen is equal to one mole of hydride or a proton. So we would go ahead and fix this by expanding our reactant side And bouncing the hydrogen by adding six moles of our hydride. Our next step is to recognize the charges here. So right now we have a net charge on the reactive side of six times plus one, which is plus six and then minus one. So we have a net charge of plus five here, whereas on our product side we just have neutral compounds. So we have a net charge of zero. So we need to cancel out this net charge of plus five on the reacting side here. And in order to do so, we're going to go ahead and expand our react inside again By canceling out that Plus five charge by adding five electrons. So now we have balanced charges for this for both of our first two half reactions. But we're also going to have another half reaction because we have the electrolysis of an Aquarius solution, meaning that we're going to have the electrolysis of water occur as well. And so that means that water we should recognize is going to form hydrogen gas as a product. Now we want to go ahead and make sure that even though our hydrogen we can see are balanced, we need to make sure our oxygen's are balanced. And so we would balance out oxygen in this case with hydroxide. So if we add hydroxide to the product side, we would now have one mole of oxygen but now three moles of hydrogen. And so we would instead add two moles of hydroxide which would give us now two moles of or four moles of hydrogen rather and two moles of oxygen. And so we would go ahead and bounce us out by placing a coefficient of two in front of our water, meaning we would need two moles of water. So our last step is to make sure that this is balanced as far as the charges. So on our reactant side we have a net charge of zero. But on our product side we have two times this minus one charge, meaning we have a net charge of -2. And so we need to go ahead and gain a net charge of -2 on the reactant side here. So we're going to expand our react inside here And gain a net charge of -2 by adding two electrons here. And so we have a net charge of -2 on both sides of this equation, meaning that this half reaction is also balanced as far as the charges. And the atoms. Now, what we should recognize is that we've added electrons on all of the reacting sides of these half reactions, meaning that all of these half reactions are reductions. So we can say that these are all reduction reactions, all reduction half reactions. And we would recognize that this makes sense because when we look at the oxidation state of lead to I date, we see that it has an oxidation state Or oxidation number of Plus five. And that is due to the fact that it's really difficult to oxidize lead to I date. And so that is why we ended up with all reduction half reactions here so far. Now we should recall that our reduction reactions will all occur at the cathode of our voltaic cell. So right now, we've determined three possible half reactions that occur at our cathode, which I'm highlighting here in yellow. So, these are our first three answers for our possible half reactions that will occur at our cathode. And now we want to determine what would possibly occur at the anodes now because we have water involved in this reaction. We can say for our possible an ode reaction, we would use water. So we have our water which should oxidize too oxygen gas. And now we need to make sure that this half reaction is balanced. So as far as our oxygen atoms, we do not have a balance here. So we would go ahead and because we have two oxygen atoms on the product side and just one here on the one oxygen atom on the reactant side, we're going to add a coefficient of two in front of our water, which will now give us two oxygen's on both sides of this reaction. But now we have four hydrogen on the reactant side. And so to balance out hydrogen, we're going to go ahead and recall that we can use hydride to balance out hydrogen. So we would add for four moles of hydrogen on the product side, we would add four moles of hydride. So now we have our four atoms of hydrogen on both sides of our equation. So all of our oxygen and hydrogen atoms are balanced. But now we need to make sure the net charge for this half reaction is balanced. So on our reactive side we have a net charge of zero because we have a neutral compound of water. However, on our product side we have a net charge of this coefficient of four times plus one, which would give us a net charge of plus four on the product side here. So we need to go ahead and cancel out this net charge of plus four so that we have a net neutral charge for both sides of our reaction. And so in order to cancel out this net charge of plus four, We're going to recall that we're going to add four electrons here to the product side. And this would make sense that this would be our possible an odd reaction because we would recall that when we add electrons to the product side, this half reaction occurs as an oxidation. And recall that our oxidation half reaction occur at the anodes. So, for our fourth final answer, we're going to have this equation here, which is are possible half reaction that occurs at our anodes. So everything highlighted in yellow represents our possible half reaction equations that can occur at the anodes as well as our cathode. So I hope that everything I explained was clear. If you have any questions, please leave them down below. Otherwise, I will see everyone in the next practice video