A disproportionation reaction is an oxidation–reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (b) MnO₄^2-(aq) → MnO₄^-(aq) + MnO₂(s) (acidic solution)

Video transcript

welcome back everyone when the same substance is oxidized and reduced in an oxidation reaction, it is referred to as a disproportion ation reaction, complete and balanced the following disproportion ation reaction in an acidic solution. So we're given our diet to two minus an eye on producing our thio sulfate to minus an ion and are by sulfate one minus an ion correction by cell fight one minus an eye on. So what we need to do based on this overall reaction is to write out two half reactions. So we're going to begin with our diet tonight to minus an ion. So S 2042 minus which produces as a product, our sulfate to minus an ion. So S 2032 minus. This is our first half reaction. And for our second half reaction we have our deo sulfate an ion. So S 2032 minus and correction, We begin with our di thio night two minus an eye on again as our second half reaction because it's our only reactant where this time our product is going to be our by sulfite and ion. So we need to come up with a new overall equation by adding these two together. But we need to make sure that the atoms as well as charges are balanced. So beginning with balancing out our atoms, we recognize for our first half reaction that we have two sulfur atoms on both sides. However, we have four oxygen atoms on the reacting side and three on the product side. And we want to recall that we balance oxygen using water. So for this first half reaction, we're going to go ahead and add one mole of water to the product side, which will now give us four moles of water on the product side to match with the react inside. But now we've introduced hydrogen and so we want to recall that we can balance hydrogen using our hydride cat ion. And so because we have now two moles of hydride on the product side, we're going to expand our react inside here and add two moles of hydride. And so now we have our atoms balanced on both sides of our first half reaction. Let's see our second half reaction. If the atoms are balanced, we see we have two moles of sulfur on the reacting side here, where we have one on the product side. So we're going to place a coefficient of two in front of our by sulfite and ion on the product side where now we have two of our sulfur atoms on both sides. We need to bounce out the oxygen here. So because we have three on the product side, we're going to add, sorry correction. We have three times two, which gives us six moles of oxygen on the product side. So we're going to fix that by expanding our reacting side here and adding two moles of water on the product on the reactant side. Where by doing so we can see that we now have a total of six moles because we would have four plus two here of our oxygen. We now have two moles of our sulfur on both sides. And now we have a total of four moles of hydride on the reactant side here. And so we call that to balance out hydrogen, we will add two moles of hydride on the product side, where we will now have four moles of hydrogen on both sides of our reactions for the second half reaction. So now we have all atoms balanced in both of our half reactions. And now we need to see if the charges are balanced in both of our half reactions. So for our first half reaction, we have an overall charge on the reactant side being minus two plus one, which would be minus one here on the react inside. Where on the product side we have an overall charge of minus two. And sorry correction. So we have a coefficient of two here. So that would be a plus. To canceling out the -2 years, meaning we have a net charge of zero on the reactant side where we have a net charge of -2 on our product side. So we need to cancel out Or rather introduce a charge of -2 on the reactant side. And to do so we're just going to add electrons to our reactant side as a third reactant. So we'll add three electrons. Or sorry, two electrons Two electrons to give us an overall charge of -2 on both sides of our reaction. And now focusing on our charge of our second half reaction. We have a net charge of -2 on the reactant side. Where on the product side we have a net charge of two times negative. One being -2 there plus two here, which will cancel out to zero. So to go from that zero to a minus two charge on this side, we're going to add two electrons to our product side here which will now give us a minus two net charge, meaning our charges are balanced for both of our half reactions. Now that we have balanced charges on both half reactions as well as identical number of electrons. We cannot add up these half reactions together to come up with an overall reaction. Now adding them up, we're going to cancel out our electrons here on the react inside and here on the product side. Also canceling out whatever else we have. Similar, we can get rid of our two moles of hydride here on the reaction side where we have two moles of hydrogen on the product side, in our second half reaction and we can cancel out one of the two moles of water here, leaving us with one mole of water with this one mole of water here in the first half reaction. And now we'll carry down everything that we're left with. So we're left with when we add our one mole of our finite an ion from the first half reaction to the one mole of die tonight. An ion from the second half reaction that gives us a total of two moles of our diet tonight. An ion, this is our only or rather we have a second reactant being our one mole of H 20. So that we were left over with plus H +20 liquid as our second reactant. And now going to our product side with what we're left with, were left with our one mole of RTO sulfite or sulfate an ion. So as to oh +32 minus Plus our two moles of our by sulfate and ion. And so this is going to complete this example as our complete and balanced overall reaction here from our disproportion ation reaction in acidic solution. I hope everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video