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Balance each of the following equations according to the half-reaction method:(a) $\mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2}(g)(\text { in acid })$(b) $\mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{NH}_{3}(a q)(\text { in base })$(c) $\mathrm{CuS}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{S}(s)+\mathrm{NO}(g)(\text { in acid })$(d) $\mathrm{NH}_{3}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)(\text { gas phase })$(e) $\mathrm{Cl}_{2}(g)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}_{3}^{-}(a q)(\text { in base })$(f) $\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{O}_{2}(g)(\text { in acid })$(g) $\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}_{2}^{-}(a q)(\text { in base })$(h) $\mathrm{Fe}^{3+}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{I}_{2}(a q)$

(a) $52 \mathrm{n}(s)+2 \mathrm{NO}_{3}^{-}(a q)+12 \mathrm{H}^{+}(a q) \longrightarrow 5 \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$(b) $4 \mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{Zn}^{2+}(a q)+\mathrm{NH}_{3}(a q)+9 \mathrm{OH}^{-}(a q)$(c) $2 \mathrm{CuS}(s)+\mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Cu}^{2+}(a q)+2 \mathrm{S}(s)+\mathrm{NO}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$(d) $4 \mathrm{NH}_{3}(a q)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$(e) $13 \mathrm{Cl}_{2}(a q)+24 \mathrm{OH}^{-}(a q) \longrightarrow 4 \mathrm{ClO}_{3}^{-}(a q)+12 \mathrm{H}_{2} \mathrm{O}(l)+22 \mathrm{Cl}^{-}(a q)$(f) $5 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow 5 \mathrm{O}_{2}(g)+5 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)$(g) $2 \mathrm{I}^{-}(a q)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{Fe}^{2+}(a q)$

01:41

Sisi G.

Chemistry 101

Chapter 4

Stoichiometry of Chemical Reactions

Chemical reactions and Stoichiometry

Cole H.

March 5, 2021

This video does not help you learn how to solve it at all.

University of Kentucky

Brown University

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Mhm. This next question contains multiple parts and each part is quite lengthy in each part. You're asked to balance the following redox reactions using the half reaction method. So let's just get started with the first one. We have zinc reacting with nitrate producing zinc ion and nitrogen gas. So let's separate the zinc parts into a half reaction. Will have zinc solid going to zinc two plus, will balance the sinks, their balanced. If there were hydrogen and oxygen would balance them accordingly. We don't have any, so we just balance charges, no charge two plus. So we need to make the right hand side have no charge. Also with the addition of the two electrons. The second part of this reaction is nitrate going to nitrogen gas nitrate is N. 03 minus going to nitrogen gas and to will balance the nitrogen. Everything about hydrogen and oxygen to do that. We need to put it to in front of the nitrate, then we'll balance our oxygen's with water. We've got six oxygen's. So we need six waters on the right hand side that just gave us 12 hydrogen. So we need 12 hydrogen ions. We balance hydrogen with hydrogen ions on the left hand side. Then we can balance charges. We have 12 plus minus two, we have 10 plus and we have no charge. So to get rid of the 10 plus on the left hand side, we need 10 electrons. Now we need to send these two reactions together and had the electrons cancel. We've got to we've got 10 and we've got to So if we multiply this top reaction by five, that will get rid of our electrons. Now, when we sum them up, we'll have five sinks producing or reacting with 12 hydrogen and to nitrates producing and two and our six waters and then our five times two plus five zinc buses. Now let's make sure it's balanced five sinks five sinks 12 hydrogen is 12 hydrogen to nitrogen to nitrogen. Six oxygen's six oxygen's. What about the charges? We've got 12 plus minus two. That's 10 plus five times two. That's 10 plus. So it's balanced. Next we'll go to the second half reaction. This is similar, but now in basic solution. So the first half reaction is essentially the same zinc going to zinc two plus plus R two electrons. The second half reaction is nitrate going to ammonia, ammonia NH three. So will balance our nitrogen is their balanced will balance our oxygen's with water. I need three waters on the right hand side. We now have six plus three. So that's nine hydrogen. So we need nine hydrogen ions on the left hand side. Now we'll balance charges. We've got nine minus one, that's eight plus and no charge. So we need eight electrons on the left hand side to make sure the electrons cancel when we some these two half reactions. We need to multiply the top reaction by four so that the eight electrons will cancel. This will then give us our four sinks plus nine hydrogen ions, plus the nitrate going to the ammonia and the three waters plus our four zinc two pluses. But this, however, is in basic solution. So we need to convert our hydrogen ions into hydroxide ions to do that. We had the same number of hydroxide to each side as we have hydrogen ions, the hydrogen ions and the hydroxide is combined to form water. The nine waters here cancel the three waters on the right hand side, giving us just six waters on the left hand side. Then we sum everything together. Let's make sure it's balanced. We've got four sinks four sinks one, nitrogen one nitrogen 123 plus six. We've got nine oxygen's nine oxygen's and then we've got 12 hydrogen 123 plus nine is 12 hydrogen. Now let's look at charges, we've got one minus and we've got eight plus minus nine, that's one minus. So it's balanced. The next reaction is copper to sulfide reacting with nitrate producing copper two plus sulfur and nitrogen monoxide. So they have reactions to consider are the ones with copper and sulfur and another one with nitrogen and oxygen. So the half reaction with copper and sulfur would be the copper to sulfide, going to copper two plus and sulfur. The second half reaction would then be the nitrate going to the nitrogen monoxide. To balance the first one, we have one copper one, copper one, sulfur one, sulfur. Everything's balanced accepted charges. So we need to put it to minus here because of the two plus. So we'll add two electrons to the right hand side. For all of these half reactions, you should recognize that when you have two half reactions comprising an entire reaction, one will be an oxidation where the electrons are on the right and one will be a reduction where the electrons are on the left. If not, then you did something wrong. So we'll balance the nitrogen is their balanced, then we'll balance the oxygen's with water. We have three oxygen's one oxygen. So we need to waters. So we have three oxygen's on both sides. Now we have four hydrogen is that we introduce? So we need to add four H plus is to the left hand side. Now we will balance the charges, we have four plus and a minus. So that's three plus and no charge. So to make both sides be no charge, we're going to add three electrons to the left hand side. Now we're going to sum them up and make sure that the electrons cancel. To have the electrons cancel. We need to multiply this bottom reaction by two and the top reaction by three. So we get six electrons that will cancel. Well then get three copper to sulfides with two times 48 hydrogen to nitrates producing to Eno's and two times 24 waters. And then up here three coppers and three suffers. Then let's make sure that it's an acidic solution, which it is. So we are done. If it were in basic solution, there would be more steps. It's always best to verify that it is balanced. We've got three coppers, three coppers, three Soul furs three suffers eight hydrogen, eight hydrogen to nitrogen to nitrogen. And then we've got six oxygen's two and 46 oxygen. So it's balanced. What about the charges? We've got eight plus and two minus. So that's six plus. We have six plus Over here, the next reaction is ammonia with oxygen producing nitrogen dioxide gas. So there's two reactions here, One that contains nitrogen and the other one with just oxygen by itself. I'll get to that one in a minute. So we have ammonia, going to nitrogen dioxide, one nitrogen, one nitrogen. We have two oxygen's. So we need to waters. Then we introduced four hydrogen is there are three here, so that's seven hydrogen. So we need seven H plus on the right hand side. Now we need to make sure that the charges are balanced. We have no charge. We have seven plus, so we need seven electrons here, so that it will have no charge on the right hand side. What about the next half reaction? All we have left is oxygen. Well, oxygen is going to go to something. This is where we begin balancing with waters. So we'll start the half reaction this way with just oxygen. Then we need to waters on the right hand side because we have to oxygen's on the left. We introduced four hydrogen, so we'll add four H plus is to the left. Now we'll balance the charges. We have four plus and no charge. So we need four electrons on the left hand side. So to get this to balance, we're going to have to multiply the bottom one by seven and the top one by four. When we do that, we'll get are electrons will cancel and we will get for ammonia. As you'll notice that the four times seven gives us 28 hydrogen. The seven times four gives us 28 hydrogen. So the hydrogen is end up canceling. What about? Let's see then the seven over here gives us seven oxygen's and then we've got the four times this one. Well give us four nitrogen dioxides. But then what about the waters? We have four times two. That's eight. And then we've got two times seven. So that's 14. 14 minus eight is six. So we're going to end up with a net amount of six waters on the right hand side. And this one says we're balancing it in gas phase. So you'll notice that the hydrogen have gone away and everything is balanced as is for nitrogen is for nitrogen. 12 hydrogen, 12 hydrogen 14 oxygen's. We've got eight plus six more is 14 oxygen's and then there's no charge on either side. So it's balanced. Now. For the next one we have chlorine gas reacting with hydroxide producing chloride and C L. 03 by minus the chlorate and ion. We'll separate this into two half reactions. Um I rewrote it here because the chlorine is actually going to go to the chloride is one half reaction and the chlorine again is going to go to the chlor eight. As the other half reaction, I'm not going to worry about the hydroxide because I'm going to add those in later because this will be balanced in basic solution. So one half reaction is chlorine gas going to chloride. We had to chlorine. So we need to chlorides. Now we have a tu minus charge and no charge. So we need to electrons on the left hand side to get a tu minus charge. The next half reaction then is chlorine. Going to chlorate. We have to chlorine so we're going to need to close rates. That gives us six oxygen's. So we're going to need six waters on the left hand side. That gives us 12 hydrogen. So we're going to need 12 H plus is on the right hand side. Then we have no charge. We have I guess to minus and 12 plus. So that gives me 10 plus. So I'm going to need 10 electrons so that I have no charge on both sides. Then I need to sum these up to sum them up so the electrons cancel. I need to multiply the top reaction by five. The electrons will then cancel and I will get five times. One is five plus. One more, gives me six cl two S and then I have six H two Os. And that will go to five times too. That will give me 10 chlorides. And then I've got to cielo three minuses and then I've got my 12 H plus is okay but this is basic solution. So I need to get rid of the hydrogen ions by adding the same number of hydrogen are hydroxide to both sides as they have hydrogen ions. So these 12 will combine together to form 12 waters and these 12 waters will cancel those six. So I'll be left with just six waters on the right hand side. So when I'm done I'll have my 12 hydroxide, my six cl two s. They will go to my 10 chlorides, my two close rates and then I'll have my six waters left left over. And because everything can be divided by two, if I divide all my coefficients by two. This is what I get. Ellis verify that it's balanced. I have six oxygen's. I have three plus three more. Six oxygen's. I have six hydrogen is three times to six hydrogen. I have six chlorine. Xai have five plus one more. So that's six. I have a six minus charge. I have minus five minus one more six minus. So everything is balanced. And now for F I have H 202 going to go to plus Well let's back up. This will be one half reaction. The other half reaction will be with manganese, which I'll get to in a minute. So I had to oxygen's to oxygen's. The oxygen's are balanced to balance the hydrogen with hydrogen ions. Then we have no charge. We have a two plus charge. So I'll balance the charges with electrons. So now both sides have no charge. The second half reaction coming from the full reaction will be the permanganate, going to the manganese ion. The bank bonuses are balanced. The oxygen's are not have four oxygen's on the left hand side. So I need four waters on the right. That introduced eight hydrogen will have eight H plus is on the left hand side. Now I have a two plus charge here and I have an eight minus one. I have a seven plus charge here. So to get a two plus, I need to add five electrons to the left hand side. So to get this to have the electrons cancel, I need to multiply the bottom reaction by two in the top reaction by five. Then my electrons will cancel and I'll be left with five H 202 plus. The two times there eight gives me 16 and the five times the two gives me 10. So I have 16 H plus is on the left, 10 on the right, so 10 are going to cancel from both sides. And I'll be left with just six H. Plus on the left hand side. And then I have my two permanganate. It's on the right hand side. I'll have my 502 S. And my two mm into pluses. And then the waters we have eight waters two times four is eight. And then this is balanced if it requested in acidic solution, which it did. So let's verify. We have 10 hydrogen. Well as the 10 plus six more that 16 hydrogen, we have five times two. That's 10 plus. Eight more. That's 18 oxygen's so we have 10 plus 8, 18 oxygen's We have to manganese to manganese. So we're balanced. What about charges? We have a six plus and a tu minus. So that's four plus. And then we have two times two. So that's four plus. Okay, now, for G we have two half reactions. Both will include the single reactant. No two will have no to going to nitrate and no to going to nitrate for no to going to nitrate. Are nitrogen czar balanced. We have two oxygen's three oxygen. So we need a water on the left hand side that introduced to hydrogen. So we need to hydrogen ions on the right hand side. We now have no charge and we have a plus one charge. So we're gonna need one electron on the right hand side. Then the next one will be the no to going to the nitrite one, nitrogen one, nitrogen to oxygen's two oxygen's. We just have a one minus charge. So we need to put an electron on the left hand side. Now we simply some these up that we don't need to multiply anything because the electrons will cancel as is and we get H 20 plus n +02 Two of them goes to know three minus plus two minus. And then we have our two age pluses for this one. It does say that it is occurring in basic solution. So we need to add the same number of hydroxide is to both sides as we have H plus is converting these into two waters. These two waters will cancel one of the waters on the left hand side. So we'll just be left with one water on the right hand side and we end up with the two hydroxide and the two of us go into nitrate. And I try it plus one water. Let's make sure it's balanced. We have two oxygen's plus four more. So that's six oxygen's we have three plus two, That's five plus one. So that's six oxygen's We have two hydrogen. Is we have two hydrogen is we have to nitrogen we have two nitrogen zits balanced. Okay. And now for the last one we have three plus, going to be two plus. Everything is balanced but charges. So we need to put an electron on the left hand side so it's two plus on both sides. And then we have iodide going to iodine. We need to balance the iodine so we have two of them. Now we have a tu minus charge and no charge. So we need to electrons on the right hand side to get the electrons to council. We need to multiply the first reaction by two. The electrons will then cancel and I'll be left with to F E three plus and two iodide. It's going to I two plus two. F E three plus is and this one says that it's not acidic or basic. So we're done. Let's verify. We've got two irons, Two irons to IOD icts or to iodine is to iodine. We've got a two times three that's six plus minus two. So that's a four plus. We've got a two times two so that's a four plus so it's balanced.

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