Balance each redox reaction occurring in acidic aqueous solution. b. Mg(s) + Cr3+(aq) ¡ Mg2+(aq) + Cr(s)

Video transcript

hey everyone in this example, we're given the following redox reaction and we need to bounce this under acidic conditions. So because this is a redox reaction, we need to break this up into two half reactions. Our first half reaction is going to be our solid barium which produces the barium two plus catalon. And we should recognize that we went from a net neutral charge to a plus two charge overall, which means that we lost electrons. So this would be our oxidation half of our reaction. And moving on to our second half reaction, we would have our iron three plus catalon that produces our iron as a solid atom. And so we went from a net three plus charge to a net neutral charge. So this would mean we gained electrons meaning this is our reduction half of our reaction. Our next step is to make sure non oxygen and non hydrogen atoms are balanced. So looking at our first half reaction, we have one mole of barium and one mole of barium on the reactive side. So barium is balanced. And looking at our product side, we have one mole of iron and one mole of iron on the reactive side. So iron is balanced as well. We also do not have oxygen or hydrogen present in either of our half reactions. So we would say that both of these half reactions as far as the atoms are balanced. Our next step is to make sure that the charges, the net charges are balanced for both half reactions. So looking at our first half reaction, we have a net charge of plus two on the product side. However, on the reactant side we stated that we have a net charge of zero because this is a neutral compound. So we need to go ahead and get a net charge of zero on this side by adding two electrons. So you would say Plus two electrons. So that we would have a net neutral charge on both sides of our reaction here. Now looking at our second half reaction, we have a net charge of plus three on the reactant side. And on our product side we have a net charge of zero because this is a neutral atom of iron. And so we need to cancel out this net charge of three plus here by Adding three electrons to our reactant side. So we can just make more room and say Plus three electrons here to give us a net neutral charge for both sides of our equation. Now, it's important that the electrons in both of our half reactions equal the same amount for both half reactions. So we need to go ahead and multiply both of our half reactions by a factor so that we get the same amount of electrons. So for our first half reaction, we're going to go ahead And multiply this by three. And then for our second half reaction we're going to place it in brackets and multiply the entire reaction by two. And so what we would say is that now we have two new half reactions where our first half reaction is now three moles of our solid barium producing Our now three moles of our barium to Pluskat ion plus six electrons. And then for a second half reaction we would have three moles of the iron, three plus cat ion Plus six electrons, yields two moles of our solid iron as a product. And so now you can see we have our charges balanced as well as the electrons balanced. And so now we can cancel out what appears on opposite sides of both of our reactions, but is the same. And that would be our two reactions. We have the same amount of electrons here on the product side in the first half reaction and then here on the react inside in the second half reaction. And so now we're just going to combine these two equations now that we've canceled everything out so that we now have three moles of our solid barium reacting with three moles of our iron, three plus cat ion. And on our product side we would produce three moles of R two plus barium Added to two moles of our solid iron. And just to make a quick correction, we multiply this second half reaction by two. So we should correct this so that this says two moles of our iron three plus catalon. And so that our final answer will say two moles of our iron, three plus cat ion in the redox reaction. And so now it should be clear that this here is our final answer as our bounced redox reaction under acidic conditions. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.