Oxidation and reduction deals with the transferring of electrons between reactants. The reactant that loses electrons is oxidized, while the reactant that gains electrons is reduced.
Understanding Oxidation versus Reduction
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we're gonna say chemists usually use important terminology to describe the movement of electrons. Remember, electrons are the negative subatomic particles we're going to say in Redox reactions, Redox reactions. We have the movement of electrons from one reactant toe. Another reacted. Now we're gonna use ah, funny type of sentence to help us remember what's going on. And the sentences. Leo the lion goes ger might sound corny, but trust me, it helps a lot with understanding, oxidation and reduction. So we said Leo, the lion goes girl. So Leo, lose electrons. Oxidation goes, girl gain electrons reduction. Now remember, electrons are negatively charged. So if you're losing negatively charged subatomic particles, you're becoming more positive. So lose electrons means you're becoming mawr positive. Okay, if you're gaining negative electrons, you're gaining negative things. So you are gonna become more negative. Okay? And then here's the word thing. If you've been oxidized, if you want to go oxidation, you're called the reducing agent. Go, Go. Okay, so you're the reducing agent. If you've been reduced, then you're the oxidizing agent
When in doubt about oxidation or reduction just remember the phrase:"LEO the lion goes GER."
Rules for Oxidation Numbers
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Now that we know the distinction between oxidation and reduction, it's important to set down some ground rules toe help us figure out what's the particular oxidation state of a particular element. So I break rules for assigning oxidation numbers, which we're gonna abbreviate O n so oxidation numbers o n I break it down into two sets of rules. We have our two general rules, and then we have arm or specific rules. Now, the two general rules, we're gonna say, for an Adam in its elemental form, what is elemental for me. While elemental form means two things, it means that the metal is by itself, such as sodium, sodium metal is by itself or the element is connected to itself. Oxygen here is connected to another oxygen sulfur here is connected to a bunch of sulfur is just like itself. So if you're by yourself or you're connected to copies of yourself and you have no charge, that is your elemental form. So let's write that down. Elemental form means you are by yourself or connected two copies of yourself and you have no charge. So if you fit these types of categories, then you're oxidation numbers equal to zero. Not the second general rule is for an eye on whatever the ions charges. That's it's oxidation number, so sodium here is by itself, but it has a plus one charge, so it's oxidation. Number is plus one. Calcium is by itself, but it has a plus to charge. So it's oxidation numbers plus two. Now No. Three minus. That's our nitrate ion, our Polly atomic ion. The oxidation number for the entire compound is negative one. Now, if we want to find the oxidation state of each of the individual elements, that's when we move on to our specific rules. Now are specific rules. We're gonna say Group one A plus one group to eight plus two, based on the charge distributions that we learned earlier. But we're going to save the hydrogen. Hydrogen is in Group one A. But hydrogen is not always plus one. We're gonna say it's plus one when it's connected to non metals. So if it's connected to si, ahl or perk loit or in water and these three examples water is connected. Toa Onley nonmetal, so it's plus one. But if hydrogen is connected to ah, metal or blond, it's minus one. So if we had any H or B H three than in those cases, hydrogen would be negative one. We're gonna say Florida is always minus one, no matter what. Oxygen, on the other hand, can be a pain because oxygen has various oxidation states, and it's all based on what kind of compound it's in. So we're gonna say for oxygen. When it's it's ah peroxide, it's gonna be negative one. Now, what exactly is a peroxide? Ah, peroxide has the formula X two zero two X equals a group one a element. So hydrogen, lithium, potassium, all those air group won eight medals. So a good example. Here you could have hydrogen peroxide or sodium peroxide. Both are peroxide because both have the formula X two o two. In this case, oxygen would be minus one. Now oxygen will be negative. A half when it's a super oxide. Ah, super oxide has the formula X o two X again is a group on a element. So, for example, you could have sodium super oxide or lithium super oxide. Now we're gonna basically say if oxygen is not a peroxide or super oxide weaken. Say that it's minus two. And this is the normal state of oxygen because we sell them. See peroxide or super oxides. But if we do not see one of those to remember oxygen, we minus two now groups seven eight elements are halogen. They're gonna be minus one. Except when they're connected to oxygen when they're connected. Toa oxygen. We're gonna have to use some mathematical knowledge in order to do it. Simple algebra. We'll get to see how this works when we do practice questions dealing with this concept. So remember, we have our two general rules, and then we have arm or specific rules.
We use these oxidation number rules whenever we are asked to determine the oxidation number of an entire compound or the individual elements within a compound.
What is the oxidation number of each underlined element
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Now, Now that we've seen this, let's take a look at the example. It says, What is the oxidation number off each underlying element For the first one, we have I to so remember which rules should we look at? Well, here we have nonmetal. We have an element connected to a copy of itself, and it has no charge. Remember, when we see this, the oxidation numbers automatically equal to zero. Come for the next one. We have C s to go to remember, this has the formula X to go to. And cesium is a group one element. So this would be a peroxide, and therefore the oxidation number of oxygen would be minus one. Okay. And then finally, the last one here. I don't specify which one we wanna find. I'm gonna tell it to you now. It's carbon. I want us to find carbon. Now, I'm gonna take myself out of the shot guy. So we have more room to work this out together. Okay, so we want to find the oxidation number off carbon. The thing is, we never came up with a rule for carbon. It's nowhere in the specific rules. So because of that we don't know what it is. And therefore it's X hydrogen. We do have a rule for it. When it's connected to non metals, it's plus one. This compound doesn't take the form of ah peroxide. It doesn't look like this. It doesn't take on the form of a super oxide. It doesn't look like this. Therefore, oxygen must be minus 2/4 oxidation state. Now what we're gonna do here is we're gonna use simple algebra to help us software the missing variable. So we have one hydrogen, that is plus one. There's one carbon, which is X plus. There are three oxygen's. Each one is minus two. This equation will equal the charge of the compound. Now the charge of the compound is minus one. Now, we're just gonna solve for the missing variable. So we're gonna have one plus x three times. Negative two gives me negative six equals minus one. The positive one in the negative six will combine to give me negative five. We still want to find X, so we add five to both sides. So the oxidation state of carbon would be plus four
Instead of being asked who is oxidized or reduced, sometimes you may be asked to determine the reducing and oxidizing agents.
In the following reaction identify the oxidizing agent and the reducing agent
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Welcome back, guys. In this new video we're gonna practice on How do we identify the oxidation numbers of certain compounds based on the rules that we learned about redox reactions. So let's take a look. We say this example in the following reaction identified the oxidizing agent and the reducing agent. So let's take a look here. We're gonna first try to identify the easiest portions of this we're gonna say, Here we have 02 We have an element connected to a copy of itself, and it has no charge, so the oxidation number of oxygen here will be zero. Next, we're gonna look at the hydrogen. Hydrogen here is connected to a non metal, which is carbon. Therefore, it's oxidation. State is plus one. Now, we don't have a rule for carbon here, So we're just gonna say that carbon here is X and we're gonna use a mathematical type of strategy in order to solve for that missing variable. Remember, when we're looking for the oxidation state of a particular element, ignore the coefficient coefficient is not important. So we're going to say we have six carbons, so six x plus six hydrogen is each one is plus one equals the charge of the compound. The charge of the compound is zero. I don't say anything about it being positive or negative in this space here. So we just say zero now six X plus six equals zero. We're gonna say Subtract six from both sides. So six X equals negative six. Divide by six X equals negative one. So the oxidation state of carbon, you ought to be minus one. Next, we're gonna go to the other side. Here we know that hydrogen is connected to oxygen, which is a non metal. So therefore it's plus one. This structure is not ah peroxide or a super oxide. Remember, we still need to know that so oxygen, who will be minus two Co two? This is not a super oxide because remember, super oxide does have the form of X 202 But acts needs to be a group. When a element carbon is not a group on a element. It's a group for a element. Therefore, oxygen here is minus two. Now we have to solve for carbon again. So it's X and remember it's X because we don't have any type of role for it. We don't have a general rule or specific role for it. Okay, so this right here, ignore the coefficient. So now we just saw for we're gonna say X because there's only one carbon plus two oxygen's. Each one is minus two equals that charge the compound, which is zero x minus four, equals zero at four to both sides. X equals plus four. Now we have to identify the oxidizing agent and the reducing agent. Remember to do that, we have to remember Leo the lion goes ger. Okay, so we're looking to see who became more positive and who became more negative where we go from its reacting form to its product form. So if we take a look at carbon, we're gonna say Carbon starts out as negative one when it's reacted. Then when it becomes a product, it's oxidation state goes to plus four. So we're gonna say carbon became or positive because it goes from negative 12 plus four. Since it became more positive, it was oxidized. If you were oxidized, then you're the reducing agent. So that's the type of mindset you have to have when doing these types of questions does the oxidation state become more positive or more negative? As you go from being a reactant to becoming a product, who else changes where we're going to say that Oxygen 02 goes from being zero to being negative too? It doesn't matter which oxygen we look look at for products because they're both minus two for their oxidation states. So we're gonna say oxygen goes from zero to negative two. It became more negative. Therefore it was reduced. And if it was reduced, its the oxidizing agent Mhm. So that's the pattern you have to follow in order to solve this type of question. Remember, it's all based off of Leo. The lion goes ger Leo, lose electrons, oxidation, ger, gain electrons reduction. Go back to the previous concept video and look at how we manage that chart. And just use it to solve this question. Okay, guys, the next to set of questions on the same page, our practice. So I want you guys to attempt on doing it on your own. Then come back and look at the video where explained how we solve both of these practice questions. Good luck
Try your skills at determing the oxidation number of elements within molecules or polyatomic ions.
What is the oxidation number of each underlined element?
a) P4 b) BO33– c) AsO42– d) HSO4–
a) -3 b) +3 c) -2 d) -2
a) -2 b) 0 c) -6 d) -6
a) 0 b) +3 c) +6 d) +6
a) 0 b) +3 c) -6 d) -6
Just remember if you've been oxidized then you're the reducing agent and if you've been reduced then you're the oxidizing agent.
In the following reaction identify the oxidizing agent and the reducing agent:
so recall that oxidation reduction reactions or redox reactions involves the transferring of electrons from one reactant toe another. Now here in our equation, we have lithium solid reacting with chlorine gas, and in this process we produce lithium ion as well as to chloride ions. Now, remember, when it comes to Redox reactions, we have oxidation and reduction to help us remember key concepts in terms of those two words we use Leo, the lion goes ger. Remember, Leo means that we're losing electrons and therefore represent oxidation. When you're losing negatively charged electrons, the species or element in this case is becoming more positive. That's because it's losing something negative now. What exactly about that element is becoming more positive? Well, it's oxidation. Number is becoming more positive. As a result. We're gonna say here that if you are being oxidized, then you represent the reducing agent or the reductive. In this case, we have lithium solid in its natural state, so it's oxidation. Number is equal to zero now, as an ion, it's oxidation. Number is tied to its charge, so here it be plus one. So for lithium, it goes from being zero to plus one so it's oxidation number has increased. Therefore it has been oxidized. Therefore, it is the reducing agent or the reductive on the other side. When we gain, electrons were gaining negative electrons. That means the species or the element in this case is becoming more negative. As a result of this, the oxidation number decreases here. If you are being reduced, then you represent the oxidizing agent or the Occident. In this case, C L two gas it's in. It's natural or elemental state, so it's oxidation numbers equal to zero. Now, as an ion, it's oxidation numbers tied to its charge. Each chloride ion is minus one, so each chloride ion is minus one in terms of oxidation. Together they would be minus two. We would say here that we said lithium was being oxidized, so it's losing electrons. It loses. In this case, technically would lose two electrons and chlorine itself would pick up those two electrons because each chlorine needs to gain an electron so each chlorine becomes a chloride ion. As you go deeper, deeper into redox, reactions will talk about concepts such a self potential, the use of electrochemical cells as off as well as other concepts related to charge or voltage. Now that you've gotten down the basics of Redox reactions, click on the next video and see the different types of variables tied to the transferring of electrons between reactions.
Basic Redox Concepts 2
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with the concept of redox reactions. We have the transferring of an electron from one reacting toe another. When we combine this idea with the use of electrochemical cells, we can have the transferring of an electron from one cell to another. This transferring of electrons helps in the generation of voltage. It also introduces terms such as current electrical charge as well as work. Now, with these concepts, we have essential formulas that we're gonna have to use and apply to get the answers that we want. Now here, when we talk about electrical charge, realize that charge. Use the variable que and the units for electrical charge are measured in cool lumps, so Capital C would represent our columns. Now we're going to say here that connected to charge connected to cool, um, is Faraday's constant. So to figure out Faraday's constant, we have the charge of an electron, which is 1.602 times 10 to the negative. 19 cool ums times avocados, number by moles in verse. At the end, That gives us Faraday's constant, which is 9.647 times 10 to the four Coolum over one mole of electrons so This is the charge for one mole of electrons. Now here we're gonna say Q equals charge. It equals n times. Faraday's constant. So here we have moles of electrons, times Faraday's constant one mole of electrons here on the bottom. Your most of electrons would cancel out so that at the end you're charge would have units of cool apps here. We could also talk about electrical current for current use. The variable I, the unit for electrical current, is in amperes or amps, so we can say a He represents the units for current. Now we're gonna stay here. That current equals charge, which we said the units would be cool ums, Okay, divided by time here, the units for a time would be in seconds. So what this is telling me? It's telling me that current is in units off and piers and an MP represents columns per second. So if you are given 25. amps, that would translate into 25 cool ums per one second. Next, we have electrical voltage. Now, with electrical voltage, we have a series of equations we can use. So here we can say the relationship between work and voltage can be expressed as work equals voltage, which is e times their charge. Q. Now here, when it comes to voltage, the units for voltage, our energy in terms of jewels divided by columns. And we already said earlier that charge uses the units of columns so here columns would cancel out, so work at the end would have units off jewels. Now, besides this equation, we can say that work, which is w equals force times distance here force would be in units of Newton's, which is n and here. Distance would be meters. Now we can say that one Newton is equal to kilograms times, meters over second squared. So here we have kilograms times, meters over second squared distances meters, so that would come out 2 kg times meters squared over seconds squared. All those units combined together equal one jewell. So whether we're utilizing this equation here for work or this equation here for work, both give us jewels as the units for work. Next, we have the relationship between Gibbs Free Energy, which is Delta G and electrical or electric potential. So here are electric potential, which is B e again, our voltage so here would say, Gibbs, free energy equals negative end, which is your most of electrons times Faraday's constant times their voltage. So here we'd have Delta G equals moles of electrons, times, Fridays constant, which remember, is cool. Um, over most of electrons and voltage. Remember, we just said in the previous example dealing with work that voltage is Jules over Cool apps. So most of electrons would cancel out. Coolum should cancel out. And you'd have jewels as your final units for Gibbs Free Energy. Here is well, now we also have homes, Law. OEMs law, we're gonna say uses units for resistance and that would just be homes or Omega here we'd say that current again, we're dealing with current, so realize current can be found in more than one place. We have current here and we have current here. So here we say that our current, which is I equals your voltage over your resistance. So that's this is just yet again, another equation we can utilize in order to help us determine what our current will be. Remember at the end, current would have units off columns per second. Now power finally here, power represents work done per unit of time. We'd say that the unit for power are in Watts. So capital w here. So power equals voltage times current. We could also say that power equals work over time because we just said it's work done per unit of time work uses units of jewels. Time seconds. So a one watt is equal to Jules per second. Here, power also equals voltage times current. So that be jewels over seconds for voltage times their current, which we said was columns per second. So those who can slot and we'd get Jules per second so we could utilize this equation, help us determine power. Or we could utilize this equation here to help us determine power. So with some of these, we have the same variable with more than one method to get to the answer for it. So just keep in mind that with Redox reactions were talking about the transferring of electrons from point A to point B. This movement of electrons helps with the creation of voltage or electricity or charge. And with these concepts, we have different formulas we can utilize to help give us numerical values. So keep in mind all the concepts we learned in terms of these equations and how they relate to Redox reactions. As we move further into determining the potential differences in electrochemical cells, these equations will come into play.