Basic Concepts - Online Tutor, Practice Problems & Exam Prep

Redox reactions involve the transfer of electrons, where oxidation is the loss of electrons and reduction is the gain. In electrochemical cells, this transfer generates voltage, with key concepts including electrical charge (q, in coulombs), current (I, in amperes), and voltage (E, in volts). Important equations include $q=nF$ and $P=VI$. Balancing redox reactions requires systematic steps, especially in acidic or basic solutions, ensuring charge and mass conservation.

Basic Redox

Recall that oxidation-reduction reactions, or redox reactions, involve the transferring 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".

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. 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'd be +1. For lithium, it goes from being 0 to +1 so 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, Cl₂ 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 -1, so each chloride ion1 ion is -1 in terms of oxidation. Together, they would be -2.

We would say here that we said lithium was being oxidized, so it's losing electrons. It loses in this case, technically, 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 you 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 transferring of electrons between reactants.

Basic Redox

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 transfer 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.

Now here when we talk about electrical charge, realize that charge uses the variable \( q \). And the units for electrical charge are measured in coulombs. Capital C would represent our coulombs. Now we're going to say here that connected to charge, connected to coulombs is Faraday's constant. So to figure out Faraday's constant, we have the charge of an electron which is \( 1.602 \times 10^{-19} \) coulombs times Avogadro's number, by moles inverse.

F = 96447 ∕ C/mol

At the end, that gives us Faraday's constant, which is \( 9.647 \times 10^4 \) coulombs over 1 mole of electrons. So this is the charge for 1 mole of electrons. Now here, we're going to say \( q = n \times F \). So 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. What this is telling me, it's telling me that the 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 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 \( e \times q \). 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.

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 \( \Delta g \) and electrical potential. So here, our electric potential, which is \( e \) again, our voltage. Here, it's Gibbs free energy equals negative \( n \), which is your moles of electrons times Faraday's constant times your voltage. Here, we'd have \( \Delta g = n \times F \times e \).

Δ G = − n F e

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. So that's 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 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. 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. So 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, 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 Redoxs Calculations 1

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? 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 would be volts 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. 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. So those are the real world applications of changing 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

So here it says 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 asked to determine resistance, so \( R \). 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 \( I = \frac{E}{R} \).

Rearranging this to isolate \( R \) by itself, we see that \( R = \frac{E}{I} \). So 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. So, that would be our answer in terms of this example. So just remember, all we're doing here is using the 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.

• A. 0.075 C/s
• B. 7.68x103 C/s
• C. 13.3 C/s
• D. 7.50 C/s

Solve for the missing variable in the following circuit.

• A. 66.6 V
• B. 150 V
• C. 0.015 V
• D. 1.50x103 V

Basic Redoxs Calculations 2

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 ΔG of reaction under standard conditions is ΔG=−RTlnk.

This involves utilizing an equation that we saw in the past to connect our equilibrium constant k to ΔG. Here, R is -8.315 here or 8.314472 joules per Kelvin per moles. The temperature needs to be in Kelvin, so we're going to add 273.15 to 25 degrees Celsius to give us 298.15 Kelvin. Then we're going to have ln of k which is 0.130. Here, our Kelvins cancel out.

Our ΔG value comes out to −322.3J per mol. That's how we get our ΔG in the beginning. Now we're going to have to form a connection between ΔG 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 ΔG.

−322.3J per mol equals moles of electrons transferred, which is 4 moles of electrons, times Faraday's constant, which is 9.647×104 coulombs per mol 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. So negative 4 moles of electrons times 9.647×104 coulombs per mol of electrons on both sides. So when we work that out, we have our final answer really in terms of joules per coulombs. So here, this comes out to 8.35×10−4J per C. And remember, joules per coulombs really just means volts.

So, this represents our cell potential. 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.

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 2 question.

Basic Redoxs Calculations 2

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 it would take for 525 milligrams 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.

Here, we're just going to figure it out in terms of seconds. Now remember, an ampere represents the unit for current. And 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. One 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 1 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 9.647×104 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 3.7×102 seconds involved. This is a typical electrochemical question where we're tying charge and 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?

• 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

Basic Redoxs Calculations 3

Here it states, in the following reaction, 1 mole of zinc solid plus 1 mole of copper sulfate react to produce 1 mole of zinc sulfate plus 1 mole of copper solid. It asks, what is the maximum energy produced when 15 grams of zinc is completely reacted in a zinc copper electrochemical cell with an average cell potential of 1.10 volts? They are asking for the maximum amount of energy. This is equivalent to asking for the maximum amount of work that can be done.

They are asking us to figure out ΔG (delta G). Here, they provide the mass of zinc and the cell potential. The formula connecting delta G to E_cell is ΔG = -n × F × E_cell. First, we need to determine the number of electrons transferred.

Zinc starts with an oxidation state of 0, but when paired with sulfate, which has an oxidation state of -2, zinc adopts an oxidation state of +2, indicating that 2 electrons are transferred. Similarly, copper starts with an oxidation state of +2 and changes to 0, confirming that 2 electrons are involved in the process. Thus, n equals 2.

We will use Faraday's constant, 96470 Coulombs per mole of electrons, and multiply it by the voltage, where a volt equals joules per coulomb, resulting in 1.10 joules per coulomb. With units of electrons and Coulombs cancelling out, the calculation yields -2.12 × 10⁵ joules per mole. This result corresponds to exactly 1 mole of zinc solid.

However, we are dealing with 15 grams of zinc, less than 1 mole. Convert 15 grams of zinc to moles using the molar mass of zinc, 65.409 grams per mole. Multiplying the moles of zinc obtained by the energy per mole gives us -4.87 × 10⁴ joules.

Thus, we find that 4.87 × 10⁴ joules of energy is produced as a result of reacting 15 grams of zinc. Understanding the variables provided and their connection via formulae allows us to deduce the energy released with less than a mole of zinc. Now proceed to example 2 and attempt to answer it. If you encounter difficulties, refer back to this approach to see how I tackled example 1.

Basic Redoxs Calculations 3

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 2 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 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 2 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. Our mass here is 110 pounds, but we need that to be in kilograms.

So we're going to say here, 110 pounds times for every 1 kilogram, it's 2.205. So that's 49.8866 kilograms. Gravitational acceleration is 9.8 meters over seconds squared. So that's gonna 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 2 together, that's gonna give you 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 gives me as my answer here, 234.7 watts. And when we have joules over seconds, that comes out to be watts. So she generates 234.7 watts of power by going from the 1st floor to the 2nd floor. Now 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 that power equals work over time. We were already given work, we were already given 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 the second equation, work equals force times distance. Distance was given to us already.

We had to determine what our force is. 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 chemist. 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

So in this example, it says, determine the amount of time in minutes needed to produce 1.7 times 10 to the 2 watts from 1500 joules of work committed. All right. 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 going to say here power is in units of watts and a watt is equal to joules per second.

So that's 1.7×102 joules per second. And then we're going to say here our work is 1500 joules. 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.

Multiply both sides by x. So 1.7×102 joules/seconds times x equals 1500 joules. Divide both sides here by 1.7×102 joules/seconds. Joules cancel out. We'll 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 0.147 or 0.15 minutes. So that's the amount of time it would take.

So just remember all the different formulas that utilize, in some way, 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 in example 2. So keep in mind the different formulas that you will have to utilize at some point, whether it's asking you to solve for power, for current, or energy.

Balancing Redox Reactions

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 are a lot more steps that are necessary to get our final 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 do first.

First, we write the equation into two half reactions. Step 2, we balance elements that are different from oxygen and hydrogen first. Then 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 different numbers 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 electrons in both.

From there, we combine the half reactions and cross out the reaction intermediates. Now, when it comes to balancing in our basic solution, we follow the first six steps as we would above but then we add a seventh step. In this seventh step, we say 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. So, we're just going to 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 going to 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 Br₂ and MnO₄^- gives us Mn²⁺. First, we balance elements different from oxygen and hydrogen. We have 1 Br here but 2 Br's here, so we're going to put a 2.

Next, we have 1 manganese and 1 manganese, so that's already tied and that's balanced. Next, we balance oxygens 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⁺ here. Now we have to balance overall charge. Oops. And actually we put the 2 here, to give us 2 Br's on both sides.

We balance out 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. 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 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 Br₂ plus 10 electrons.

And then here we're going to have 2 So everything I multiply 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.

This here represents our balanced chemical equation in an acidic environment. Just remember, when it comes to balancing a redox reaction, there are 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

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⁺, 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 balance it within basic solution. At the end, the OH^- would reappear.

So here, we have chromium connects with chromium, and then we have this oxygen connected to this oxygen. So, we're going to have Cr₂O₇^2- giving us Cr³⁺ for one half reaction, and H₂O₂ giving us O₂ for another half reaction. Remember, we first balance elements different from oxygen and hydrogen. Here we have two Cr. Here, we only have one, so we put a 2 here.

Here, we don't have anything different from oxygen and hydrogen, so we just leave this alone for now. Next, we're going to balance oxygens by adding water. Here we have seven oxygens, so I have to put seven 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 that 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 so that this side is 0, just like this side is 0 overall for the charge. The number of electrons doesn't match, so I have to multiply this 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. Cancel out intermediates, so the electrons must always totally cancel out. All six 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 the acidic steps, the first six 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.

Cr₂O₇^2– (aq) + SO₃^2– (aq)  →  Cr³⁺ (aq)  +  SO₄^2– (aq)

Here we're going to balance this redox reaction in a basic environment. We're going to first break it up into its half reactions. So we have chromium with chromium, sulfur with sulfur. So that's going to give us our two half reactions initially. Now, we balance elements different from oxygen and hydrogen first. We have 2 chromiums here but only 1 here, so we put a 2. Then we have high, oxygens by high, oxygens by adding water. So we have to add 7 waters here. Here we have 3 oxygens and here we have 4, so we put one water here. Next, we're going to balance out waters by adding hydrogens by adding H+. So we have 7 times 2 is 14 H+ and then we have 2 hydrogens here so that's 2 H+. Balance overall charge, so plus 14 minus 2 gives me plus 12 overall. And then here, this is 2 times plus 3, so that's plus 6. Here, this would be minus 2 overall for the charge. And over here, it's 2 times plus 1 is plus 2. Minus 2 is 0. So I'm going to put 6 electrons here so that both sides are plus 6. And then I have to put 2 electrons here so that both sides are minus 2. The number of electrons are not the same, so I multiply this here by 3. So what I'm gonna get now is I'm gonna get Cr2O72- + 14H+ + 6e- gives me 2Cr3+ + 7 H2O. Then we have 3SO32- + 3H2O gives me 3SO42- + 6H+ + 6e-. Now, cross out the intermediates. So this cancels out with this. This cancels out leaving us with 8 here. Bring down everything. So this will represent our balanced equation in an acidic environment. So we're not quite done yet because we need to balance it in a basic environment. The 7th step is we look at how much OH-, how much H+ is left and we add the same number of OH- to both sides. And then also we have 7 H2O. So, these here combined together to give me 8 waters. So we add everything. Now, here you can have water on both sides, so all 7 of these cancel out with 7 from here leaving us with one left. And that represents our balanced equation in a basic environment. We just lost 7 of these so we have one left. So this would be our basic solution with its balanced redox reaction. So remember, balancing it in this type of solution is almost the same as balancing it in an acidic environment except we have that last additional step where we add equal amounts of OH- to both sides based on the amount of H+ remaining. When H+ and OH- combine, they give us water molecules. We can't have water on both sides of the equation, so they're going to cancel out and we'll see what is remaining. When it comes to balancing redox reactions, remember the steps necessary in order to get your final answer.

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 (Cl₂), 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 H₂O 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.