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Basic Redox Concepts

1

concept

Basic Redox

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So recall that oxidation reduction reactions or redox reactions involves the transferring of electrons from one reactant to 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 two 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 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 its oxidation number is equal to zero. Now, as an eye on its oxidation number is tied to its charge. So here would be plus one. So for lithium it goes from being zero to plus one. So its 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, we're 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 oxidant. In this case cl two gas. It's in its natural or elemental state, so its oxidation number is equal to zero. Now, as an ion, its oxidation number is tied to its charge. Each chloride ion is -1, so each chloride ion is -1 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 and deeper into redox reactions will talk about concepts such as self potential, the use of electrochemical cells as of 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 react ints.

2

concept

Basic Redox

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7m

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with the concept of redox reactions. We have the transferring of an electron from one reacting to 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, um 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 uses the variable Q. And the units for electrical charge are measured in columns. So capital C would represent our columns. Now we're going to say here that connected to charge connected to Coolum 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 columns 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 columns 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 multiple electrons would cancel out. So that at the end you're charge would have units of columns here. We could also talk about electrical current for current. Use the variable I the unit for electrical current is in amperes or amps. Okay, so we can say a here represents the units for current. Now we're gonna stay here. That current equals charge. Which we said the units would be cool. Ums divided by time here, the units for a time would be in seconds. So what this is telling me is telling me that current is in units of amperes and an MP represents columns per second. So if you are given 25.0 amps, that would translate into 25 columns per one second. Next we have electrical voltage now with electrical voltage and 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 are 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 of jewels. Now, besides this equation, we can say that work, which is W equals force times distance. Here, force would be in units of newtons, 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 to kilograms times meters squared over seconds squared. All those units combined together equal one jewel. 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 moles of electrons, times Faraday's constant, times your voltage. So here we'd have delta G equals moles of electrons times friday's constant which remember is cool um over moles of electrons and voltage. Remember we just said in the previous example dealing with work, that voltage is jules over columns. So most of electrons would cancel out. Cool um would cancel out. And you'd have jewels as your final units for gibbs. Free energy here as well. Now we also have Homes Law. OEMs Law. We're gonna say uses units for resistance. And that would just be OEMs 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 of columns per second. Now, power Finally here power represents work done per unit of time. We'd say that the units 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 jewels per second. Here. Power also equals voltage times current. So there'd be jewels over seconds for voltage times our current which we said was cool um per second. So those would cancel and we 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.

Basic Redox Concepts Calculations

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example

Basic Redoxs Calculations 1

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here it states what happens to the current in a circuit? If a three volt battery is removed and replaced by a one volt battery? Now from the previous formulas that we've gone over, we know that current which is I equals your charge E divided by your resistance. R. Remember here that the units for our charge here would be cute and our resistance here would be in homes. So if we think about this, we have current Here. It's 3V Were not given a resistance which means our resistance is being kept the same in both cases. Um so here let's say our resistance is .50 OEMs. So when we worked that out, that would give us a current of six. Remember the units for current are cool items per second And now we've replaced it with one volt. Our resistance can stay the same. So 0.50 homes. And now that's two Cool ems per second. So we can see that our current has gone down and if this current were attached to a light bulb because the current has decreased, that light bulb would become dimmer because we've replaced it um from a three volt battery to a one volt battery. So those are the real world applications of changing the voltage connected to any type of electrical circuit. Now that we've seen this basic question. In terms of currents, let's take a look at example, two in the next question, attempt to do it on your own based on the formulas that I showed on the previous page. If you get stuck, don't worry, just come back and see how I approach that same exact example question.

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example

Basic Redoxs Calculations 1

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So here it says if the voltage of a T. E. Series enhanced balance has a 240 volt battery, what is the resistance in the circuit? If the current is 2400.80 amperes. Alright so we're asking to determine resistance. So are we have amperes which is just our current. So that is I. And then we have voltage here which is E. So our equation as before is I equals E. Over R. Rearranging this to isolate our by itself. We see that our equals my voltage divided by my current. So here we'll plug in our voltage of 240V divided by our current of 0.80 amperes. So that gives me 300 gems as my resistance. So that would be our example. Well that would be our answer in terms of this example. So just remember all we're doing here is using the law and just rearranging it to solve for the missing variable which in this case is resistance. Move on to practice question one where we're shown an image of a circuit but again, we're still basing this all off of homes law. So it's just solving for the missing variable based on the information provided

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Problem

Problem

Solve for the missing variable in the following circuit.

A

0.075 C/s

B

7.68x103 C/s

C

13.3 C/s

D

7.50 C/s

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Problem

Problem

Solve for the missing variable in the following circuit.

A

66.6 V

B

150 V

C

0.015 V

D

1.50x103 V

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example

Basic Redoxs Calculations 2

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here. It says to calculate gibbs free energy under standard conditions for our reaction as well as our self potential. Under standard conditions for redox reaction with N equals four. That has an equilibrium constant of K equals 40.130 at 25 degrees Celsius. Alright, so here remember that N represents our moles of electrons. So we have four moles of electrons. And we're going to say here that the connection between our equilibrium constant K and r delta G of reaction under standard conditions is delta G of reaction equals negative R T L N K. So this involves utilizing the equation that we saw in the past to connect our equilibrium constant. K two delta G. So here R is negative 8.315 here or 8.314472 jewels over k times moles Temperature needs to be in Kelvin. So we're gonna add 73.15-25°C To give us 2 98 0.15 Calvin. And then we're gonna have Ln of K which is 0.130. Here are kelvin's cancel out. So our delta G value comes out to negative 3, 22.3 jewels over moles. So that's how we get our delta G in the beginning. Now we're gonna have to form a connection between delta G and our cell potential. So here the equation that connects them is delta G of our reaction equals negative N times f times E cell. So we're gonna input the value that we just got in terms of delta G, so negative 3, 22.3 jewels over moles Equals moles of electrons transferred which is four moles of electrons Times Faraday's constant, which is 9.647 times 10 to the four columns over moles of electrons. And we don't know what? S all is, that's what we're looking for. So divide out this portion here to isolate our sl so negative four moles of electrons And then 9.64, 7 times 10 to the four columns over moles of electrons on both sides. So when we work that out, we have multiple electrons cancel out. So here, what we're gonna have at the end for S. O. Is really in terms of jewels over coolants. So here this comes out to 8.35 times 10 to the negative four jules over columns. Remember jules over columns. Really just means volts. So this represents our self potential. So remember this question utilized questions um formulas that we've seen in the past in order to form a connection between these three different variables of your equilibrium constant. K Your gibbs free energy of the reaction. Under standard conditions and the cell potential of the reaction. Under standard conditions. Now that we've taken a look at this example, move onto the next one here, we're gonna apply stoke eom a tree to the whole idea of current were given 4.3 amps and some mass of copper. And from that we need to determine the amount of time that has elapsed. So attempt to do this on your own. But if you get stuck, don't worry, come back and see how I approach that same example to question.

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example

Basic Redoxs Calculations 2

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3m

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copper can be electro plated at the cathode of an electrolysis cell by the half reaction of copper two plus ion plus two electrons produces one mole of copper solid. Here we were asked, how much time would it take for 525 mg of copper to be plated at a current of 4.3 amperes. Alright, so we're looking for time, it doesn't specify if we want to find it in minutes, seconds, years etc. So here we're just gonna figure it out in terms of seconds. Now, remember an amp or an ampere represents the unit for current. And remember an ampere is equal to um charge per over time. So charge or columns over s for seconds. So 4.3 amperes represents 4.3 columns per second. We're gonna first start out with 525 mg of copper. Were they going to convert those milligrams of copper into grams of copper? So one mg of copper is equal to 10 to the negative three g of copper. So milligrams cancel out now we have grams of copper. We're next going to change those grams into moles. So one mole of copper, according to the periodic table weighs 63.546g for copper. And now that I have moles of copper, I can say for my equation that for every one mole of copper solid, we have two moles of electrons involved. So for every one mole of copper, from my balanced equation there are two moles of electrons involved, moles of electrons is really a part of Faraday's constant. So we're gonna say we have one mole of electrons and according to Faraday's constant, that's equal to 9.64, 7 times 10 to the four columns. Finally we have columns which can cancel out with these cool lumps, leaving us with seconds at the end. So we put 4.3 columns on the bottom, one second on top. So now we're gonna have seconds remaining Here, we're gonna get 3.7 times 10 to the two seconds involved. So this is a typical electrochemical question where we're tying charge current two story geometry in some way. Using the balanced equation, we see that the ratio is for every one mole of our solid metal, we have two moles of electrons involved. That comparison there is what allows us to find a way of connecting current to store geometry. Now that we've seen this example. Move on to the final question on this page where we have to determine molar mass of the metal from the question. Now realize what are the units for molar mass. That is key to answering the question correctly. Once you've attempted it, come back and see if you can match your answer with what I get. So good luck guys

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Problem

Problem

A metal forms the salt MCl _{3}. Electrolysis of the molten salt with a current of 0.700 A for 6.63 h produced 3.00 g of the metal. What is the molar mass of the metal?

A

8.6610 g/mol

B

103.9323 g/mol

C

51.9662 g/mol

D

5.77 g/mol

E

25.98 g/mol

10

example

Basic Redoxs Calculations 3

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So here it states in the following reaction, one mole of zinc solid plus one mole of copper, two sulfate react in order to produce one mole of zinc sulfate plus one mole of copper solid here? It asks, what is the maximum energy produced when 15 g of zinc is completely reacted in a zinc copper electrochemical cell that has an average cell potential of 1.10 volts. Alright, so they're asking us for the maximum amount of energy here. This is equivalent to them asking what's the maximum amount of work that can be done? So what they're asking us to figure out is delta G. And here they give us the mass of zinc and they're giving us self potential. So they're giving us sl realize that the formula that connects delta G to e cell is delta G equals negative N times f times sl here we have to determine the number of electrons transferred. Well, here's ink is zero, but when it's here with sulfate it's plus two because sulfate is minus two. So that's two electrons transferred At the same time. Copper here has to be plus two and here at zero. So we have two electrons involved that are being transferred. So that means N is two. So we have two moles of electrons. We're gonna multiply it by Faraday's constant 9.647 times 10 to the four columns over moles of electrons. And then we're gonna multiply by my voltage. Remember Vault is equal to jules over columns. So that's equal to 1.10 jewels per cool. Um so what happens here is moles of electrons cancel out? Cool um cancel out. So all have jewels of electrons. So that comes out to negative 2.12 times 10 to the five jewels per mole. Now here's the thing, this is the answer when we're dealing with exactly one mole of zinc solid, but in the question, we're not dealing with exactly one mole of zinc solid. What we're dealing with is 15 g of zinc, which is less than one mole. So we're gonna do here is we're gonna take those 15 g, we're gonna convert them into moles. So we're gonna stay here for every one mole of zinc, The Masses, 65.409 g of zinc. So grams of zinc cancel out. Now I have moles of zinc And we're gonna stay here for every one mole of zinc. We found this value here. So it's negative 2.12 times 10 to the five jewels. So this council out with this. So we get at the end is negative 4.87 times 10 to the four jewels. So you can say here that we produce 4.87 times 10 to the four jewels as a result of dealing with just 15 g of zinc. Remember in this type of question, realize what variables are they giving us? And how are they connected together in terms of a formula. Knowing that part allowed us to determine what the energy released when it comes to one mole of zinc, realizing the question that we're dealing with less than a mole of zinc. So what would the new amount of energy be involved Now that you've seen this example? Move on to example, to look to see if you can answer it if you get stuck. Don't worry, come back and see how I approach example, too.

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example

Basic Redoxs Calculations 3

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So here it states, a chemist weighing 110 lb, takes her NMR sample from the first floor to the second floor, which is 12 m up in 25 seconds. How much power has she generated? Alright, so power has two formulas involved. Power equals work divided by time. Or power equals voltage times current. Now, within this question, we make no mention of voltage. We don't talk about current at all. So we can't use the second formula. We're gonna have to utilize the first formula here. We're gonna say here that power equals work over time. We already know what our time is. It's 25 seconds. So we need to do now is we need to determine what our work will be. Work also has two formulas we can utilize. Work equals voltage times charge or we can say that work equals force times distance. Again, we don't make any mention of voltage. We don't make any mention of charge. So we can't utilize the first formula. We're gonna have to utilize this formula here. Work equals force times distance. So, bringing that down force will be in units of newtons and distance will be in meters. When you multiply those two together. That will give us jules as the units. All right now, in this question, I don't give us Newtons directly, but I do give us distance. We say that she's traveled 12 m up. So that's our distance traveled. What we need to do now is we need to determine what are newton's will be. Well, remember that one. Newton is equal to kilograms times meters over seconds squared. And we're gonna say here using Newton's second law of motion. We can say that force equals mass in kilograms times gravitational acceleration. So our mass here is 100 and 10 lbs. But we need that to be in kg. So we're gonna say here £110 times for every one kg, it's £2.205. So that's 49.8866 kg. Gravitational acceleration is 9.8 meters over seconds squared. So that's gonna give me 488.889 kg times meters over second squared. So that force represents my newtons. So that's 4 88.889 newtons. When we multiply those two together, that's gonna give me my jewels. So that comes out to being 586, jewels. So we just determined our work, take that and plug it in to find power. So 58586, jewels over 25 seconds. Gives me as my answer here. 2:34.7 and then we have jewels over seconds. That comes out to being wants. So she generates 234.7 watts of power by going from the first floor to the second floor. Now, this question was a bit tricky. It really, really involved us using different formulas in order to find our final answer. So realize here, we had to first utilize this formula that power equals work overtime. We already given work on already given time in 25 seconds. So that's the easy part. The hard part was determined what our work is to determine our work. We use this second equation. Work equals force times, distance, distance was given to us. Already, we have to determine what our forces force. To figure that out. We used Newton's second law of motion to figure out our force. Once we found that force in newtons, we plugged it in to determine our work. Once we determined our work, we plugged it into the power equation to figure out the watts of power generated by our chemists. So keep in mind the interconnectedness of these different formulas and how we can utilize them to help us determine a lot of different things in particular power for this particular question. Now that you've seen this example move on to the last example on this page attempted on your own. Once you do come back and see if your answer matches up with mine

12

example

Basic Redoxs Calculations 3

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So in this example it says determine the amount of time in minutes needed to produce 1.7 times 10 to the two watts from 1500 jewels of work committed. Alright, so they're giving us watts and we're talking about work, we know that those two variables are connected to power. So power equals work over time. We're gonna say here power using units of watts and a watt is equal to jewels per second. So that's 1.7 times 10 to the two jewels per second. And then we're gonna say here our work is 1500 jewels and we don't know what our time is, that's our X. All we have to do here is isolate X, which will give us seconds. So multiply both sides by X. So 1.7 times 10 to the two jewels over seconds, times X equals 1500 jewels, Divide both sides here by 1.7 times 10 to the two jewels per second Jules Counselor will have seconds. So x equals 8.82 seconds and then you just have to change seconds into minutes. So one minute Is equal to 60 seconds. So that gives me .147 Or .15 minutes. So that's the amount of time it would take. So just remember all the different formulas that utilize in some way. Um questions dealing with energy with current with charge. They are separate from each other but they are connected to each other in one way or another. As we saw an example too. So keep in mind the different formulas that you will have to utilize at some point, whether it's asking to solve for power for current or energy.

Balancing Redox Reactions

13

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Balancing Redox Reactions

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So balancing a typical chemical reaction is pretty simple. We look at the number of elements on each side of the chemical reactions and then we place new coefficients in front of each compound. Now, when it comes to balancing a redox reaction though, there's a lot more steps that are necessary to get our final balanced reaction. So here we're gonna learn how to balance redox reaction in acidic solutions as well as basic solutions. Now, if we're in an acidic environment, these are the steps that we do 1st. 1st we write the equation into two half reactions. Step two we balance elements that are different from oxygen and hydrogen. First then we balance oxygen's by adding water and balance hydrogen is by adding H plus. At this point we need to balance the overall charge by adding electrons to the more positive side of each half reaction. Then both half reactions must have an equal number of electrons if they have different number of electrons in each half reactions, you may have to multiply one or both. Half reactions by values so that they have the same number of electrons in both. From there we combine the half reactions and cross out the reaction intermediates. Now, when it comes to balancing in a basic solution, we follow the first six steps as we would up above. But then we add 1/7 step. In the seventh step, we say we balance remaining H plus by adding an equal number of o h minus ions to both sides of the overall chemical reaction. Now that we've looked at these rules, we'll take a look at the example that we have down here. So we're just gonna go straight into this example guys and we have to balance this redox reaction in an acidic environment. So we're going to say here that we're gonna break it up into two half reactions. So bromine goes with bro, mean and manganese goes with manganese. So our two half reactions are B R minus, gives us B. R. Two, and M. N. 04 minus gives us M. N. Two plus. First we balance elements different from oxygen and hydrogen. We have one br here but to be ours here so we're gonna put it to next we have one manganese and one manganese. So that's already tied and that's balanced. Next we balance oxygen by adding water. We have no oxygen in this first half reaction here here we have four oxygen's and on this side we have none. So I'm gonna add four waters. Then we balance out hydrogen by adding H Plus here we have no hydrogen at all on either side here we have four times two which gives me eight hydrogen. So I put eight H plus here. Now we have to balance overall charge. And actually we put the two here uh to give us two BRS on both sides, we balance out overall charge. So this is two times negative one which is negative two Over here, this is neutral. So this is zero. Alright then we have eight times plus one which is plus eight minus one Gives us Plus seven overall for this side of the half reaction. Then water here is neutral. So don't so just ignore it And then this is plus two. Alright, we balance overall charge by adding electrons to the more positive side. So on this first half reaction, this side that has an overall charge of zero is more positive. I have to add enough electrons to this side here so that it has the same overall charge as this side here. So I'd have to add two electrons. So now both sides are -2 Over here. This side is more positive at Plus seven. I need to add five electrons so that it has an overall charge of plus two. Just like this side here has an overall charge of plus two. Notice your electrons do not equal each other, one is two and one is five. We say that the lowest common multiple that they share is 10. So that means I multiply this here by five And this here by two. So what I get at this point is 10 BR gives me five br two plus 10 electrons. And then here we're gonna have to permanganate ions plus 16 H plus plus 10 electrons Gives Me two, Manganese, two ions plus eight waters. So everything that multiplied by to cancel out intermediate things that look the same except once a product. And when the reactant, your electrons are supposed to always completely cancel out from this, that those are our only intermediates that we have, that we can cancel out. So bring down everything else. So this here represents our balanced chemical equation in an acidic environment. So just remember when it comes to balancing a redox reaction, there's a lot of steps necessary to do that. Remember the sequence that we took in order to balance any redox reaction you come face to face with. So keep practicing, you'll be able to do this really quickly, as long as you remember the sequence of steps necessary.

14

example

Balancing Redox Reactions Calculations

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So here it says to balance the following redox reaction in acidic solution. Now first of all, when we take a look at this reaction, we have the presence of water realize here that if in your redox reaction you already begin with H plus O. H minus or water. That must mean those were added at some point while balancing the reaction. So it's best to just remove them. They'll get added back in later on as we're balancing the redox reaction. If they don't tell you what type of solution it's in and you see that H plus is present. That must mean that it's being balanced in an acidic solution. You would still ignore that H Plus and then proceed to balance it in an acidic solution. If it doesn't tell you what solution is it again? But you see the presence of O. H minus. That means it was balanced at some point within basic solution. So again you would remove the O. H minus and then balance it within basic solution. At the end the O H minus would reappear. So here we have chromium goes with chromium and then we have this oxygen connected to this oxygen. So we're gonna have cr 2072 minus giving us cr three plus for one half reaction and H 202 Giving us 02 for another half reaction. Remember we first balanced elements different from oxygen and hydrogen here we have two chromium here we only have one so we put it to here here we don't have anything different from carbon and from oxygen and hydrogen here So we just leave this alone for now. Next we're gonna balance oxygen's by adding water here we have seven oxygen's so I have to put seven waters here. Over here we have two oxygen's and two oxygen, so we're fine. Next balance out hydrogen by adding H plus so we have seven times two gives me 14 H Plus here I have two hydrogen is here. So I have to put two H plus here. Now balance our overall charge. So this is 14 times plus one is plus 14 -2. Gives me a plus 12 charge overall for this site. Then here water is neutral, so ignore. Then it's two times plus three is plus six overall for this side over here, it's neutral Here, it's two times plus one. So it's plus two add electrons are the more positive side and add enough electrons so that it has the same overall charge as the other side. So here we have plus 12, I add six electrons. So that this side becomes plus six overall, just like this side is plus six overall here, I put two electrons so that this side is zero just like this side is zero overall for charge. The number of electrons don't match. So I have to multiply this by three so that I have six electrons in this half reaction and six electrons in this half reaction. Now rewrite both half reactions and then cross out intermediates. So writing down everything And now here everything is getting multiplied by three. Now we can cancel out of intermediates. So the electrons must always totally cancel out all six of these, cancel out with six from here but we're still left with eight at the end and those are all of our intermediates. So bring down everything. So here would be are balanced redox reaction in an acidic environment. So that would be our final answer. So again remember they tell us to balance in an acidic environment so we balance it in a cynic environment. If H plus H minus or water already is already present, remove them because they'll get added back in once we finish balancing the redox reaction. If H plus is present from the very beginning, that means we're balancing in an acidic environment If o H minus is present from the very beginning, that means we're dealing with a basic environment. Remember these fundamentals to help you balance any redox reaction that you see now that you've seen this example, attempt to do the practice question. We haven't done a basic one yet. So you can at least balance it up to at least the acidic steps. The first six steps and then come back and see how I finish it off with step seven. Help balance it in a basic environment

15

Problem

Problem

Balance the following redox reaction in basic solution.