Practice: Given the following standard reduction potentials,
Hg22+(aq) + 2 e– 2 Hg (?) E° = +0.789 V
Hg2Cl2(s) + 2 e– 2 Hg (?) + 2 Cl-(aq) E° = +0.271 V
determine Ksp for Hg2Cl2(s) at 25 °C.
Subjects
| Sections | |||
|---|---|---|---|
| Redox Reaction | 32 mins | 0 completed | Learn Summary |
| Balancing Redox Reaction | 31 mins | 0 completed | Learn |
| The Nernst Equation | 11 mins | 0 completed | Learn |
| Faraday's Constant | 10 mins | 0 completed | Learn |
| Galvanic Cell | 85 mins | 0 completed | Learn |
| Batteries and Electricity | 9 mins | 0 completed | Learn |
| Additional Practice |
|---|
| Cell Notation |
| Standard Hydrogen Electrode |
| Ion Selective Electrodes |
| Cell Potential |
| Electroplating |
| Electrolysis of Water & Mixture of Ions |
| Additional Guides |
|---|
| Nernst Equation |
Concept #1: Galvanic versus Electrolytic Cells
Transcript
Hey guys, in this new video, we're going to take a look at the difference between a galvanic cell versus a voltaic cell. The first thing we're going to talk about before we talk about the differences between these two cells is first we're going to say with redox reactions, we will now deal with a new variable. It's called the reaction’s cell potential. It uses the variable E. We have E without the circle that means it's under non-standard conditions. Then we have E with a circle, which means we are under standard conditions.
Standard conditions usually refer to us having a molarity of 1 or having a pressure of 1, or having a pH that's close to neutrality, so pH equals 7. When we have these conditions, we are talking about cell potential under standard conditions. We're going to say here, the greater the variable, then the more likely reduction will occur. The smaller this variable, then the more likely oxidation will occur. Remember also that if we know about reduction and oxidation, then we know which electrode we're talking about. Remember, reduction predominantly happens at the cathode and oxidation happens at the anode.
Now, let's first talk about the galvanic cell. In a galvanic cell, we have the creation of electricity. So galvanic cells basically, they either produce electricity or they discharge electricity over time. Basically, a galvanic cell can also be called a voltaic cell. Both of them because they produce electricity, they are just batteries. They are a spontaneous cell because they create this electricity on their own. The opposite of a galvanic or voltaic cell is called an electrolytic cell. An electrolytic cell does not produce or discharge electricity, instead it uses electricity to happen. Here it consumes electricity. We’d say here that an electrolytic cell, it is non-spontaneous. It requires an outside energy source in order for it to occur.
What we have here is an example of a basic galvanic cell, galvanic or voltaic cell. It's based off of these two equations here. We need room to work this out guys so let me take myself out of the image.
Now we're going to say generally speaking, the larger your E value, again reduction is more likely to occur so it's the cathode. Right here, this is the cathode. The smaller your cell potential, the more likely oxidation occurs so it's the anode. Based on these two equations, we're saying here that the nickel solid is our cathode. When it comes to a galvanic or voltaic cell, the positive electrode is the cathode. So we have nickel here, nickel solid, and then here cadmium solid represents my anode. Based on these two half-reactions, the nickel has and Ni2+ ions floating around in solution. Then here the cadmium has Cadmium two plus ions floating around in solution.
We're going to say here in a galvanic or voltaic cell, the anode is negative. Now what's happening here is oxidation remember means we're losing electrons. Over time, electron start to leave the anode so they're heading this way towards this electrode here. Here it's producing electricity, so right here we have this something called a voltmeter which we'll talk about how much voltage is being produced later. Here there are two periodic trends you need to keep in mind when it comes to this process. We have ionization energy and we have electron affinity. Remember, cadmium is the anode. It’s undergoing oxidation. Here, ionization energy is the energy to remove an electron. We have the cadmium solid losing electrons to become cadmium 2+. Here go the two electrons that's lost.
This is a spontaneous reaction which means the electrons come off the anode very easily. The amount of energy required is very low. On the opposite end, electron affinity means the energy involved in adding an electron. In this case, we have nickel 2+ ion gaining two electrons to become nickel solid. Here, again this is a spontaneous reaction, so this happens naturally. We're going to say here the propensity or the likelihood of nickel gaining those electrons is very high. The electron affinity here would be high. Nickel wants to gain those electrons.
What's happening here over time what we should realize is this cathode metal here is gaining electrons. Over time, the surface of the cathode electrode is becoming more and more negative. Imagine if the surface is becoming more negative, what do you think those positive ions dissolved in the solution are going to do? They're going to start to connect themselves to this negative surface. Positive ions and negative surface combine together. What's going to start to happen is they're going to in crossed themselves onto this cathode. Over time, the cathode is going to get fatter. It's going to get bigger. We're going to say here we're going to have the cathode plates out. Meaning, it's going to get bigger because those positive ions that are dissolved in solution will connect to the surface of the cathode metal.
At the same time, the anode is losing more and more electrons. Over time, the anode piece of metal gets smaller and smaller. More and more of it is going to dissolve away. The anode overtime is going to shrink that's why we say the anode dissolves away. But this process can't happen without the use of this structure here. This is called your salt bridge. Your salt bridge contains inert ions. It contains usually chloride ions and nitrate ions. They're usually connected to maybe sodium or potassium. Remember, we say that chloride ion and nitrate ion are inert negative ions. Why are they inert? Because they come from strong acids. Remember, if you come from a strong acid, you're a very, very weak conjugate base so weak that you're neutral.
What's happening here is we have these negative ions basically traveling from the cathode solution towards the anode solution. What's happening here is those negative ions are connecting with the ions in the solution and neutralizing those cadmium2+ ions. We have cadmium maybe combining with the chloride ions if they're in there to form cadmium chloride. Or we have the cadmium ions combining with the nitrate ions to become cadmium nitrate. This is important because the salt bridge, it neutralizes the positive ions in the anode chamber. But more importantly, they help to complete the circuit. For those of you who have taken physics, remember we have to have the movement of light charges in opposite directions to complete an electrical circuit.
We have negative electrons traveling this way from the anode to the cathode. But at the same time from the cathode side to the anode side, we have those negative ions. I'm just putting it as NO3 minus but it could also be chloride ions. You need these same charges, negative charges, to flow in opposite directions to complete the circuit. The bridge has two things that it's doing. It’s neutralizing the positive ions in the anode chamber and at the same time helping to complete the circuit so that the galvanic or voltaic cell helps to produce electricity.
Here talking about producing electricity, how could we help to make more electricity happen? Remember, I told you that we want to neutralize these positive ions here in the anode chamber because if they get too high, too large, it's going to disrupt the electrical flow. You want to make sure that the concentration of the anode ions is low. At the same time, we need electrons to flow from the anode to the cathode. The electrons are more likely to flow towards the cathode if there's a buildup of positive ions within the solution. You want to make sure that the cathode’s concentration is high, thereby attracting more electrons to that side. This helps with the like charges to move in opposite directions. This is the major points that come to dealing with the galvanic or voltaic cell.
What's the difference between this and electrolytic cell? Pretty much a lot of things. We're going to say here for example, remember an electrolytic cell is non-spontaneous. An electrolytic cell we're going to say here we still have the movement of electrons from the anode to the cathode, but in an electrolytic cell, the anode is actually positive and the cathode is actually negative. Think about it. Why is it non-spontaneous? Negative electrons are traveling from the anode to the cathode still, but think about it. Why would negative electrons want to travel to a negative electrode? They wouldn't want to go that way that's why we need the use of a battery. Instead of having a voltmeter up here, it would have actually a battery, so you’d place a battery on top. That battery would help to force the electrons to leave the positive anode and head towards the negative cathode. That's one huge difference between a spontaneous cell versus a non-spontaneous cell.
How do we figure out the amount of voltage being produced in this galvanic or voltaic cell? Remember, cell potential which is E cell equals cathode minus anode. Here, we’re going to say that your cathode is negative 0.25 volts minus a minus 0.40 volts. Remember, minus of a minus is really a positive. That's point 0.15 volts. You’re going to place 0.15 volts in here. Here we just found out that a galvanic or voltaic cell is spontaneous and it has a positive cell potential.
What kind of effect does that have? That's where we move down to this list right here We have four variables we’re looking at. We’re looking at the entropy of the universe, which deals with the second law of thermodynamics. We're looking at Gibbs free energy. We're looking at your equilibrium constant and we have your cell potential. When we have them all in these signs, positive and negative and greater than 1 and positive, it is a spontaneous reaction. If you're a spontaneous reaction, what kind of cell are you? You’re again a galvanic or voltaic cell. Then here we're going to say if you're the complete opposite, you’re non-spontaneous and so you are an electrolytic cell.
Then let's say you're at 0, 0, 1, and 0, here you're not spontaneous, you're not non-spontaneous. You are at equilibrium. What do you represent then? You represent a dead battery in this case. Going through all the different things that we've seen, remember a galvanic and voltaic cell is spontaneous, ionization energy is low. For the anode, electron affinity is high. For an electrolytic cell, it’s the exact opposite. For an electrolytic cell, ionization energy would be high because electrons don’t want to leave a positive anode. Electron affinity would be low for an electrolytic cell. Why? Because electrons don't want to go to a negative electrode, such as the cathode when we're dealing with an electrolytic cell.
Basically, if you can remember the steps for a galvanic/voltaic cell, the electrolytic is almost the exact opposite. The only place that they agree is that anodes are always oxidized. They always lose electrons to help reduce the cathode. That's the only thing they really have in common with one another. Remember, this whole process is only possible with the help of the salt bridge which has negative ions flowing from the cathode portion to the anode portion to help neutralize those ions. Also, we do have some leakage and some seeping out of these positive ions that can help to go towards the cathode side as this surface becomes more and more negative over time. Remember that the anode dissolves away as the cathode builds up or plates out. Of course, remember cell potential is cathode minus anode to help you find your cell potential.
Concept #2: An electrolytic cell presents an electrochemical cell that is nonspontaneous.
Example #1: A certain electrochemical cell involves a five electron change and has a value of Keq = 3.0 x 1016 at 298 K. The value of ΔHo for the reaction is -68.3 kJ/mol. Calculate the values of ΔGo, ΔEo, for a standard electrochemical cell constructed based on this reaction and also ΔSo for the reaction.
Practice: Given the following standard reduction potentials,
Hg22+(aq) + 2 e– 2 Hg (?) E° = +0.789 V
Hg2Cl2(s) + 2 e– 2 Hg (?) + 2 Cl-(aq) E° = +0.271 V
determine Ksp for Hg2Cl2(s) at 25 °C.
Example #2: The cell notation for a redox reaction is given as the following at (T = 298 K):
Zn (s) Ι Zn2+ (aq, 0.37 M) ΙΙ Ni2+ (aq, 0.059 M) Ι Ni (s)
a) Write the balanced half-reactions occurring at the anode and the cathode.
b) Write out the complete balanced redox reaction.
c) Determine the Eo cell.
d) Calculate the maximum electrical work that can be produced by this cell.
e) Calculate the reactant quotient, Q, for this cell and the cell potential under non-standard conditions.
+0.53 V
Example #3: Answer each of the following questions based on the following half reactions:
HALF REACTIONS Eo (V)
Cl2 (g) + 2 e – 2 Cl – (aq) + 1.36
l2 (g) + 2 e – 2 l – (aq) + 0.535
Pb2+ (aq) + 2 e – Pb (s) - 0.126
V2+ (aq) + 2 e – V (s) - 1.18
a) Which is the strongest oxidizing agent?
b) Which is the strongest reducing agent?
c) Will I – (aq) reduce Cl2 (g) to Cl – (g)?
Example #4: Electric current, or flow of electrons is measured in Amperes (A). One Ampere is the delivery of one coulomb (C) of charge per second. What mass of Zinc (in g) is oxidized (to Zn2+) by a dry cell battery that supplies 125 mA of current for two hours (recall that Faraday's constant is the charge in coulombs on a mole of electrons)?
Ignore this portion dealing with 2 hours. 1 hour would be 60 x 60, which gives you 3600 seconds not minutes. The actual setup to find the mass of zinc will be shown below.
Transcript
Okay guys, let's take a look at the following question. So, here it says, electric current of flow electrons is measured in amperes or a, we're going to say one ampere is the delivery of 1 Coulomb of charge per second, what mass of zinc in grams is oxidized to zinc 2+ by a dry cell battery that supplies 125 milli amperes of current for two hours, recall that Faraday's constant is the charge in coulombs on a mole of electrons. So, we need to realize here, is first of all, when we say amperes a, a means coulombs per second, like they said here. So, it's 1 Coulomb charge per second. So, when they give up 125 milli amperes, we change that into just amperes. So, one milli is 10 to the negative 3 so this is 0.125 amperes, which really means we have 0.125 coulombs per one second. Now, they're telling us, we're running a current for two hours. So, two hours equals how many seconds? Well, two hours is sixty minutes times 60 minutes, which is 3600 minutes, right? And then here, remember we're going to say, let's actually do it over here. So, two hours, we're going to say here, for one hour you have 16 minutes and we're going to say here, for every one minute we have 60 seconds. So, what happens here, this and this cancel out this and this cancel out, we have seconds now, and guys let me take myself out of the image so we have more room to work with. So, we're going to say, now we have seconds and how can we cancel out those seconds? Well, we can cancel out those seconds with these seconds. Now, how do we cancel out the coulombs? Well, we make reference to Faraday's constant and Faraday's constant has coulombs in it. So, it's 96,485 coulombs per one mole of electrons, then we're going to say here, okay? For every one mole of zinc, how many moles of electrons are involved, how do we figure that out? Well, here we're talking about solid zinc and then all of a sudden it becomes 2+. So, how many electrons do we lose, we lost two electrons. So, just remember, that's just a transition from being zinc neutral to zinc 2+ and then if we look at the periodic table one mole of zinc weighs about 65.39 grams. So, if we look all the units cancel out and we're isolating grams like we wanted. So, multiply everything on the top and divide by the bottom, when you do that you should get 0.3 grams of zinc. So, here there's a lot that goes into this type of calculation but what's essential is what does an amp mean, amperes or current, all mean the same thing, they all mean coulombs per second and then we have to make a connection between amperes with Faraday's constant, this allows us to transition from time and current to grams or even moles, knowing that is what you need to do in order to solve questions like this.
Example #5: Consider the combustion of formaldehyde CH2O and the data below for the following five questions.
I. CH2O (g) + O2 (g) → CO2 (g) + H2O (l)
II. 4H+ + 4 e - + O2 (g) → 2 H2O (l) ΔE° = 1.23V
Which compound is reduced in reaction I?
{C}a. {C}CH2O
{C}b. {C}H2O
{C}c. {C}O2
{C}d. {C}CO2
{C}e. {C}None of these
What is the change in oxidation number of the carbon in reaction I?
{C}a. {C}-4
{C}b. {C}-1
{C}c. {C}0
{C}d. {C}1
{C}e. {C}4
Example #6: The purpose of a galvanic cell is to:
a. Transduce chemical energy to electrical energy.
b. Purify solids.
c. Allow for oxidation without reduction.
d. To consume electricity.
Transcript
Hey guys, let's take a look at the following question. So, here it says, the purpose of a galvanic cell is to, so remember in a galvanic cell, what do we have, we have one anode container and in a galvanic cell or voltaic cell, same thing, we have the anode, which is negative and over here we have the cathode, which is positive and remember, we have our electrodes our anode electrode and our cathode electrode and remember what happens here is that negative electrons are leaving the anode and going to the cathode and remember, this is a natural process because it makes sense that negative electrons would want to go to an electrode that is positive. So, make sense for them to travel to that and remember during this process, we have a voltmeter here, which reads the electricity being generated, that's because a galvanic or voltaic cell is a spontaneous cell that creates electricity, now, so the answer here because it's creating electricity, we're going to say galvanic or voltaic cell basically takes a chemical reaction in this case redox reaction where one element is gaining electrons and another one is losing it, so it takes a redox reaction and converts the chemical process from that redox reaction and harnesses it to create electrical energy. So, galvanic cell or voltaic cell same thing it is a spontaneous cell that uses a redox reaction to produce electricity. So, here we're going to say a is the answer, here that used to purify cells at all, allow for oxidation without reduction note both occur and here consume electricity, no, if you're consuming electricity you're a non spontaneous chemical cell, so that means that you are an electrolytic cell, an electrolytic cell consumes electricity, a galvanic or voltaic cell produces electricity. So, remember the difference.
Example #7: During the process for electrolysis of water a current is passed through water and produces hydrogen gas and oxygen gas. Which of the following statements is true?
a. O2 gas is produced at the anode.
b. H2 gas is produced at the cathode.
c. In the reaction, H2 moles are twice the O2 moles.
d. All of the following are correct.
Transcript
Hey guys take a look at the following question. So, here I say during the process of electrolysis of water a current is passed through water and produces hydrogen gas and oxygen gas, which of the following statements is true. So, what I'm saying here is we have water and we're passing electricity through it, which produces H2 gas because remember, hydrogen exists as H2 gas and O2 gas, here it's not balanced, balancing it really doesn't matter too much. Alright, so now it's balanced. So, we're going to say here that hydrogen is connected to oxygen, which is a nonmetal. So, hydrogen would have an oxidation number of 1+, oxygen is not a peroxide or a superoxide so its oxidation number is 2-. Now, here they're by themselves as products. So, their oxidation numbers here would be 0. Now, let's talk about what's happening, we're going to say that H in H2O goes from being 1+ to 0. So, its oxidation number decreased or is reduced. So, reduction occurred and remember, reduction happens at the cathode, which would mean that H2 is being produced at the cathode. Now, let's look at oxygen, O in H2O, what happens to it, it goes from 2- to 0, it's becoming more positive, its oxidation number is increasing, so it was oxidized and you're oxidized that means that you are at the anode where oxidation occurs. So, what does that mean? that means O2 gas is being produced at the anode and then here, this is true and this is true, in the reaction H2 moles are twice the O2 moles, yes that's true, because look for every two moles of H2, we only produce one mole of O2. So, there's twice as many H2 moles as there are O2 moles. So, C would also be true. So, A, B and C are true, so the answer is all of the following are correct. So, remember, it's not only important to know about oxidation and reduction but it's also now important to tie those two are different types of electrode and naming our cathode and our anode, knowing this was key to answering the question, and knowing how to balance the chemical reaction in producing the chemical reaction was also key to getting this correct.
Example #8: Which statement is false?
a. Reduction occurs at the cathode.
b. A reducing agent will lose electrons.
c. Cations migrate to the cathode in both electrolytic and electrochemical cells.
d. Li (s) is the strongest oxidizing agent; F2 is the strongest reducing agent.
Transcript
Hey guys, let's take a look at the following question. So, here it says, which statement is false. So, reduction occurs at the cathode. So, remember, we say an ox, an ox and red CAT. So, a node equals oxidation and reduction equals cathode. So, here reduction occurs at the cathode, that is true, next a reducing agent will lose electrons, remember, if you're the reducing agent that means you've been oxidized and if you've been oxidized yes you are losing electrons, so this is true, cations migrate to the cathode in both electrolytic and electrochemical cells. So, remember what's happening here is we have the anode and we have the cathode and remember what happens here, naturally is, naturally and unnaturally. So, because in both the galvanic and the electrolytic cell, this is always true, electrons always move from the anode to the cathode whether that's natural like in the galvanic cell or if they're forced to go there in the electrolytic cell and then here, remember at the same time we have electron, we have negative ions from the salt bridge traveling from the cathode side to the anode side, this helps to complete the circuit, if you haven't taken physics yet, that's just called completing the circuit, you want charges to basically move in opposite directions to close the circuit so that electricity can be generated. So, what's happening here is the surface of the cathode is becoming more and more negative because it's gaining these negative electrons and here, remember, the cathode is in a solution just like the anode is in a solution, and what's happening here is the surface of the cathode is becoming more and more negative because the electrons are just going there and inside of the solution we have positive ions and positive ions will be attracted to the electrons in the surface of the cathode. So, what's happening here is the cathodes are going to, the cations are going to connect to the electrons, become neutral and that's why the cathode grows in size, when in this reaction, so this is true, cations do migrate to the cathode surface because the cathode is placing more and more electrons on its surface, which is going to attract the positive metal ions in the solution of the cathode, then lastly here, lithium solid is a strongest oxidizing agent, F2 is the strongest reducing agent. So remember, if you do oxidising agent that means you are being reduced and if you have a reducing agent that means you're being oxidized, but think about it, this doesn't make any sense because remember, who's the most electronegative element on the periodic table, who wants more electrons than any other element on the periodic table, the answer would be fluorine. So, fluorine is not the strongest reducing agent, it's the strongest oxidizing agent, it wants electrons more than any other element therefore it wants to be reduced, it wants to gain electrons. So, if it's going to gain electrons, it's being reduced, which will mean it's the oxidizing agent, is the oxidizing agent here. So, d is false for that reason but if you could reason through A, B and C, D is the last one, so it has to be the false one and just remember fluorine is the most electronegative element on the periodic table. So, it wants to gain electrons not lose them, so it doesn't want to undergo oxidation and wants to undergo reduction, making it the strongest oxidizing agent.
Example #9: Which of the following reactions may occur at the anode?
a. Ga3+ (aq) + 3 e – → Ga (s)
b. Cu2+ (aq) + 2 e – → Cu (s)
c. 2 Cl– (aq) → Cl2 (g) + 3 e –
d. Co (s) + e – → Co+ (aq)
Transcript
Hey guys. Let's take a look at the following question. So, here it says, which of the following reactions may occur at the anode, so the key to this question is realizing this, what kind of reaction occurs at the anode, we're going to stay at the anode whether you are a spontaneous cell like the galvanic or voltaic cell or non spontaneous cell like the electrolytic, this is always true, at the anode there is always oxidation, there is always oxidation at the anode. So, you can approach this question in two different ways, you can realize here that here we're going from plus 3 for gallium and here it's in its natural state so it's 0, you can say here that the oxidation number of gallium goes from plus 3 to 0 therefore its oxidation number is decreasing or being reduced and reduction happens at the cathode there. So, we can do that for each one, another technique that we can use and just realize on what side are the electrons, for these 3 the electrons are reactants. So, remember electrons as reactants equals reduction, R goes with R and if you're reduction you are the cathode and then the opposite of that you can say here electrons as products equals oxidation, which we mean it's the anode. So, these odd one out would be C, which would make sense because here chlorine has a charge of minus one so its oxidation number is minus one, then it goes to its natural state. So, it's 0 now, it goes from negative one to 0. So, its oxidation number increased, which means it underwent oxidation, which means it's the anode. So, two different approaches to get the same exact answer, either look at the change in terms of the oxidation number and we can do that in these examples because it's pretty simple but if you were to get a more complicated type of reaction just look on what side to get electrons are, electrons as reactants equals reduction equals cathode, electrons as products equals oxidation equals anode. So, remember those two techniques you'll get the same exact answer.
Example #10: Define a salt bridge.
A) A pathway, composed of salt water, that ions pass through.
B) A pathway in which no ions flow.
C) A pathway between the cathode and anode in which ions are reduced.
D) A pathway between the cathode and anode in which ions are oxidized.
E) A pathway by which counterions can flow between the half-cells without the solutions in the half-cell totally mixing.
Transcript
Okay guys, let's take a look at the following question. So, here it says, define a salt bridge. So, let's first draw an image of what's going on inside of a galvanic cell or electrolytic cell. So, we're going to say here we have, let's say we're looking at the galvanic cell, the spontaneous one. So, we have two jars, one has one electrode the other one has another electrode, and here our salt bridge is this right here, okay? So, we have our solution and we'll see that this is metal m and this is metal n. So, metal m produces metal ions floating around in solution and metal n also has ions floating around in solution. Now, remember, we're going to say that this is a galvanic cell. So, here this is negative and this cathode here is positive, electrons naturally move from the anode to the cathode, okay? So, they're naturally moving that way. Now, what's going to happen here is the whole point of a salt bridge is to help neutralize these positive metal ions in here, because if their concentration gets too high then the left side will become more positive and the more positive that side gets the less likely electrons are to leave, okay?
Because electrons are going to want to stay where there's a opposite charge to them to counteract these positive ions from building up, that's why salt bridge is created, in this salt bridge we have neutral ions neutral negative ions like Cl minus or NO3 minus, these are neutral negative ions because they come from strong acids, in this case HCl and HNO3, because they come from strong acids they're very weak negative ions. So weak that they're neutral, meaning they're not going to take place, take part in any type of reaction. So, we don't have to fear them. So, these guys are naturally flowing from a Cathode side towards the anode side, and remember, we do this to complete the circuit, because remember, you want your electrons flowing this way and then you have negative charges, negative ions flowing the opposite way, light charges moves opposite directions close the circuit, which helps us generate electricity. So here, again, salt bridge helps to basically neutralize the anode side, get rid of all those positive ions. So, they don't build up. So, here a pathway composed of salt water that ions pass through. So, salt water is NaCl, the thing is we only have chloride ions, we don't have sodium ions, positive ions floating in there. So, this wouldn't work, here a path pathway in which no ions flow, no, we just set that chloride ion the nitrate ion flow, a pathway between the cathode and the anode in which ions are reduced, no reduction happens at the cathode, oxidation happens at the anode. So, both C and D are wrong, so the answer has to be E, a pathway by which counter ions can flow between the half cells without the solutions and the half cells totally mixing. So, yes. So ions, negative ions are formed from the cathode side to the anode side but we make sure that none of the n positive ions somehow find their way onto the m plus ion solution. So, only negative ions are flowing, none of those positive and positive ions are flowing. So, basically salt bridge helps to neutralize positive ions on the anode side from building up too much, too much of a buildup of positive ions on the anode side would stop the reaction from occurring, that side would become too positive, which means that electrons would be less likely to leave the anode and go towards the cathode. So, here option E is your answer.
Example #11: What statement is NOT true about standard electrode potentials?
A) E°cell is positive for spontaneous reactions.
B) Electrons will flow from more negative electrode to more positive electrode.
C) The electrode potential of the standard hydrogen electrode is exactly zero.
D) E°cell is the difference in voltage between the anode and the cathode.
E) The electrode in any half-cell with a greater tendency to undergo reduction is positively charged relative to the standard hydrogen electrode and therefore has a positive E°.
Transcript
Okay guys, let's take a look at the following question. So, here it says, which statement is not true about standard electrode potential. So, E cell. So, here E cell or your standard electrode potential standard because we're under standard conditions. Remember, standard conditions means that you have one molar, means you have 25 degrees Celsius, means you have one atmosphere, these are your standard conditions, that's what that little circle means. So, E cell under standard conditions is positive for spontaneous reactions, that is true, if your cell potential is greater than 0 you're a spontaneous reaction, meaning that you are a galvanic or voltaic cell, electrons will fall from more negative electrode to more positive electrode, this is true, if you are a spontaneous cell, negative electrons naturally want to go somewhere that's more positive. So, in a galvanic or voltaic cell, same thing, negative electrons leave the negative anode and travel to the positive cathode, the electrode potential of the standard hydrogen electrode is exactly 0, the standard hydrogen electrode is referred to as the S.H.E. electrode, it is the reference electrode. So, when you see a half reaction have a positive Y cell, that's in reference to the S.H.E. it's that much more likely to be reduced than the S.H.E. electrode, when you see a half reaction with a negative E cell value, that means that it is is that much more likely to be oxidized compared to the S.H.E. So, yes, the S.H.E. is the reference electrode, which is equal to 0, so it has a 0 voltage, E cell is the difference in voltage between the anode and the cathode, no, that's not true because cell potential is the difference between the cathode and the anode, it's cathode minus anode. So, technically d is false, here the electrode in any half cell with a greater tendency to undergo reduction is positively charged relative to the standard hydrogen electrode and therefore has a positive e value, that is true, if something is going to be reduced more than the S.H.E. then it's going to have a positive value compared to the S.H.E. if something is going to be oxidised compared to the S.H.E. then it will have a negative e value, So, e is true, so the only thing here that's false is d, it's not E cell equals anode minus cathode, it's E sell equals cathode minus anode, knowing that gets us the correct cell potential.
Example #12: Use the standard reduction potentials below to determine which element or ion is the best reducing agent.
Pd2+ (aq) + 2 e– → Pd (s) Eo = + 0.90 V
2 H+ (aq) + 2 e– → H2 (g) Eo = 0.00 V
Mn2+ (aq) + 2 e– → Mn (s) Eo = - 1.18 V
a) Pd (s) b) H+ (aq) c) Mn2+ (aq) d) H2 (g)
Transcript
Hey guys. Let's take a look at the following question. So, here it says, use the standard reduction potentials below to determine which element or ion is the best reducing agent. So remember, we're saying the best reducing agent and realize if you're a reducing agent, that means you've been oxidized, if you've been oxidized that means you're the anode, if you're the anode that means that you have the smallest e value, and remember, oxidized means that you are away from the electrons because you're losing electrons, you're not going to be near them. So, if we look at the options, this is our smallest e value. So, looking at this reaction here, but here's an issue, this is the element that is away from the electrons, this magnesium 2+ is next to the electrons so it wouldn't be our answer. So, magnesium solid it's supposed to be the best reducing agent but I don't give you magnesium solid as a choice, so that means we're going to ignore this reaction and go to the next reaction with the smallest e-value. So, next smallest value will be this one at 0 and who's away from the electrons? the hydrogen gas is away from the electrons. So, H2 gas will be our answer here. So, the option d. So, just remember this, this Web, how reducing agent expands out to other things. Now, what if they wanted the strongest or best oxidizing agent, oxidizing agent would mean you've been reduced, which means that you're the cathode, which means you have the largest e value, which means that you're next to the electrons. So, if you understand one the other one is the exact opposite in terms of trends and just realize that this is the approach you need to take when you're trying to connect the ideas of oxidation and reduction to your different types of electrodes, meaning your anode and your cathode.
Example #13: Consider an electrochemical cell where the following reaction takes place:
Na2O (aq) + Ba (s) → 2 Na (s) + BaO (aq)
What is the cell notation for this cell?
What is the ratio of oxidizing agent to reducing agent?
Transcript
Hey guys. let's take a look at the following question. So, here it says, consider an electrochemical cell where the following reaction takes place, here I say what is the cell notation for this cell. So, cell notation was a way of them coming up with a quick way of describing what's going on in electrochemical cell without have to draw all the parts to it. So, remember that in the cell notation, we have our phase boundary, which separates the solid form the non-charged form usually of the element versus its charge form.
Now, we have a physical boundary and then we have another phase boundary. So, remember here on this side we're going to have the lower oxidation state, which I'm just going to say ox state. So, lower oxidation state of the element and the lower oxidation state is usually a solid or a liquid or even a gas and then on the inside we have the higher oxidation state of the element. So, that's usually the ion, same here the higher oxidation state is in here where it's an ion again and then here we have the lower oxidation state. Alright. So, now we're going to say sodium here is with oxygen. So, it's plus 1, group is plus one Barium here's by itself neutral. So, it's 0, sodium here is 0 and here Barium is with oxygen. So, it's plus two, oxygen is not changing its oxidation number so we don't worry about it. So, we're going to say here that Na plus goes from being what? it goes from being plus 1 to 0, so it was reduced, its oxidation number decreased, because it was reduced it is the cathode and you're going to say here that barium solid Ba goes from 0 to plus 2, its oxidation number increased. So, it's oxidized therefore it is the anode. Now, remember cell notation is easy as A, B, C. So, some notation is as easy as A, B, C, A for anode, B for the physical break between the anode jar and the cathode jar and then C is for cathode, so the anode here, remember, we're going from the lower oxidation state, which is the barium solid, to the higher oxidation state, which is barium 2 plus, and on the other side we have the higher oxidation state. So, Na plus and then here its lower oxidation state Na solid, so that will be our cell notation. Now, what is the ratio oxidizing agent to reducing agent? Well, here we're going to say that we said Na plus goes from being plus 1 to 0, so that deals with one electron, barium goes from 0 to plus 2, so that deals with 2 electrons, the number of electrons don't match, that's why we have to put, that's why there's two sodium's there, so the ratio of oxidizing agents. So, oxidizing agent, means what's been reduced, versus reducing agent, what's been oxidized. So, what's been reduced as a sodium ion and there are two of them because I put it two in front of it and then here the barium is being oxidized, how many of them are there? there's just one. So, this would be a 2 to 1 ratio, 2 to this one barium. So, for every 2 sodium ions that are reduced one barium atom is oxidized, which makes sense because barium loses two electrons and those two electrons go to each one of the sodium ions. So, it'll be a two to one