Balance the redox reaction in basic solution. What is the coefficient on the hydroxide ion, and on which side of the equation does it appear? (a) 2 OH^- in reactants (b) 4 OH^- in products (c) 4 OH^- in reactants (d) 3 OH^- in reactants

Video transcript

welcome back everyone balance the following redox reactions in basic solution. What is hydroxide or the hydroxide ions coefficient. And where does it appear in the equation. So we have an unbalanced reaction here, we have nitric acid reacting with business to produce nitrogen monoxide gas and our biz mutate catalon. And we're going to begin by recognizing that with redox reactions. To balance them, we need to use the half reaction method. So we have our first half reaction, which is going to be nitric acid, which is going to produce nitrogen monoxide gas. We're going to begin by bouncing non hydrogen and non oxygen atoms first. So looking at nitrogen, we have one mole of nitrogen on both sides of the equation. So now we can move on to bouncing oxygen recall that we bounce oxygen using water. So looking at our equation, we have two moles of oxygen on the react inside and one mole of oxygen on the product side. So we're going to expand our product side and add one mole of water as a product. But now with adding one mole of water as a product, we've introduced two moles of hydrogen to the product side where on the reacting side we only have one mole of hydrogen. So we're going to an extra call that we balance hydrogen using protons H plus. And so we're going to expand our react inside And add one mole of protons to the reactant side. That now gives us two moles of hydrogen on both sides of the reaction so that we have a bounced or we have our hydrogen atoms balanced. Now with Adams balanced, we need to balance net charge. We have a net charge of plus one on the reactant side and a net charge that is neutral on the product side. So recall that we add electrons to the more positive side. And so we would expand our reaction side and add one electron as a reactant. Now with Adams and charges balanced, we can move into our second half reaction. In our second half reaction, we have bismuth solid which produces our bismuth eight plus one carry on. So beginning by balancing non oxygen and non hydrogen atoms First we begin by bouncing bismuth but we see that we have one mole of bismuth on both sides. So bismuth is balanced. Moving on to oxygen. We have one mole of oxygen on the product side and none on the reactant side. So we're going to expand our react inside and as before we use water to balance oxygen. So we would add one mole of water by adding this one mole of water. We've now introduced two moles of hydrogen to this half reaction. And so now we would Add two moles of protons to the product side. So two moles of h plus That gives us two moles of hydrogen on both sides of the reaction. And now we're going to bounce the net charge. So we have a net charge of plus three on the product side and a net charge that is neutral on the reactant side as we did before we add electrons to the more positive, more positive side of the equation. So we would add three electrons to cancel out that plus three charge on the product side. And now we have atoms and charges balanced for the second half reaction. Now we need to add these two half reactions together, but before doing so, we need to make sure the electrons match. We have one electron in the first half reaction and three in the second half reaction. So we're going to take the entire first half reaction and multiply it by a coefficient of three. This is going to change our first half reaction to now. three moles of nitric acid Plus three moles of protons Plus three electrons yields three moles of nitrogen monoxide gas, plus three moles of water. Now, with electrons balanced, we can straightforwardly add these two half reactions by beginning by canceling out the three electrons here on the product side and then here on the react inside. We next want to see that we can cancel out this water with the three moles of water in the second half reaction, which now would leave us with two moles of water, sorry, three moles of water from the first half reaction. Now we're left with two moles of water and we can also cancel out the two moles of H plus in the second half reaction with the three moles of H Plus Now in the first half reaction, that leaves us with one mole of H Plus in the first half reaction. Now with as much as possible canceled out, we are going to bring down everything to come up with our overall equation. This leaves us with Bismuth solid. one mole of business solid Reacting with three moles of nitric acid Plus one mole of protons yields three moles of nitrogen monoxide gas plus one mole of bismuth eight caddy on Plus two moles of water. Recall that from the prompt? We are told that this reaction is occurring in a basic solution. And so because this is a basic solution and we added a proton to the react inside that made our solution a bit more acidic. And we need to counter this by adding an equivalent number of hydroxide. And so we're going to now have an overall reaction, that is one mole of bismuth solid plus three moles of nitric acid plus one mole of protons, adding an equivalent of hydroxide. We add one mole of hydroxide to the reactant side. And now we would need to add that to the product side because what we do to one side of the equation, we must balance out with the other side. So we have three moles of nitrogen monoxide gas plus one mole of bismuth eight caddy on Plus two moles of water. And just to make more room We have plus one mole of hydroxide on the product side as well. Next we want to recognize that hydroxide and protons will combine to form water. And so this gives us the next, the next simplification of our overall equation where we would have one mole of bismuth solid plus three moles of nitric acid plus one mole of water. When we combine one mole of protons with one mole of hydroxide forms three moles of nitrogen monoxide gas. I'm sorry this arrow here should be better. And then plus one mole of this Canyon Plus two moles of water Plus one mole of hydroxide. So just to be clear, this water here was formed from the combination of protons and our one mole of protons and our one mole of hydroxide. So now that we have water on both sides of the equation, we need to recognize that we can cancel it out so we can cancel out this one mole of water with the two moles of water on the product side, leaving us with one mole of water on the product side. And now we have our simplified overall reaction, which we can write out as one mole of bismuth solid plus three moles of nitric acid plus or yields since we canceled out that water, three moles of nitrogen monoxide gas Plus one mole of our business Plus one Mole of Water. And then plus one mole of hydroxide. And so going back to the prompt were asked hydroxide coefficient and where it appears in the equation, and we can clearly see from our overall balanced reaction, We have one mole of hydroxide on the product side, and so this is going to correspond to choice C as the correct choice. So choice C is the correct choice to complete this example. one mole of hydroxide is on our product side. After we bounce out our net redox reaction. I hope that everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.