Fundamentals of Electrolysis - Online Tutor, Practice Problems & Exam Prep

Electrolysis involves passing an electrical current through a substance to induce chemical changes, making it a non-spontaneous process. For example, passing current through water generates hydrogen and oxygen gases. The current (I) is measured in amperes (A), where 1 A = 1 coulomb/second. The moles of electrons can be calculated using the equation: $n=\frac{I}{F}t$, where F is Faraday's constant. Factors like ohmic potential, overpotential, and concentration polarization affect the overall cell potential, making it more negative and requiring more current to drive the reaction.

Fundamentals of Electrolysis

So when we talk about electrolysis, we're going to say that electrolysis deals with passing an electrical current through a substance in order to produce chemical changes. Now because we're using outside energy, we're going to say that processes dealing with electrolysis are non-spontaneous. Remember non-spontaneous things do not occur naturally and therefore require this outside energy in order to drive the reaction forward. Now here we're going to say a good example of electrolysis. We have the passing of an electric current through water. This helps to generate the standard components of water. So here we have water as a liquid. We drive electricity through it and as a result, we produce oxygen gas and hydrogen gas. This is not a natural thing. We need that outside energy in the form of the current in order to force the reaction to occur. Some of these concepts, we've seen before. But just to revisit, remember, we deal with electrical currents. We're going to say here the units for electrical currents are in amperes or amps. So if we are given 15 amps, we'd say an amp is equal to Cs. So this would be equal to 15 coulombs per second. And remember, when we're talking about current, we can use the variable I. I here, which is current, equals qt. So q here would be our charge. Here, this would be our time. If we're talking about moles of electrons, we can say the moles of electrons within a reaction are determined by molesofelectrons=current×timeFaraday'sconstant. Remember, current is coulombs per second. Time here will be in seconds. Faraday's constant has the units Cmoles of electrons. So if you see, coulombs cancel out. Seconds cancel out. That is how you are left with moles of electrons at the end. Now, with non-spontaneous processes, we should realize that our cell potential will be negative. It'll be less than 0. And we have to take into account that with non-spontaneous processes, our current is not negligible. It's a necessary part of the whole process. Because of this, we're going to have to take into account 3 other terms. We're going to have ohmic potential which we're going to say is E. It's the voltage necessary to overcome resistance R when the current I is flowing. Remember here, we'd say I can also equal ER. So I again is our current. Resistance. We're going to also have another term, overpotential. This is the voltage required to overcome the activation energy for reaction on the given electrode. It's a non-spontaneous process, so the activation energy has to be overcome in order to drive the reaction forward. Next, we have concentration polarization, occurs when there is a difference in the concentration of reactants on the surface of the electrode when compared to the solution itself. So the concentration of ions could be a very different number when it's close to the metal surface of our electrode or inert electrode as opposed to within the bulk solution itself. Now we're going to say here, electrolysis is made more difficult by these different values because here, your ohmic potential here is subtracting from our total and our overpotentials are subtracting from our total. Our concentration polarization is found here between these two. All these concepts help to make our overall cell potential more negative. And remember, the more negative and smaller your cell potential becomes, the more non-spontaneous your process, the more current you need to supply in order to drive the reaction forward. So remember, when we're talking about electrolysis, we're basically running current through a substance to elicit a response. We have to do this because the whole process in itself is not a spontaneous one.

Fundamentals of Electrolysis Calculations 1

So here it states that aluminum can be electroplated at the cathode of an electrolysis cell by the half-reaction Al3++3e-→Al(s). Here it says, how much time would it take for 825 milligrams of aluminum to be plated at a current of 4.1 amperes. So here, we're using current to drive electrons to precipitate a solid. Since a current is being used in electrolysis, this is a non-spontaneous process. Here, we're going to have to figure out time. I don't specify what units time will be in, so we'll just opt for seconds. What we're going to do first is realize that 4.1 amperes really means that we have 4.1 coulombs per one second. So, it's our goal to isolate seconds here at the end. That means I'm going to have to find a way of canceling out coulombs by using these grams these milligrams. So, we're going to start out with 825 milligrams of aluminum. And the first thing we're going to do is convert those milligrams into grams. So, 1 milligram is 10^-3 grams. So milligrams cancel out. Now we have grams. Next, I'm going to change those grams into moles. So 1 mole of aluminum is 26.982 grams. Then we're going to say here, according to my equation, for every 1 mole of aluminum solid, we have 3 moles of electrons involved. So 1 mole of aluminum involves 3molesofelectrons. Moles of electrons are part of Faraday's constant, so it's 96,485 coulombs per 1 mole of electrons. Now that we have these coulombs, we can finally use the amps that we had originally. So, take those 4.1 coulombs and put them on the bottom and one second on top. So here, at the end, what we'll get is we'll get 2.2∙103s. So that's the approximate time it would take to basically plate that many milligrams of aluminum with a current of 4.1 being applied. Now that we've seen this example, move on to example 2. You could attempt it on your own, but if you get stuck, don't worry. Come back and see how I approach this same exact question.

Fundamentals of Electrolysis Calculations 1

Alright. So in this question, it says, in the electrolysis of molecular iodine to iodide ions for 0.15 molar of sodium iodide solution containing 4.2 × 10^-4 molar iodine at a pH of 6 with a partial pressure of oxygen gas equal to 1.25 bar. Calculate the voltage needed to drive the reaction. Alright. So, we have 2 half reactions given to us and remember because this is part of electrolysis, this represents a nonspontaneous process. We're going to say, normally under a spontaneous process, the cathode, which is positive here, should be the larger cell potential. But again, because we're under nonspontaneous conditions, everything is reversed. Now, the cathode will have the smaller cell potential and the anode will have the larger one. So that way, when we do cathode minus anode at the end, we'll get a cell potential that's negative, illustrating that we have a nonspontaneous process that's happening. Alright. So what we're going to do here is we're going to calculate. First, let's calculate the standard cell potential, the potential cell potential of the cathode is. So that equals the standard cell potential of the cathode minus 0.05916 volts divided by the number of electrons transferred times the log of products over reactants. So here, my product here is a liquid. So we're going to ignore that. So that's going to be 1 over. Here we have a gas and it has a pressure in bar, so that's the pressure of O₂. Okay. And then we're going to say here times H⁺ and the coefficient here is 4 to the 4th. Plug in our values, so that's -0.41 volts minus 0.05916 volts divided by the number of electrons transferred. Alright. So the number of electrons transferred has to be the same in both half reactions. The second one is only a 2, so we have to multiply this by 2. OK. So we're going to multiply that by 2. Alright. So now it's still 4 here. So it's 4 electrons, 4 moles of electrons times the log of 1 over Here, we want the pressure in bars. So that's 1.25 times H⁺ to the 4th. Since you know pH, you know what H⁺ is because H⁺ is 10 to the negative pH. So that's 10^-6 to the 4th. So here, that would be -0.41 volts. And then if you do this all correctly, it'll give you -0.35 volts. So that comes out to -0.76 volts. Now we do the same thing for the anode. So now it's the standard cell potential of the anode minus 0.05916 volts divided by n times the log of we multiplied everyone by 2, 2, so the number of electrons would be the same in both half reactions. So now that's actually I^- to the 4th divided by I₂ to the second. So that's 0.54 volts minus 0.05916 volts divided by 4 electrons times the log of I^- to the 4th. So, I^- is 0.15 molar to the 4th divided by the concentration of I₂, so 4.2 × 10^-4 squared. So that's going to give me 0.54 volts minus 0.05 volts, which gives me positive 0.49 volts. Now, we just say that my cell potential now equals cathode minus anode. So that's -0.76 volts minus 0.49 volts equals -1.25 volts. So, since the cell potential is negative, it's a nonspontaneous process, which makes sense because electrolysis is a nonspontaneous process. This means that I have to run a current that high in order to drive the reaction forward. So remember, under nonspontaneous conditions, we have to use outside energy in order to have the reaction occur. In this case, we need to apply or use 1.25 volts of outside energy so this whole process can occur. Now that we've seen this example, take a look at the practice question left at the bottom. If you get stuck, don't worry. Come back and see how I approach that same practice question.

During electrolysis the concentration of I₂ increases to 8.3 x 10 ^-3 M, while all other concentrations remain unchanged. If the electrical resistance is 1.8 ohms, the current is 71 mA, the anode overpotential is 0.013 V and the cathode overpotential is 0.115 V, what is the voltage needed? 

• A. -1.48 V
• B. -1.04 V
• C. -1.54 V
• D. -0.0228 V
• E. -0.846

Electrolysis Calculations 1

Hey guys. So here we're talking about the concept of electrical current. So when we say electrical current, another term for it would be amps or amperes. We are going to say here current or amperes. It is really just charge over time. And we are going to say here, charge, we use c to represent coulombs, and time here would be in seconds. So knowing this is key to get answering these questions.

So for the first one it says, Gold can be plated out of a solution containing gold 3 based on the following half reaction. So here we have gold 3 ion reacting with 3 moles of electrons to give us 1 mole of gold solid. Here we are asked what mass of gold is plated out by a 41-minute flow of 6.8 amperes, or current. Alright. So they are asking us to figure out the mass of gold. Realize here when they say 6.8a, that represents amperes. So this really represents 6.8 coulombs per one second. Because again, amperes amps is coulombs per second. Now, because this has seconds in it, that means that we have to convert the minutes given to us into seconds as well. So we're gonna start out with the minutes. We have 41 minutes. We're going to say here for every 1 minute, it comes out to 60 seconds. Now that we have seconds we can cancel out the seconds in the amps. So it's 6.8 Coulombs per one second. And where do we see coulombs? We see coulombs with Faraday's constant, which is f. Faraday's constant is 96,485 coulombs per 1 mole of electrons. So we want to cancel out coulombs. We are going to put 96,485 coulombs on the bottom so they can cancel out. And that's equal to 1 mole of electrons.

Now if we take a look at this equation, for every 1 mole of gold there are 3 electrons involved. So we're gonna say for every 1 mole of gold, solid, there are 3 moles of electrons involved. So moles of electrons cancel out. Now that I have moles of electrons I can change that into grams. So for every 1 mole of gold, its mass on the periodic table is 196.97 grams for gold. So what we have at the end is grams of gold, which comes out to 11.38 grams of gold. So that will be our final answer here. Remember, current is just charge over time, and other words for it, amps, amperes, or just a.

Electrolysis and Current Calculations 1

Here it states a solution of Manganese(II) is used to plate out manganese in an electrochemical cell. If a total of 1.13 grams of manganese is plated out and a total time of 1600 seconds, what was the electrical current used? Remember, electrical current is the same thing as amps or amperes. Remember here this is just coulombs per second. We need to find coulombs and divide them by the total number of seconds. We already know the total number of seconds in the question. We have 1600 seconds total. So 1600 seconds over on the bottom. All we have to do now is figure out the amount of coulombs we have. To do that, we're going to start out with the 1.13 grams of manganese given. So we have 1.13 grams of manganese. We're going to say here for every 1 mole of manganese, we're told that its mass is this, 54.94 grams. So those cancel out. Now we are going to say here for every 1 mole of manganese, how many moles of manganese cancel out, now we have 5 moles of electrons.

So moles of manganese cancel out, now we have 5 moles of electrons. And now that we have moles of electrons we can use Faraday’s constant. So that's 1 mole of electrons is 96485C. So moles of electrons cancel out. Now I have coulombs. And when I punch this in this gives me 9,922.47 coulombs. So take that and plug it up here. So when we divide that by 1600 that gives us 6.20 A or amperes. So that will be our final answer for the amount of electrical current.

Electrolysis Calculations 1

Hey guys. Here these sets of questions might seem more suited for a physics class, but electrochemistry is a relationship that chemistry and physics have in common. So let's take a look at the question. Here it says, if a steady current of 15 amperes is provided by a stable voltage of 12 volts for 600 seconds, answer each of the following questions. So before we start, let's break down what these values are saying. Amperes or amps. So 15 amps is really saying we have 15 coulombs per one second. And here, they are telling us about 12 volts. So when we say volts, the units for a volt are joules per coulomb. So 12 volts is really saying we have 12 joules per 1 coulomb. Those are going to come in handy later on when we start answering each one of these parts. So I can be less distracting, I'm going to take myself out of the image, guys.

Here we're first asked to calculate the total charge that passes through the circuit in this time. So just realize that when they're saying total charge, they're really saying total coulombs or c. We're going to say here charge equals time times current. So our time is 600 seconds, which they told us over here. And current is the same thing as amperes or amps. So that's 15 coulombs per one second. Seconds cancel out, so we have coulombs left, which is 9,000 coulombs.

For the next part, they're asking us to calculate the total number of moles of electrons that pass through the circuit in this time. So moles of electrons equals charge divided by Faraday's constant. So here the charge that we just found, 9,000. And we're dividing by Faraday's constant, so we put the 96,485 coulombs here on the bottom, and 1 mole of electrons on top. So coulombs cancel out, we'll have moles of electrons at the end which comes out to 0.093 moles of electrons.

Now they want us to calculate the total amount of energy. Now typically, energy is in joules or kilojoules. And we're going to say here that energy equals charge times volts. So remember our charge we found in the very beginning, 9,000 coulombs. And we said we had 12 volts, remember volts are Joules over coulombs like we said. So we have 12 Joules per 1 coulomb. So that's 108,000 joules. Or if you wanted to change it into kilojoules it'll come out to 108 kilojoules.

And then finally, power. Calculate the power that the battery provides during this process. So power equals energy over time. So the energy we just found is 108,000 joules, and then time in the very beginning was told to us as being 600 seconds. So what do we have here? We have 180 joules per second. And what are joules per second also equal to? Joules per second is also equal to watts. So I know this seems a lot like Physics, but again there is a relationship that Chemistry and Physics have in common when it comes to electrochemistry. So just keep in mind some of the equations that we talked about here because they'll not only help you in understanding electrochemistry better, but they'll also help you if you do take physics and you get to the section called circuits. Because in that section you'll see all about watts and powers and charge. So these are the steps we have to take for each one in order to get the correct answer.

Here’s what students ask on this topic:

What is electrolysis and how does it work?

Electrolysis is a process that involves passing an electrical current through a substance to induce chemical changes. This process is non-spontaneous, meaning it requires external energy to occur. For example, when an electric current is passed through water (H₂O), it decomposes into hydrogen gas (H₂) and oxygen gas (O₂). The current (I) is measured in amperes (A), where 1 A = 1 coulomb/second. The moles of electrons involved in the reaction can be calculated using the equation: $n=\frac{It}{F}$, where F is Faraday's constant. Factors like ohmic potential, overpotential, and concentration polarization affect the overall cell potential, making it more negative and requiring more current to drive the reaction.

What are the units of electrical current in electrolysis?

The units of electrical current in electrolysis are amperes (A). One ampere is defined as one coulomb per second (1 A = 1 C/s). This unit measures the flow of electric charge through a conductor. In the context of electrolysis, the current (I) is crucial for driving the non-spontaneous chemical reactions by providing the necessary energy to overcome the activation energy and other resistances in the system.

How do you calculate the moles of electrons in an electrolysis reaction?

The moles of electrons in an electrolysis reaction can be calculated using the equation: $n=\frac{It}{F}$, where $I$ is the current in amperes, $t$ is the time in seconds, and $F$ is Faraday's constant (approximately 96485 C/mol). This equation helps determine the amount of substance that undergoes oxidation or reduction at the electrodes during the electrolysis process.

What factors affect the overall cell potential in electrolysis?

Several factors affect the overall cell potential in electrolysis, making it more negative and requiring more current to drive the reaction. These factors include:

• Ohmic Potential: The voltage necessary to overcome the resistance (R) when the current (I) is flowing.
• Overpotential: The voltage required to overcome the activation energy for the reaction on the given electrode.
• Concentration Polarization: Occurs when there is a difference in the concentration of reactants on the surface of the electrode compared to the bulk solution.

These factors contribute to the non-spontaneous nature of the process, necessitating additional energy input to achieve the desired chemical changes.

Why is electrolysis considered a non-spontaneous process?

Electrolysis is considered a non-spontaneous process because it requires an external source of energy to drive the chemical reaction. In a non-spontaneous process, the cell potential is negative, meaning the reaction does not occur naturally and needs an external electric current to proceed. For example, the decomposition of water into hydrogen and oxygen gases through electrolysis does not happen on its own; it requires an electric current to break the chemical bonds in water molecules. This external energy input is necessary to overcome the activation energy and other resistances in the system.