Balance each redox reaction occurring in acidic aqueous solution. a. PbO₂(s) + I^–(aq) → Pb²⁺(aq) + I₂(s) b. SO₃^2–(aq) + MnO₄^–(aq) → SO₄^2–(aq) + Mn²⁺(aq) c. S₂O₃^2–(aq) + Cl₂(g) → SO₄^2–(aq) + Cl^–(aq)

Question

Accepted Answer

hey everyone in this example, we need to balance the following redox reaction under acidic conditions. So what we should recognize is that because this is redox, we need to break into two half reactions where our first half reaction is going to be our an ion which produces as a product, the blooming liquid. And we should recognize that we went from a net negative charge to now a neutral compound. And so this would be our gain of electrons, meaning this is our reduction half reaction. Whereas for our second half reaction, we have our 10 as a two plus kati on Which produced 10 as a 4-plus cat ion as a product. And here we went from a net two plus charge on the reactant side to a net four plus charge on the product side, meaning we lost we lost electrons. So this would be the oxidation half of our reaction. Now our next step is to go ahead and make sure non hydrogen and non oxygen atoms are balanced. So looking at our first half reaction, we have bromine, we have two atoms of roaming on the product side and one atom of bromine on the reactant side. So we would balance bromine by placing a coefficient of two here and looking at our second half reaction, we have one mole of tin and one mole of tin on the reactant side. So 10 is already balanced in the second half reaction. So now our next step is to focus on balancing oxygen. So looking at the first half reaction we have because we have this coefficient to here, we have now six moles of oxygen on the reactive side. But we don't have any oxygen on the product side. And we should recall that to balance out oxygen, we're going to say that one mole of oxygen is equivalent to one mole of water. So we can use water to balance out oxygen, meaning that we would go ahead and add to our first half reaction. Six moles of water. So we would have two moles of our bromate, an ion producing are brimming liquid. Plus, we would add six Molds of water. And so now through adding these six moles of water to our first half reaction, we've now added hydrogen into our equation. So we have six moles or sorry, 12 moles of hydrogen on the product side. And we have no hydrogen on the reacting side. So we should recall that to balance out hydrogen we would use for one mole of hydrogen that is equivalent to one proton or one mole of H plus. And so we would use H plus to balance out hydrogen. So we would say that we can now expand our reaction side by saying our two moles of our bromate, an ion are going to be reacting with and we'll use purple here, one mole or sorry, we need 12 moles of H plus. So we would say plus 12 H plus. And so now our oxygen and hydrogen are balanced for our first half reaction. And looking at our second half reaction, we also do not have hydrogen present. So we would just maintain the second reaction as is so we would have 10, 2 plus producing 10 4 plus. So now our next step now that all of our atoms are balanced is to go ahead and look at the net charges of each of our reactions. So for our first reaction, looking at the product side first, we should recognize that both of our compounds are neutral. So we have a net charge of zero on the product side. However, on the reactant side here we have a coefficient of 12 in front of this plus one charge, meaning we have a plus 12 charge. And then from our coefficient of two in front of our bromate and I and we have two times negative one. So we would have plus 12 minus two. And that would leave us with a net charge of plus 10 on the reactant side. So we want to cancel out this net charge of plus 10. So that we have it being neutral on the to match with the product side here, meaning we would need to add electrons to get rid of those of that plus 10 net charge. So we would go ahead and say that We would want to add 10 electrons to this first half reaction. So we would now have our two moles of our bromate, plus 12 moles of H plus plus 10 electrons to cancel out that Plus 10 charge there on the reactant side, yields are blooming liquid Plus six moles of water. Now looking at our second half reaction, we would see that we went from a net charge of plus two to a net charge of plus four. So we want the charges to bounce out here for the second half reaction by getting a net charge of plus two on the product side, meaning we need to cancel out two electrons or we need to add two electrons here, which would give us a net charge of plus two now. So we would go ahead and rewrite the second equation so that we have one mole of 10, sorry. So that should just be 10 Yielding one mole of 10, 4 plus plus two electrons. To get that net charge of plus two on both sides. And so now that we have our electrons balanced for both of our half reactions, our next step is to make sure that we have the same amount of electrons in both of our half reactions, we currently have 10 electrons here and two electrons in our second half reaction. So now we want to go ahead and multiply the second half reaction by a factor of five. So that we have 10 electrons. And so this would give us now we would have five moles of our 10, 2 plus carry on, yields five moles of our 10 4 plus cat ion plus 10 electrons for the second half reaction. And so now we can go ahead and cancel out the second half reaction with the same things that appear on opposite sides of both of our half reactions. And so that would apply to our 10 electrons. So we can cancel out the 10 electrons here and here because it appears on the react inside in the first half reaction and on the product side in the second half reaction. And now we're just going to combine these two reactions now that we've canceled everything out to get our net balanced redox reaction. And what we would have for our final net redox reaction is now five moles of the two plus 10 catalon Plus two moles of the bromate. An ion Plus 12 moles of H plus yields five moles of the 10 4 plus cat ion Plus one mole of our blooming liquid, Plus six moles of water. And so this would be our final answer for our balanced 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

Balance each redox reaction occurring in acidic aqueous solution. a. PbO2(s) + I–(aq) → Pb2+(aq) + I2(s) b. SO32–(aq) + MnO4–(aq) → SO42–(aq) + Mn2+(aq) c. S2O32–(aq) + Cl2(g) → SO42–(aq) + Cl–(aq)

Verified Solution

Balance each redox reaction occurring in acidic aqueous solution. a. PbO₂(s) + I^–(aq) → Pb²⁺(aq) + I₂(s) b. SO₃^2–(aq) + MnO₄^–(aq) → SO₄^2–(aq) + Mn²⁺(aq) c. S₂O₃^2–(aq) + Cl₂(g) → SO₄^2–(aq) + Cl^–(aq)