Recall that oxidation-reduction reactions, or redox reactions, involve the transfer of electrons from one reactant to another. Now here in our equation, we have lithium solid reacting with chlorine gas. In this process, we produce lithium ion as well as 2 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. So remember, LEO means that we're losing electrons and therefore representing 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, its oxidation number is becoming more positive as a result. We're going to say here that if you are being oxidized, then you represent the reducing agent or the reductant. In this case, we have lithium solid in its natural state so its oxidation number is equal to 0. Now, as an ion, its oxidation number is tied to its charge. Here it would be plus 1. For lithium, it goes from being 0 to plus 1. Its oxidation number has increased. Therefore, it has been oxidized. Therefore, it is the reducing agent or the reductant. 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_{2} gas, it's in its natural or elemental state so its oxidation number is equal to 0. Now as an ion, its oxidation number is tied to its charge. Each chloride ion is minus 1, so each chloride ion is minus 1 in terms of oxidation. Together, they would be minus 2. We would say here that we said lithium was being oxidized so it's losing electrons. It loses in this case, technically, it would lose 2 electrons, and chlorine itself would pick up those 2 electrons because each chlorine needs to gain an electron so each chlorine becomes a chloride ion. As we go deeper and deeper into redox reactions, we'll talk about concepts such as cell potential, the use of electrochemical cells, 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 transfer of electrons between reactants.

- 1. Chemical Measurements1h 50m
- 2. Tools of the Trade1h 17m
- 3. Experimental Error1h 52m
- 4 & 5. Statistics, Quality Assurance and Calibration Methods1h 57m
- 6. Chemical Equilibrium3h 41m
- 7. Activity and the Systematic Treatment of Equilibrium1h 0m
- 8. Monoprotic Acid-Base Equilibria1h 53m
- 9. Polyprotic Acid-Base Equilibria2h 17m
- 10. Acid-Base Titrations2h 37m
- 11. EDTA Titrations1h 34m
- 12. Advanced Topics in Equilibrium1h 16m
- 13. Fundamentals of Electrochemistry2h 19m
- 14. Electrodes and Potentiometry41m
- 15. Redox Titrations1h 14m
- 16. Electroanalytical Techniques57m
- 17. Fundamentals of Spectrophotometry50m

# Basic Concepts - Online Tutor, Practice Problems & Exam Prep

## Basic Redox Concepts

### Basic Redox

#### Video transcript

### Basic Redox

#### Video transcript

With the concept of redox reactions, we have the transferring of an electron from one reactant 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, electrical charge, as well as work. Now with these concepts, we have essential formulas that we're going to have to use and apply to get the answers that we want. When we talk about electrical charge, realize that charge uses the variable *q*. And the units for electrical charge are measured in coulombs. So capital *C* would represent our coulombs. Now we're going to say here that connected to charge connected to coulombs is Faraday's constant. To figure out Faraday's constant, we have the charge of an electron which is 1.602×10-19 coulombs times Avogadro's number. At the end, that gives us Faraday's constant which is 9647×104 coulombs over 1 mole of electrons. Now here, we're going to say *q* equals charge. It equals *n* times Faraday's constant. Here, we'd have moles of electrons times Faraday's constant. One mole of electrons here on the bottom. Your moles of electrons would cancel out so that at the end, your charge would have units of coulombs.

Here, we could also talk about electrical current. For current, use the variable *I*. The unit for electrical current is in amperes or amps. We can say *A* here represents the units for current. Now we're going to say here that current equals charge, which we said the units would be coulombs divided by time. Here, the units for time would be in seconds. This tells us that current is in units of amperes and an ampere represents coulombs per second. If you are given 25.0 amps, that would translate into 25 coulombs per one second.

Next, we have electrical voltage. Now with electrical voltage, we have a series of equations we can use. Here we can say the relationship between work and voltage can be expressed as work equals voltage, which is *e* times our charge *q*. Now here when it comes to voltage, the units for voltage are energy in terms of joules divided by coulombs. We already said earlier that charge uses the units of coulombs. Here, coulombs would cancel out. Work at the end would have units of joules. 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 1 newton is equal to kilograms times meters over seconds squared. So here we have kilograms times meters over seconds squared. Distance is meters. So that would come out to kilograms times meters squared over seconds squared. All those units combined together equal 1 joule. So, whether we're utilizing this equation here for work or this equation here for work, both give us joules as the units for work.

Next, we have the relationship between Gibbs free energy which is *Δg* and electrical or electric potential. So here our electric potential, which is *e* again, our voltage. Here, it would say Gibbs free energy equals negative *n*, which is your moles of electrons, times Faraday's constant times your voltage. Here, we'd have *Δg* equals moles of electrons times Faraday's constant, which remember is coulombs over moles of electrons. Voltage, remember we just said in the previous example dealing with work that voltage is joules over coulombs. So moles of electrons would cancel out, coulombs would cancel out and you'd have joules as your final units for Gibbs free energy here as well.

We also have Ohm's law. Ohm's law, we're going to say uses units for resistance and that would just be ohms or omega. Here, we'd say that our current, which is *I* equals your voltage over your resistance. This is just yet 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 coulombs 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 joules times seconds. So a *1 watt* is equal to joules per second. Here, power also equals voltage times current. So that'd be joules over seconds for voltage times their current which we said was coulombs per second. Those would cancel out and we'd get joules per second. We could utilize this equation to help us determine power or we could utilize this equation here to help us determine power.

With some of these, we have the same variable with more than one method to get to the answer for it. Just keep in mind that with redox reactions, we're 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've 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

### Basic Redoxs Calculations 1

#### Video transcript

Here it states, what happens to the current in a circuit if a 3 volt battery is removed and replaced by a 1 volt battery? 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 would be \(q\) and our resistance would be in ohms. If we think about this, we have current. Here, it's 3 volts. We're not given a resistance which means our resistance is being kept the same in both cases. Here, let's say our resistance is \(0.50\) ohms. When we work that out, that would give us a current of \(6\). Remember, the units for current are coulombs per second. Now, we've replaced it with 1 volt. Our resistance can stay the same, so \(0.50\) ohms. And now that's \(2\) coulombs per second. 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 from a 3 volt battery to a 1 volt battery. 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 2 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.

### Basic Redoxs Calculations 1

#### Video transcript

So here it states, if the voltage of a series enhanced balance has a 240 volt battery, what is the resistance in the circuit if the current is 0.80 amperes? Alright. So we're asking to determine resistance, represented by R. We have amperes which represent our current, denoted as I, and then we have voltage here which is denoted as E. So our equation as before is I=ER. Rearranging this to isolate R by itself, we see that R=EI. Here we'll plug in our voltage of 240 volts divided by our current of 0.80 amperes. So that gives me 300 ohms as my resistance. That would be our answer in terms of this example. So just remember all we are doing here is using Ohm's Law and just rearranging it to solve for the missing variable, which in this case is resistance.

Move on to practice question 1 where we're shown an image of a circuit, but again, we're still basing this all off of Ohm's Law. So it's just solving for the missing variable based on the information provided.

Solve for the missing variable in the following circuit.

7.68x103 C/s

Solve for the missing variable in the following circuit.

1.50x103 V

### Basic Redoxs Calculations 2

#### Video transcript

Here it says to calculate Gibbs free energy under standard conditions for our reaction, as well as our cell potential under standard conditions for a redox reaction with n=4 that has an equilibrium constant of k=0.130 at 25 degrees Celsius. Alright. Here, remember that n represents our moles of electrons. We have 4 moles of electrons. We're going to say here that the connection between our equilibrium constant k and our Δ_{g} of reaction under standard conditions is Δg=−RTlnk. This involves utilizing the equation that we saw in the past to connect our equilibrium constant k to Δ_{g}. Here, R is 8.314472 joules per kelvin per mole. Temperature needs to be in Kelvin, so we're going to add 273.15 to 25 degrees Celsius to give us 298.15 Kelvin. And then we're going to have ln(0.130). Here, our Δ_{g} value comes out to −3223.3 joules per mole. That's how we get our Δ_{g} in the beginning. Now we're going to form a connection between Δ_{g} and our cell potential. Here, the equation that connects them is Δg = −nFE_{cell}. We're going to input the value that we just got in terms of Δ_{g}. −3223.3 joules per mole equals moles of electrons transferred, which is 4 moles of electrons, times Faraday's constant, which is 96485 coulombs per mole of electrons. And we don't know what E_{cell} is. That's what we're looking for. So divide out this portion here to isolate our E_{cell}. Then, calculating this, we have E_{cell}=0.0835V. Remember, this question utilized questions, 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 on to the next one. Here we're going to apply stoichiometry to the whole idea of current. We're given 4.3 amps and some mass of copper, and from that, we need to determine the amount of time that has elapsed. 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 2 question.

### Basic Redoxs Calculations 2

#### Video transcript

Copper can be electroplated at the cathode of an electrolysis cell by the half-reaction of copper 2+ ion + 2 electrons produces 1 mole of copper solid. Here, we were asked how much time would it take for 525 milligrams of copper to be plated at a current of 4.3 amperes? All right. We're looking for time. It doesn't specify if we want to find it in minutes, seconds, years, etcetera. Here, we're just going to figure it out in terms of seconds. Now remember, an amp or an ampere represents the unit for current. Remember, an ampere is equal to charge per over time. So charge or coulombs over s for seconds. So 4.3 amperes represents 4.3 coulombs per second. We're going to first start out with 525 milligrams of copper. We're then going to convert those milligrams of copper into grams of copper. So 1 milligram of copper is equal to 10-3 grams of copper. Milligrams cancel out. Now we have grams of copper. We're next going to change those grams into moles. 1 mole of copper according to the periodic table weighs 63.546 grams for copper. Now that I have moles of copper, I can say for my equation that for every 1 mole of copper solid, we have 2 moles of electrons involved. So for every one mole of copper from my balanced equation, there are 2 moles of electrons involved. Moles of electrons are really a part of Faraday's constant. So We're going to say we have 1 mole of electrons and according to Faraday's constant, that's equal to 9.647104 coulombs. Finally, we have coulombs which can cancel out with these coulombs leaving us with seconds at the end. We put 4.3 coulombs on the bottom, one second on top. Now, we're going to have seconds remaining. Here, we're going to get 3.7102 seconds involved. This is a typical electrochemical question where we're tying charge, current to stoichiometry in some way. Using the balanced equation, we see that the ratio is for every 1 mole of our solid metal, we have 2 moles of electrons involved. That comparison there is what allows us to find a way of connecting current to stoichiometry. Now that we've seen this example, move on to the final question on this page where we have to determine the 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. Good luck, guys.

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?

### Basic Redoxs Calculations 3

#### Video transcript

Here it states, In the following reaction, 1 mole of zinc solid plus 1 mole of copper sulfate react in order to produce 1 mole of zinc sulfate plus 1 mole of copper solid. Here it asks, what is the maximum energy produced when 15 grams of zinc is used in a 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. This is equivalent to asking what's the maximum amount of work that can be done. What they're asking us to figure out is ΔG. Here they give us the mass of zinc and they're giving us cell potential. They're giving us E_{cell}. Realize that the formula that connects ΔG to E_{cell} is ΔG = -nF E_{cell}. Here we have to determine the number of electrons transferred. Well, zinc is initially 0 but when it's combined with sulfate, it's +2 because sulfate is -2. So that's 2 electrons transferred. At the same time, copper here has to be +2 and here it's 0. So we have 2 electrons involved that are being transferred. So that means n is 2. So we have 2 moles of electrons. We're gonna multiply it by Faraday's constant, 96470 C/mol of electrons. Then we're gonna multiply by my voltage. Remember, a volt is equal to joules/coulombs, so that's equal to 1.10 joules/coulomb. What happens here is moles of electrons cancel out. Coulombs cancel out, so you're left with joules. So that comes out to -2.12 × 10^{5} joules/mole.

Here's the thing. This is the answer when we're dealing with exactly 1 mole of zinc solid. But in the question, we're not dealing with exactly 1 mole of zinc solid. What we're dealing with is 15 grams of zinc which is less than 1 mole. What we're going to do here is we're going to take those 15 grams. We're going to convert them into moles. We're going to say here for every 1 mole of zinc, the mass is 65.409 grams of zinc. Grams of zinc cancel out. Now I have moles of zinc. And we're going to say here for every 1 mole of zinc, we found this value here. So it's 2.12 × 10^{5} joules. So this cancels out with this. What we get at the end is -4.87 × 10^{4} joules. So you can say here that we produce 4.87 × 10^{4} joules as a result of dealing with just 15 grams of zinc. Remember, in this type of question, realize what variables they are giving us and how they are connected together in terms of a formula. Knowing that part allowed us to determine what the energy released when it comes to 1 mole of zinc. Realizing in the question that we're dealing with less than a mole of zinc, we then determined what the new amount of energy involved would be. Now that you've seen this example, move on to example 2. Look to see if you can answer it. If you get stuck, don't worry. Come back and see how I approach example 2.

### Basic Redoxs Calculations 3

#### Video transcript

Here it states, A chemist weighing 110 pounds takes her NMR sample from the first floor to the second floor, which is 12 meters 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 going to have to utilize the first formula here. We're going to say here that power equals work over time. We already know what our time is. It's 25 seconds. What 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. We can't utilize the first formula. We're going to have to utilize this formula here. Work equals force times distance. Bringing that down, force will be in units of newtons and distance will be in meters. When you multiply those two together, that'll give us joules as the units. In this question, I don't give us newtons directly but I do give us distance. We say that she's traveled 12 meters up, so that's our distance traveled. What we need to do now is we need to determine what our newtons will be. Well, remember that 1 newton is equal to kilograms times meters over seconds squared. And we're going to 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 a 110 pounds but we need that to be in kilograms. We're going to say here, 110 pounds times for every 1 kilogram, it's 2.20 pounds. So that's 49.8866 kilograms. Gravitational acceleration is 9.8 meters over seconds squared. That's going to give me 488.889 kilograms times meters over seconds squared. So that force represents my newtons. So that's 488.889 newtons. When we multiply those two together, that's going to give me my joules. So that comes out to being 5866.67 joules. So we just determined our work. Take that and plug it in to find power. So 5866.67 joules over 25 seconds just that comes out to be watts. So she generates 234.7 watts of power by going from the first floor to the second floor. This question was a bit tricky. It really involved us using different formulas in order to find our final answer. So realize here we had to first utilize this formula P=Wt. We were already given work, we're already time in 25 seconds, so that's the easy part. The hard part was to determine what our work is. To determine our work, we use this second equation, W=F⋅d. Distance was given to us already. We have to determine what our force is. 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. 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. Attempt it on your own. Once you do, come back and see if your answer matches up with mine.

### Basic Redoxs Calculations 3

#### Video transcript

So in this example it says, "Determine the amount of time in minutes needed to produce 170 watts from 1500 joules 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. Thus, power equals work over time. We're going to say here power is in units of watts, and a watt is equal to joules per second. So that's 170 joules per seconds. Our work is 1500 joules. 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. Multiply both sides by x. So 170joulessecondsx equals 1500 joules. Divide both sides by 170joulesseconds. Joules cancel out. We'll have seconds. So x=8.82 seconds. And then you just have to change seconds into minutes. One minute is equal to 60 seconds. So that gives me 0.147 or approximately 0.15 minutes. So that's the amount of time it would take.

Always remember the different formulas utilized in questions dealing with energy, current, and charge. These topics are separate from each other but are connected in one way or another, as we saw in this example. Keep in mind the different formulas that you will have to utilize at some point, whether it's asking you to solve for power, current, or energy.

## Balancing Redox Reactions

### Balancing Redox Reactions

#### Video transcript

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. When it comes to balancing a redox reaction though, there are a lot more steps that are necessary to get our balanced reaction. Here we're going to learn how to balance a redox reaction in acidic solutions as well as basic solutions.

If we're in an acidic environment, these are the steps that we follow first:

- We write the equation into two half-reactions.
- We balance elements that are different from oxygen and hydrogen first.
- We balance oxygens by adding water and balance hydrogens by adding H
^{+}. - 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 a different number of electrons in each half-reaction, you may have to multiply one or both half-reactions by values so that they have the same number of to it comes to balancing in a basic solution, we follow the first six steps as we would up above but then we add a seventh step. In this seventh step, we balance remaining H^{+} by adding an equal number of OH^{-} 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. We're just going to go straight into this example guys, and we have to balance this redox reaction in an acidic environment. We're going to say here that we're gonna break it up into two half-reactions. So bromine goes with bromine, and manganese goes with manganese. So, our two half-reactions are:

- Br
^{-}gives us Br_{2} - MnO
_{4}^{-}gives us Mn^{2+}

First, we balance elements different from oxygen and hydrogen. We have 1 Br here but 2 Br's here, so we're gonna put a 2. Next, we have 1 manganese and 1 manganese so that's already tied and that's balanced. Next, we balance oxygen by adding water. We have no oxygens in this first half-reaction here. Here we have 4 oxygens, and on this side, we have none, so I'm going to add 4 waters. Then we balance out hydrogens by adding H^{+}. Here we have no hydrogens at all on either side. Here we have 4 times 2 which gives me 8 hydrogens, so I put 8 H^{+}.

Now we have to balance the overall charge. So this is 2 times negative one which is negative 2. Over here, this is neutral so this is 0. Alright. Then we have 8 times plus 1 which is plus 8. Minus 1 gives us plus 7 overall for this side of the half-reaction. Then water here is neutral, so just ignore it, and then this is plus 2. We balance the overall charge by adding electrons to the more positive side. So on this first half-reaction, this side, that has an overall charge of 0, 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 2 electrons. So now both sides are minus 2. Over here, this side is more positive at plus 7. I need to add 5 electrons so that it has an overall charge of plus 2 just like this side here has an overall charge of plus 2. Notice your electrons do not equal each other. One is 2 and one is 5. We say that their lowest common multiple that they share is 10. So that means I'd multiply this here by 5 and this here by 2.

So what I get at this point is 10 Br^{-} gives me 5 Br_{2} plus 10 electrons. And then here we're gonna have 2 permanganate ions plus 16 H^{+} plus 10 electrons gives me 2 manganese, 2 ions, plus 8 waters. So everything I multiplied by 2. Cancel out intermediates, things that look the same except one's a product, and one's a reactant. Your electrons are supposed to always completely cancel out. From this, 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.

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. Keep practicing. You'll be able to do this really quickly as long as you remember the sequence of steps necessary.

### Balancing Redox Reactions Calculations

#### Video transcript

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, H^{+}, OH^{-} or water, that must mean those were added at some point while balancing the reaction. 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^{+} is present, that must mean that it's being balanced in an acidic solution. You would still ignore that H^{+} and then proceed to balance it in an acidic solution. If it doesn't tell you what solution it's in again, but you see the presence of OH^{-}, that means it was balanced at some point within basic solution. So again, you would remove the OH^{-} and then Here, we have chromium goes with chromium. So, here we have chromium goes with chromium and then we have this oxygen connected to this oxygen. So, we're gonna have Cr_{2}O_{7}^{2-} giving us Cr^{3+} for one half-reaction and H_{2}O_{2} giving us O_{2} for another half-reaction. Remember, we first balance elements different from oxygen and hydrogen. Here we have 2 chromiums. Here we only have 1 so we put a 2 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 going to balance oxygens by adding water. Here we have 7 oxygens, so I have to put 7 waters here. Over here we have 2 oxygens and 2 oxygens so we're fine. Next, balance out hydrogens by adding H^{+}. So we have 7 times 2 gives me 14 H^{+} here. I have 2 hydrogens here so I have to put 2 H^{+} here. Now balance our overall charge, so this is 14 times plus 1 is plus 14 minus 2 gives me a plus 12 charge overall for this side. Then here water is neutral, so ignore. Then it's 2 times plus 3 is plus 6 overall for this side. Over here it's neutral. Here it's 2 times plus 1 so it's plus 2. Add electrons to 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 6 electrons so that this side becomes plus 6 overall just like this side is plus 6 overall. Here, I put 2 electrons 6 electrons in this half-reaction. By 3 so that I have 6 electrons in this half-reaction and 6 electrons in this half-reaction. Now, rewrite both half-reactions and then cross out intermediates. So, writing down everything And now here, everything's getting multiplied by 3. Now, we cancel out of intermediates so the electrons must always totally cancel out. All 6 of these cancel out with 6 from here but we're still left with 8 at the end And those are all of our intermediates. So bring down everything. So here would be our balanced redox reaction in an acidic environment. So that would be our final answer. So again, remember, they tell us to balance it in an acidic environment so we balance it in an acidic environment. If H^{+}, OH^{-}, or water is already present, remove them because they'll get added back in once we finish balancing the redox reaction. If H^{+} is present from the very beginning, that means we're balancing in an acidic environment. If OH^{-} 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 6 steps and then come back and see how I finish it off with step 7 to help balance it in a basic environment.

Balance the following redox reaction in basic solution.

Cr_{2}O_{7}** ^{2–}** (aq) + SO

_{3}

**(aq) → Cr**

^{2–}**(aq) + SO**

^{3+}_{4}

**(aq)**

^{2–}#### Problem Transcript

### Here’s what students ask on this topic:

What is a redox reaction and how does it involve electron transfer?

A redox reaction, short for reduction-oxidation reaction, involves the transfer of electrons between two species. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. For example, in the reaction between lithium (Li) and chlorine (Cl_{2}), lithium loses electrons (oxidized) to form Li^{+}, and chlorine gains electrons (reduced) to form Cl^{-}. The mnemonic 'LEO the lion says GER' helps remember that Loss of Electrons is Oxidation (LEO) and Gain of Electrons is Reduction (GER).

How do you balance redox reactions in acidic and basic solutions?

Balancing redox reactions involves several steps. In acidic solutions, first, split the reaction into two half-reactions. Balance elements other than O and H, then balance O by adding H_{2}O and H by adding H^{+}. Balance the charge by adding electrons. In basic solutions, follow the same steps but add OH^{-} to balance H^{+} ions. Finally, ensure both half-reactions have the same number of electrons and combine them.

What is Faraday's constant and how is it used in redox reactions?

Faraday's constant (F) is approximately 96485 C/mol and represents the charge of one mole of electrons. It is used in the equation q = nF, where q is the charge in coulombs and n is the number of moles of electrons. This constant is crucial in calculating the charge transferred in electrochemical reactions and in determining the relationship between electrical work and chemical energy.

What is the relationship between Gibbs free energy and electrical potential in redox reactions?

The relationship between Gibbs free energy (ΔG) and electrical potential (E) in redox reactions is given by the equation ΔG = -nFE, where n is the number of moles of electrons, F is Faraday's constant, and E is the electrical potential. This equation shows that the Gibbs free energy change is directly related to the electrical work done by the redox reaction.

How do you calculate electrical current in an electrochemical cell?

Electrical current (I) in an electrochemical cell is calculated using the equation I = q/t, where q is the charge in coulombs and t is the time in seconds. The unit for current is amperes (A), which is equivalent to coulombs per second (C/s). This relationship helps in understanding the flow of electrons and the rate at which they are transferred in the cell.