Balance each redox reaction occurring in basic aqueous solution. c. NO₂^–(aq) + Al(s) → NH₃(g) + AlO₂^–(aq)

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hey everyone in this example, we need to consider the redox reaction that is below and we need to balance this under basic conditions. So what we're going to do is create half reactions. So from our first reactant we have our di chrome eight an ion, so cr two 072 -. And for our first half reaction, this is going to produce our product being our chrome eight Catalan with a three plus charge. And then for our second half reaction, we have our second reactant methanol, so ch three ohh, producing our formic acid, H C 02 H. So focusing on our non hydrogen and non oxygen atoms, we have our chromium atoms, we have one on the product side here and two here on the react inside. And to balance out our chromium atoms, we need to place a coefficient of two in front of our product here. So now we have two on both sides. So we can say that chromium is balanced for both of our equations. However, looking at the non oxygen and non um non hydrogen atoms in our second half reaction, we have carbon here and we have one mole of carbon on the product side and one mole of carbon here. So we can say that carbon is balanced for the second half reaction. So we have carbon balanced and now we're going to focus on balancing our oxygen and hydrogen in each of these reactions. So first we're going to balance oxygen and we should recall that we can use water to balance our oxygen atoms by recalling that one mole of oxygen is equal to one mole of hydrogen. Or sorry, water. So in our first reaction we have our die crow mate and we need to balance out the oxygen's here for both sides. So this side is missing oxygen period and we have seven moles of oxygen on the reacting side. So we would go ahead and add oxygen to our product side by adding water. So we would go ahead and add seven moles of water Since we need seven moles of oxygen. So now this first half reaction has oxygen balance. So we can say oxygen is balanced and moving on to our second half reaction, we have formic acid where we have a total of just one mole of water. This is producing our product, we have formic acid, so H CO two H. And we would recognize that we have two moles of oxygen here versus one mole of oxygen in our methanol. So we need to bounce out this oxygen by using water again and to get two moles of oxygen on the reacting side. We would go ahead and add one mole of water so just plus H +20. And so now we can say we have oxygen balanced here for this half reaction. However now we need to go ahead and balance our hydrogen out. So looking at our first half reaction, we would recognize that we have 14 hydrogen because we added the seven moles of water. So we need 14 hydrogen is on the react inside here. And for our second half reaction, we have a total of six Hydrogen is here. So we need six hydrogen on our product side here because we only have two so far Meaning we would need to just add four hydrogen and we should recall that one mole of hydrogen is equivalent to one proton, so one mole of H plus. So we would go ahead and add for our first half reaction. We have die crew mate, we would now have plus 14 and sorry, let's just do this in purple. So plus 14 h plus produces our two molds of our chrome eight caddy on plus r seven moles of water. And so now we can confirm that our, sorry hydrogen is balanced and move on to our second half reaction. Where we said we need to add formals of H plus on the product side. So what we would have is CH 300 h Plus our one mole of water yields our formic acid plus we're adding plus our four moles of H plus to balance out our hydrogen on both sides of the reaction. And so now we can say by adding this, we've balanced hydrogen on both sides of this half reaction. So now our next step is to balance out our charges here and we should recognize that because we have a coefficient of two in front of our three plus chromium. We have on this product side here, a net charge of plus six. However, on the reacting side over here we have A coefficient of 14. So that would give us a plus 14 charge here, plus that negative two would give us a plus 12 charge net here. And so we need to go ahead and bounce this charge out By adding electrons to the more positive side. And we would say that because we have a coefficient of 14 in front of our plus one charge that this is the more positive side of our reaction here. And so we would go ahead and add the electrons to bounce out for a net charge of plus six to bounce out with this side. So we would have a Net charge of Plus six on both sides of our equation. And so we would get that by, as we said, adding six electrons. So for our first half reaction we should now have our die crow mate, plus now we are adding six electrons here to bounce out that net charge and then we have the same the same product side here. So looking at our second half reaction, we have on the product side here a net charge of plus four because we have a coefficient of four in front of the plus here. And then on our reactive side we have a net charge of zero. So we want to have balanced charges for both sides. And so we need to get rid of this plus four charge by adding four electrons. And we're gonna show that below. So we would have our formic acid. Sorry. So let's make more room plus water yields our so that's methanol plus water yields are formic acid. And then we would Still have our four h plus. But now to get rid of that plus four charge, we want to have a net charge of zero. So we would add four electrons here. So now that we have, our charge is balanced for both of our half reactions. We need to also make sure that our electrons match for both of these half reactions. And they currently don't because we have six electrons here and four electrons here. And so we're going to multiply by a coefficient, our entire reaction So that the electrons will balance. So we're going to multiply this entire second reaction by three and we can multiply the entire first half reaction by two. And so this is going to give us a new value for electrons for our first reaction, we're now going to have two moles of our di chrome eight Plus 28 of our H plus Plus 12 electrons now Yields four moles of our die cremate or sorry, or cremate Canyon Plus we have moles of our water. And then multiplying our second reaction by three. We would have three moles of methanol Plus three moles of water is going to yield three molds of our formic acid plus we would have 12 moles of H plus plus 12 electrons. And so as you can see, we have balanced out both the electron values for both of our half reactions. And now we want to cancel out what's opposite and the same for both half reactions. Or rather what is the same on opposite sides of both of our half reactions. So we can go ahead and cancel out the 12 electrons here because it's on the reactive side here and on the product side here, we would also see that water is on opposite sides. So we would subtract the 14 moles of water from the three moles of water on the reactant side for the second half reaction. And this would give us now 11 moles of water left. We can also get rid of the H plus because we have 28 here and here. So we would take 12 from 28. And that would give us a difference of 16 H plus is left for our first half reaction. And so now after we've canceled everything out, we're going to add these two equations up. So that now we have our overall equation which is going to be our two moles of our dicho mate plus 16 H plus. That was left over after we canceled out plus three moles of our formic acid. Sorry, our methanol here, this is methanol yielding our formals of the Chrome eight caddy on. And I'll just make more room. So this is plus are now 11 moles of water that we had left over Plus our three moles of our formic acid. And so now we need to balance our H plus because we have 16 moles of H plus on the reactant side and none on the product side. So we're going to change that by adding an equal amount of hydroxide to both sides of our equations to balance out our age plus. So we would go ahead and have instead two moles of our dick, roommate Plus 16 h plus plus 16 hydroxide Plus three moles of our methanol Yields four moles of our chrome eight cat ion Plus 11 moles of water plus three moles of formic acid. And then plus our 16 moles of hydroxide. And just so that it's all in view, just going to move this over just a bit and we should recall that H plus and hydroxide combine to form water. So this would give us a new equation where now we have water in the mix. So we have two moles of die cremate Plus because the H plus and hydroxide combined, we have now 16 moles of water Plus three moles of our methanol Yields four moles of our chrome eight caddy on Plus 11 moles of water from combining hydroxide and water or sorry hydroxide and h plus. So 11 moles of water formed from that Plus We have three moles of formic acid. And just so everything is in view. So three moles of our formic acid Plus our moles of hydroxide. And so now we can cancel out what is the same on both sides of our equation. So we can get rid of the water by taking the 11 moles of water here from R16, which would leave us with five moles of water. And so finally, for our final equation, what we should have is two moles of di chrome it Plus five moles of water Plus three moles of methanol yields four moles of our chrome eight caddy on plus, sorry, AQ. So plus three moles of our formic acid Plus our 16 moles of hydroxide. So that says hydroxide here, 16 Molds of our hydroxide. And this here would be our final answer for our redox reaction balanced under basic conditions. 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

Balance each redox reaction occurring in basic aqueous solution. c. NO₂^–(aq) + Al(s) → NH₃(g) + AlO₂^–(aq)

Verified Solution

Key Concepts

Redox Reactions

Balancing Redox Reactions in Basic Solution

Half-Reaction Method

Balance each redox reaction occurring in basic aqueous solution. c. NO2–(aq) + Al(s) → NH3(g) + AlO2–(aq)

Verified Solution

Key Concepts

Redox Reactions

Balancing Redox Reactions in Basic Solution

Half-Reaction Method

Balance each redox reaction occurring in basic aqueous solution. c. NO₂^–(aq) + Al(s) → NH₃(g) + AlO₂^–(aq)