Electrochemistry Review - Cell Potential & Notation, Redox Half Reactions, Nernst Equation

In this video we are going to talk about

electrochemistry

we are going to focus on how the voltaic

cell

works we are also going to balance

equation

s in basic acidic conditions we will identify the oxidizing agent and the reducing agent of which we are also going to talk about how to calculate the

potential

of the

cell

in standard conditions and non standard how to calculate the delta g gibbs free energy from that and also the equilibrium constant k we will also go over some conceptual examples some electrolysis problems and also some stoichiometry how to calculate the current or the mass that is deposited on the cathode , things like that, so let's do a good overview of

electrochemistry

, so let's get started, let's talk about how the cell works, let's say we have a zinc electrode on the left and a copper one. electrode on the right now the standard reduction

potential

s for zinc and copper are as follows: this is the standard reduction potential for zinc which has a negative voltage of 0.76 and here we have the standard reduction potential for copper which has a voltage positive of 0.34 now in a voltaic cell or a galvanic cell energy is produced from the cell the general reaction is spontaneous, which means that the cell potential has to be positive in an electrolytic cell energy can be used or putting energy to drag the reaction forward so the cell potential does not have to be positive, it can be positive or negative.

A good way to illustrate this is to think of a battery, when the battery powers your cell phone, it uses its own energy to do so and therefore generates a spontaneous reaction in that case, when the The battery that powers your cell phone or laptop acts like a voltaic or galvanic cell but when you connect the battery to the outlet to charge it well, the power flows from the outlet to the battery and thus it behaves like an electrolytic cell because it is receiving power from an external source , so the illustration can help you distinguish between a voltaic cell, which is a galvanic cell, and an electrolytic cell, but what you need to learn from this is that remember that we will be taking galvanic cells, the potential of the cell has to be positive , it will be a spontaneous reaction.

Now here we have a voltaic cell because there is no external energy source that is driving this system, so if the electrochemical cell is generating its own energy, it will take cell. Now here we have standard reduction potentials at any time. the electrons are on the left side, it is reduction, we need to change one of these

half

reactions

so that the overall potential of the cell is positive if we change the

half

reaction for copper, if we reverse it, the potential of the cell will change from positive to negative and if we add the two, we will get a negative cell potential, which is what we don't want for a galvanic cell, so we need to reverse the zinc half-reaction.

Notice what happens when we do that, so the potential of the cell will increase. to change from negative to positive then it is positive 0.76 volts and then at this point we can add the two half

reactions

if we add them the electrons will cancel on the left we have zinc plus copper two more and on the right we have zinc two more and metallic copper, so This is the net reaction and the cell potential for this net reaction. All we need to do is add these two numbers, so 0.76 plus 0.34 is about 1.1 volts, so this is the standard cell potential for this particular reaction now according to the standard.

It conditions the concentration of the ions such as the copper two plus and zinc two plus ions that are in the solution. The concentration has to be one molar or one mole per liter. If the concentration is different, the potential of the cell will not be 1. 1 volt, it's going to vary, it could be 1.08, it could be 1.15, it's going to change, but as long as you have a standard cell potential, that means the concentration of the ions is also standard, it's one mole per liter , but now let's talk about how this reaction works. looking at the first half of the reaction, this is an oxidation half reaction, the reason it is oxidation is because the zinc loses two electrons, it gives up two electrons and notice that the oxidation state of the zinc increases, it increases from zero to two and each time the oxidation state decreases. up or if electrons are lost it is a half reaction of oxidation and every time you see the electrons on the right side it is also oxidation whenever the electrons are on the left side it is reduction the copper ions two more are gaining two electrons and therefore therefore it is reduction and if Look at the oxidation state of copper, it goes down from two to zero, so the oxidation number is reduced or decreased, therefore its reduction.

Now let's understand what is happening in this electrochemical cell, so the zinc atoms in the zinc electrode are losing two electrons, so each atom. They are going to give away two electrons and then the electrons will flow from the zinc electrode to the copper electrode, so when the zinc loses two electrons, it will enter the solution as zinc two more and as the electrons flow towards the copper electrode , the copper ions that are in the solution, well, they are attracted to those electrons, opposites attract, so those two more copper ions will be attracted to those electrons because there is an electrostatic interaction that attracts the electrons and the positively charged copper ions together so that We are going to accelerate towards this electrode and, when the two more copper ions collect two electrodes, they will be deposited as copper atoms on this electrode.

Now the zinc electrode is called the anode and the copper electrode is known as The cathode electrons always flow from the anode to the cathode. Furthermore, oxidation always occurs at the anode. Zinc is oxidized to zinc plus two as it gives up electrons, so oxidation is the current in the anode at the cathode. The reduction always occurs at the cathode. You can see that the two more copper ions are gaining electrons. Now notice that the zinc electrode is losing mass. The zinc atoms are leaving the electrode and entering the solution, therefore the anode always loses mass, so it becomes smaller and smaller.

The cathode notices that it is gaining mass, the copper ions in the solution are depositing towards the cathode, so the cathode gains mass. Now let's talk about the saw bridge first, let's leave some space, so let's say if you have three potential compounds. to choose, say, zinc sulfate, zinc hydroxide and zinc carbonate, which of these three compounds will be the most suitable for the aqueous solution containing the zinc electrode, which one would you choose, the best one to use is zinc sulfate solution zinc and the reason why it is The fact is that zinc carbonate is insoluble, it is a solid and the same goes for zinc hydroxide, so zinc sulfate is the best option because it dissolves in water, it is watery, it's soluble, that's why you want to use it, so we'll use zinc. sulfate and copper sulfate for both solutions now the purpose of the cell bridge is to maintain the charge balance as the electrons flow from the anode to the cathode the ions will travel through the salt bridge now it turns out that zinc plus two cations or zinc two more cations are going to move towards the cathode the cations always travel towards the cathode the cations are positively charged ions the anions such as sulphate ions the anions are negatively charged ions and the anions travel towards the anode so they move to the left make sure to remember that for this particular course the cations travel towards the cathode, the anions that are negatively charged travel towards the anode, so those are the basic concepts behind the electrochemical cell, that's how it works, but now let's go back to the reaction that I wrote before the net reaction which was zinc plus copper two more which becomes zinc two more and copper metal.

The zinc and copper are on a solid face and the ions are in the aqueous phase, which means they are dissolved in water, so now how? Can we identify which is the oxidizing agent and which is the reducing agent? To do this, you first need to find a substance that is oxidized and a substance that is reduced. Notice that the oxidation number of zinc increases from zero to two more. Therefore, we could say that the zinc was oxidized and for the copper ion two more, notice that the oxidation state decreased from two to zero. The oxidation state of any pure element in its natural state is zero, so for copper it went from two to zero. was reduced, so a reduction occurred in the copper ion two more, the substance that is oxidized and reduced, it is always a reactant, it is always on the left side, so you cannot choose one of the two products on the right side, it has to be on the left side now the substance that is oxidized is also known as reducing agent and the substance that is reduced is also known as oxidizing agent the reducing agent is the one that causes the other substance to be reduced so which zinc is considered to be the reducing agent because it causes the copper two plus to reduce and the copper two plus is the oxidizing agent because it causes the other steps in the zinc to oxidize and that's why you have to reverse it so you know that metals are usually good reducing agents. and metal ions, sometimes they could be decent oxidizing agents, non-metals are usually oxidizing agents like fluorine gas and oxygen, by the way, those are oxidizing agents, so I just want to make sure you understand that particular concept, but now Let's go back to the potential of the cell.

From this reaction we now calculate that the cell potential is 1.1 volts using this information how can we calculate the delta g of this reaction? the Gibbs free energy the

equation

we need is delta g is equal to negative nfe now if we use the standard cell potential we are going to get the standard value delta g now n is the number of electrons in the equilibrium reaction. Now keep in mind that in the half reaction zinc becomes two more zinc by removing or sharing two electrons and the same is true. for copper, the two plus copper ions picked up two electrons to become a copper metal and therefore this is the final value: the number of electrons needed to balance the two half reactions, so n is 2. f is Faraday's constant, which is basically 96,485 coulombs per mole of electrons and e, which is the potential of the cell, is 1.1 volts, now volts, the unit of volts is basically joules per column, that is what a vote is, a volt is a joule times a clone, so notice that the column units cancel out and then we're going to get joules over moles, so if we multiply those numbers two times ninety-six thousand four eighty and five times one point one delta g will be equal to negative two hundred twelve thousand two hundred sixty-seven joules per mole and So this represents the electrical work that this cell can do, so this cell can do 212 joules of work using one mole of electrons and basically this is the electrical work that the cell can do and that is delta g, the Gibbs free energy, etc. that's what it's used for in this particular problem and now you know how to calculate it, but now how can we find the equilibrium constant k from this information?

But before we do that, let's talk about a few things in this example, we have a positive cell potential. As long as the cell potential is positive, the reaction is spontaneous, if the cell potential was negative, it would not be spontaneous. Note that when the cell potential is positive, delta g is negative, which is also true for a spontaneous reaction due to the fact that delta g equals negative and fe so when e is positive and you multiply it by this negative symbol delta g is going to be negative now what happens to k when the cell potential is positive that means that in equilibrium it will be the favored product because that means the reaction wants to go from left to right it wants to produce products the same thing happens when delta g is negative what happens to k whenever it has a spontaneous reaction in equilibrium does it mean that it will be the favored product and therefore for a reaction? to be a favored product it has to be significantly large, so k will be significantly greater than one, it could be like one times ten to the power of five one times ten to the power of twelve, but it is a very large number and the opposite is true if the cell potential is negative that means delta g will be positive and therefore this process will not be spontaneous so it is not spontaneous in the forward direction which means in the reverse direction it is spontaneous so Therefore, if it is not spontaneous in the forward direction it wants to go to the left, that means it wants to go towards the reactant side, it is spontaneous in the reverse direction, therefore this will be the favored reactant in the equilibrium and what Do you know about k if it is the favored reactant if it is?

The favored reaction k is significantly less than one. Note that the equilibrium constant k is the ratio of the products divided by the reactants, so if you have a high concentration of products at equilibrium, that means that k will be significantly large as the value of the numerator of a fraction, the total value of the fraction increases, sowhen you have many products, k will be big. Likewise, if you have a lot of reactants, if the reaction is favored, k will be very, very small. It could be like .001 or like one times ten to negative eight when you increase the value of the denominator, the value of the entire fraction decreases and that's what's happening here as the reactants increase k is going to be significantly smaller, but now let's go ahead and calculate k for this particular example, the equation we need to calculate k is this equation delta g equals negative rt ln k but the only problem is that we need to isolate k so let's divide both sides by negative rt and that is equal to the natural logarithm of k now the base of the natural logarithm is base e and therefore e raised everything on the left side is equal to k, this is a property of logs, this is how you can convert an equation in logarithmic form to exponential form, therefore k is equal to e raised to the negative delta g divided by rt, so let's go ahead and calculate the value of k using delta g, so in this equation, because r has units in joules, it's 8.3145 joules per mole per kelvin, delta g has to be in joules, if it's in kilojoules, you won't get the right answer, so let's plug in an e raised to the negative which was delta g for this reaction was um, it was negative 212,000 and 267 and then r is 8.3145 and let's assume the temperature is the standard 25 degrees centigrade or 298 kelvin, so now let's calculate k, if you write it correctly you should get a very large number 1.607 multiplied by 10 to the power of 37 and we expect k to be large because because the cell potential was positive and the delta g was negative, we said that the product will be favored and whenever you have a favorite product situation k is going to be very, very large and that's why it is a huge number, so now let's go back to the net reaction we had before, where the zinc reacts with the copper two more to produce the zinc ion two more and the metal copper, now using this net reaction.

How can we write cellular

notation

for this reaction? How would you do it? What we need to do is start with the oxidative part of the reaction so we know that the zinc is being oxidized to zinc 2 plus, so let's go. To start with solid zinc and then we will have a line, this line will separate the solid side from the aqueous zinc phase and then we will have a double line for the salt bridge, as you can see. The first half of the cell

notation

represents the oxidation part of the reaction, the second half is for reduction, so we will start with two more copper which is also in the aqueous phase and we will separate the aqueous phase with the solid phase , so the second side or the second part of the cell notation represents the reduction, so that is one way that you can write this cell notation for a reaction, so here we have a cell under non-standard conditions , take into account the cell potential. or the standard cell potential was 1.1 volts for this particular reaction and this would be the voltage if the concentration of the two plus copper ions and the two plus zinc ions were one mole per liter, but now notice that the concentration is not standard so, using this information, how can we calculate the cell potential for this reaction?

We know it's going to be a little different from 1.1, so the equation we need for the cell potential is equal to the standard cell potential minus 0.0591 over n times log of q this is the nurse's equation, now there's another different way to the nurse's equation, but this is a form of the nurse's equation, it is a simplified form, so using this equation, how can we calculate the potential of the cell? So, the first thing we must find is q and q is equal to the ratio of the products divided by the reactants in the same way k the equilibrium constant k are the products over the reactants the only difference is that q is associated with the initial concentrations while that k is associated with the equilibrium concentrations but the form Calculate that is the same now we cannot include solids or liquids in the expression for k or q, so in a reaction where zinc reacts with copper two more to produce zinc two more and metallic copper, we cannot include solid copper or zinc. solid in a reaction we can only include the zinc 2 plus and copper 2 plus ions because they are dissolved in the aqueous solution, so since q is the ratio of the products divided by the reactants, it will be the concentration of zinc 2 plus divided by the concentration of copper two plus, which is for copper, is ten for copper and for zinc is 0.01, so 0.01 divided by 10 is 0.001, so now let's put everything into the equation so that the potential of standard cell be 1,1 and n let's know what 2 is. and then multiply the logarithm of q which is point zero zero one, so the logarithm of point zero zero one is approximately negative three and negative three multiplied by negative point zero five nine one divided by two, that's approximately the positive point zero zero, I mean. dot zero eight eight six five and then we're going to add a dot one to that so the total cell potential you need to have for this problem is 1.19 volts so now let's understand what's going on here so let's start with the net reaction. the concentration of the reactant was very high, it was about 10 and the concentration of the product was very low, it was point zero one, so let's go back to chemical equilibrium, Shetalia's principle, what happens if you increase the amount of reactants that are present every time you increase the reagents the system will try to undo the change you impose on it, so as you increase the reagents it will try to lower them again and the only way to do that is by moving to the right as a reaction. goes to the right, the number of products increases and the number of reactants decreases again, so as long as you have a large number of reactants, it will drive the reaction forward and as you drive the reaction, it will become more spontaneous and That is why the cell potential increased from 1.1 to 1.19 because we have a large number of reactants and the same goes for the products.

We decrease the concentration of the products and that is why the system will try to raise it again. itself will try to get it back and the only way to do that is by moving to the right, as the reaction moves to the right the products will go up again and the reactants will go down, so increasing the reactants makes the cell's potential to increase and decrease in the products also causes the potential of the cell to increase, so whenever you have a large amount of reactants or a small amount of products, the potential of the cell will increase, the reaction will become more spontaneous, the delta g will become more negative. and k well k is going to be the same now the products as the concentration of the products increases will shift to the left and will be less spontaneous or more non-spontaneous, but if the products decrease, it will shift to the right, which makes the reaction is more spontaneous therefore increasing the cellular potential so make sure you understand these concepts because depending on the type of chemistry class you are taking you may be tested on this information so now let's say if we are given the cell potential, then we know that the standard cell potential will always be 1.1 volts and that is not going to change, but what if the cell potential of this reaction is the non-cell potential? standard, let's say if it is 0.97 volts with that information how?

Can you find the ratio of the products to the reactants? Basically, the ratio of zinc plus two cations to copper plus two cations. How can we do that? Basically, this is equal to q is the ratio of the products divided by the reactants. So we need to find q so starting with this equation e is equal to e zero minus 0.0591 over n log q so we need to isolate q so the first thing we need to do is um we need to subtract the standard cell potential so e minus e zero and then what The next thing we're going to do is multiply both sides by n and at the same time we're going to divide by negative 0.0591, so on the right side the .0591 will cancel out and n will also cancel out, but we.

We're still going to multiply this by what we have on the left side, so we're going to have n times e minus e zero divided by negative .0591, which is equal to log of q, so now log has the base or is in base 10, so to solve q is going to be 10 to the power everything that we see on the left is equal to q so q is equal to 10 to the power of now what I'm going to do is distribute this negative sign in this term, so if I multiply up and down by negative, one won't have the negative at the bottom, instead this e will be negative and this e nothing will be positive, so I'm going to reverse it so it's 10 to the power n times positive e zero minus e divided by positive 0.0591 okay, now that we have that, let's calculate q so q is equal to 10 to the power of n, which is two, the standard cell potential is 1.1 minus the non-standard potential which was the point 0.97 and let's go divide that by 0.0591, so you should get 25,000 and 80.

So this is the ratio of the products to the reactants, so the ratio between the zinc two plus and copper two plus ions is 25,080. now Let's say if you want to find the ratio of the reactants to the products instead of the products over the reactants, then if you want to find the ratio of copper to zinc, all you need to do is invert this number over one or less than one. so it will be one over 25 080. then it will be a very small number 3.99 times 10 to the power of negative 5. So looking at this number, 25 000 means that there are many products relative to the reactants, so this is a favorite product system , so now let's go back to our net reaction and see what's going on here.

If q is very large, that means we have many products and a very small amount of reactants, so if we increase the products, the reaction will increase. shifts to the right or to the left, if the concentration of the products increases it will shift to the left and as it shifts to the left it becomes less spontaneous and that is why the cell potential decreased from 1.1 volts to 0.97 because we have In a product-favored system, we have many products and a high ratio of products to reactants, so as long as you have a large number of products and a small number of reactants, the potential cell will decrease and the opposite is true if you have a large amount of reactants. and a small amount of products, the potential of the cell will increase, so that's all I want to say about it.

One thing I do want to mention is, notice that the cell potential, the non-standard cell potential, depends on the concentration of the reactants and the products, so if you change the concentration of the reactants or the products, the non-standard cell potential will change, not the standard cell potential, this will be relatively constant. Now the same is true for delta g, the non-standard delta g value will change. whenever the concentration of the products or reactants changes, however, the standard delta g value will be constant now k the equilibrium constant k only changes if the temperature changes if you change the concentrations of the products the equilibrium constant k does not change, remains constant So you've seen this equation delta g equals negative rt lnk, but notice that this is the standard delta g value, not the non-standard delta g value, so k will remain constant if the temperature is constant. k only depends on the temperature, so if the temperature does not change k will be the same regardless of whether the concentration of the products or the reactants changes and if the concentration of the products for the reactants changes, the potential of the standard cell will be the same and the standard Gibbs free energy will be the same, so the same k is going to be the same, however, the non-standard cell potential and the non-standard delta g value can change if the concentration of the reactants or products changes, so now let's work.

About some stoichiometry problems related to electrochemistry, here is a typical problem where you might see a current of 1.25 amperes pass through a copper sulfate solution for 39 minutes, calculate the mass of copper that was deposited on the cathode, so let's go over a few things. q is equal to i t q represents the charge and a charge is measured in coulombs i represents the current which is measured in amperes and t is the time in seconds so what you need to know is that a column is equal to one ampere times one second and The second thing What you need to know is Faraday's constant.

Faraday's constant is 96,485 coulombs per mole of electron. Now in the conversion I am not going to write coulombs. Instead, I'll use amperes per seconds instead of coulombs, so let's go. Go ahead and start, the first thing we need to do is convert minutes into seconds so there are 39 minutes and in every minute there are 60 seconds, so the unit minutes cancel out and then we're going to multiply this by the number of amps , so we have 1.25 amperes, so now that we have amperes and seconds we can convert them to moles of electrons, so there are 96 thousand 485 columns or amperes per second per mole of electrons.Now we need to convert moles of electrons into copper and we need to understand. the half reaction for copper two more, in copper sulfate we have the copper ions two more and it will pick up two electrons to become metallic copper and when that happens it can be deposited on the cathode, note the mass of the cathode increases , the mass of the anode decreases, so look at the relationship between the copper and the electrons.

It is a one to two ratio, so for every two moles of electrons that pass through the solution, we can say that one mole of copper will be deposited or coated on the cathode. so now the last thing we need to do is convert moles to grams so we need to refer to the periodic table and the atomic mass of copper is 63.55 which means there are 63.55 grams of copper for every mole of copper atoms, so let's focus on the units now we see that the units seconds cancel and the units amperes cancel as well and the moles of electrons cancel and the moles of copper cancel, so we are left with grams of copper, like this So now all we have to do is do the math, so 39 times 60 times 1.25 divided by 96, 485 times 63.55 divided by 2 is about 0.9, so that's how many grams of copper that were deposited on the cathode, so the mass of the cathode increased by 0.963 grams.

Number two, the mass of the zinc anode decreased by 1.43 grams. and 56 minutes calculate the average current that passed through the solution during this period of time now in the last exampleWe had the time in minutes and the amperes, the current and the amperes and we were able to calculate the mass in grams. Now we have the mass in grams and the time we need to find the current. So let's start with 1.43 grams of the zinc anode and let's convert it to moles, the molar mass of zinc, let me check the periodic table, is about 65.39, so there are 65.39 grams of zinc per mole of electrons, actually per mole of zinc atoms, that's what I'm going to write now that we need. to convert moles of zinc to moles of electrons and for the half reaction of zinc because we are dealing with the anode, we know what it is, we know that oxidation occurs at the anode, so it will be written this way, but the important thing is the relationship cool.

We can see that the mole ratio is one to two, meaning that for every mole of zinc that moves through the solution, two moles of electrons pass through the wire or solution. Well, technically, electrons don't really pass through. through the solution they travel through the wire, so I guess I should have rephrased this problem in a better way: it is the ions that actually pass through the solution, but the electrons travel from the anode through the wire to the cathode, so now let's use a faraday. constant to convert moles of electrons to amperes per second, so one mole of electrons is equal to 96,485 coulombs, but instead of writing coulombs I'm going to write amperes per second.

This is the part where we want to pause and think about where. We're right now, so the moles of zinc cancel out and the moles of electrons cancel out. Now we want to find the amperes, so we need to divide them by seconds, so let's convert 56 minutes to seconds. We know that all we have to do is multiply 56 by 60. So that's 3360 seconds, so if you want to get amps, we need to divide by the time in seconds, so notice that the units of seconds cancel out and we're left with amps and that will be the current that passes through the anode to the cathode through the wire, so 1.43 grams of zinc divided by 65.39 times two times 96,485 divided by 3360. then the current is 1.26 amperes and that is the answer for this problem number three , how long will it take in hours for a current of 745 milliamps to deposit 8.56 grams of chromium on the cathode using a chromium iii chloride solution, so feel free to pause the video and try this problem yourself if you want more examples, particularly multiple choice questions on chemistry, if you are preparing for like a final exam, watch two videos, one of the videos I created general chemistry 2, study guide

review

and there is another one on general chemistry, a

review

from the study guide that I also created so you can check it out and it has a lot of multiple choice problems and you can practice more with these questions if that's what you're looking for so let's go ahead and start with this problem so we're looking for time, Let's start with the dough, so there are 8.56 grams. of chromium and we need to convert it to moles, so the molar mass of chromium is 52, so one mole of chromium atoms has a mass of 52 grams, so the gram units cancel out and now we need to convert moles of chromium in moles of electrons, so write the half reaction for chromium.

Notice that chromium is in the plus three oxidation state because it has three chloride ions attached and each chloride ion has a negative charge, so this is a reduction half reaction, so chromium plus three will need three. The electrons are converted to metallic chromium, so the molar ratio is one to three, so for every mole of chromium that is deposited on the cathode, three moles of electrons move from the anode to the cathode, so now the mole units of chromium cancel out. We can convert moles of electrons to amperes per second, so one mole of electron is equal to 96,485 amperes per second.

Now, since we are looking for time, we want to cancel out not the seconds but the amperes, so what we have is the current in milliamps. 745 milliamps is the same as 0.745 amperes to convert from milliamps to amperes you need to divide by a thousand one ampere is equal to a thousand milliamps so we are going to divide by 0.745 amperes and at this point we have the unit in seconds the moles of electrons also cancel, like this which now we need to convert seconds to hours to know that there are 60 seconds in a minute and there are also 60 minutes in an hour, so now this will give us the answer, so it will be 8.56. divided by 52 times 3 times 96 485 divided by 0.745 divided by 60 and divided by 60 again then the answer you should have is approximately 17.77 hours, that is the time it will take to deposit 8.56 grams of chromium if we use a small current of 0.745 amperes consider these four standard reduction potentials.

Now I want you to think as I write these reactions. I want you to identify what species the well has, what species is the strongest reducing agent, and what is the strongest oxidizing agent in the cell. The potential or standard reduction potential for aluminum is negative 1.66 volts for fe it's about negative 44 point I think somewhere close for copper it's positive 0.34 and for silver it's about eight point zero volts, so now these four actually there are eight species here is the the four metals on the right side and the four metal ions on the left, so of these eight species, which is the strongest reducing agent and which is the strongest oxidizing agent, so if you put the cell potentials in order, the strongest reducing agent is going to be the metal on top, metals are usually reduced into nonmetallic agents and ions or metal cations, which are those on the left are oxidizing agents, so a metal can't really oxidize anything because for something to be an oxidizing agent, it wants to gain electrons and metals want to give up electrons, so by definition, metals that give up electrons are reducing agents and the metal ions that want to acquire electrons are oxidizing agents, so if you understand that the most powerful reducing agent is going to be something on the right side and the most powerful oxidizing agent is going to be one of these metal cations.

Now the most powerful reducing agent is the one that really wants to give up electrons and that will be the aluminum because if you reverse the reaction, be the most spontaneous reaction, the aluminum wants to give up three electrons and the potential of the cell is very high, it is positive 0.166, the higher the cell potential, the more spontaneous the reaction is and that is why aluminum is the strongest reducing agent of all those listed here. What about the strongest oxidizing agent? The strongest oxidizing agent. Oxidizing agents like to acquire electrons, so it will be one of the metal cations, so it will be one of these ions on the left side.

Which one wants to acquire electrons? The one that wants to acquire electrons is the one with the most positive cell potential and that is silver. The Ag Plus ion really wants to acquire electrons and has the highest cell potential when the electrons are on the left side, so Ag Plus is the strongest oxidant. agent so let's try another example problem so magnesium two plus plus two electrons becomes mg and zinc two plus plus two electrons becomes zinc and then pb2 plus plus two electrons becomes pb now for magnesium , I think these cell potentials are around negative 2.37 and for zinc negative 0.76 and for pb things like point negative 13 or 0.14 now, given the standard reduction potentials where all the electrons are on the left side, do you think?

Which of these six species is the strongest reducing agent? To look for the reducing agent, you need to look at the metals. Metals are reducing agents because reducing agents tend to give up electrons, so which of these three metals is your strongest reducing agent? Now, for it to be a reducing agent, we have to reverse the reaction. then it will be this magnesium, it will never be the middle one if you put the cell potentials in order, then it will be the top one or the bottom one, so the magnesium will be the strongest. reducing agent because when you invert it the potential of the cell will be very high it will be positive 2.37 and observe that as you invert it the reactions change from reduction to oxidation and whenever the electrons are on the right side it is oxidation, observe that the magnesium increases from zero to plus two then it is being oxidized the substance that is being oxidized is also the reducing agent and that is why we can say that reducing agents give away electrons and metals are generally the ones that give away electrons so metals are usually reducing agents so , the best reducing agent is the one that will give up electrons or the electrons will be on the right side and will have the most positive cell potential, this is how you identify the strongest reducing agent, the strongest oxidizing agent compared to all.

The reactions listed here will be one of the three metal cations but it is the most positive or the least negative so in this case it will be pb 2 plus so this is the strongest oxidized nature because it is the least negative or more positive, so it is more likely to pick up two electrons than either of the other two metal cations, so now we are going to try another question, but instead of dealing with metals and metal cations, we will deal with non-metals and non-anions. metals, so I want you to identify the strongest oxidizing agent and the strongest reducing agent.

The cellular potential for iodine is approximately 0.54 and for oxygen in acidic conditions it is 1.23 volts and then we will have chlorine and also fluorine. So at this point, as you've probably noticed, the strongest oxidizing agent and the strongest reducing agent will not be in this reaction or in this reaction, if you put the cell potentials in order, it will be in a reaction with the potential of lowest cell and not with the one with the highest cell potential, now the reducing agents like to give away electrons and the oxidizing agents like to acquire electrons, so the oxidizing agents will be the ones on the left side because these non-metals they like to acquire electrons not -Metals are usually oxidizing agents and metals are usually reducing agents.

Now on the right side, non-metal anions will be reducing agents and metal cations are oxidizing agents, so if it has a negative charge, it is usually more likely to be a reducing agent. than an oxidizing agent because if it has a negative charge it doesn't want to acquire more electrons it may want to give them away and therefore it will behave more like a reducing agent so just to review neutral metals metals are reducing agents neutral. Non-metals are oxidizing agents, metal cations, that is, metals with positive charges, are oxidizing agents and non-metal anions, which are non-metals with negative charges, are reducing agents, so what is the most powerful oxidizing agent or the stronger?, we said.

Since non-metals are oxidizing agents, then which of these has the most positive cellular potential is none other than fluorine. Now, to identify the strongest reducing agent, we must look at the non-metal anions and therefore, among the non-metal anions, the one that is the strongest. be basically on the other side, so it has to be this one, it will never be the ones in the middle, it will always be the ones at the opposite end, so iodide is the strongest reducing agent because if we reversed all the reactions , iodide Whether the most positive or the least negative, it will be negative 0.54 and negative 0.54 is less negative than negative 1.36 or negative 2.87, so iodide is the strongest reducing agent and fluorine It is the strongest oxidizing agent.

Now let's say that if they give you twobalanced the next thing we do What we need to look at is the total charge, so here this is negative one and negative six and on the right side we have a negative charge one for the chloride ion and negative six for the hydroxyl ions, so the net charge on both sides is negative seven so the atoms and charges are balanced on both sides so we know this is the correct answer but let's go ahead and try another example let's try this one let's say if we have metallic iron more permanganate and this will become two more manganese. ion plus the fe 3 plus cation so go ahead and balance this reaction first under acidic conditions and then under basic conditions so let's get started, fe will become fe3 plus and we need to balance the charges by adding three electrons to the side with the highest charge , so we are done with the oxidative part of this

redox

reaction, so now let's move on to permanganate, let's equilibrate it under acidic conditions, so we have four oxygen atoms on the left side, so we need to add four water molecules on the right side . so that the number of oxygen atoms is balanced, next let's balance the hydrogen atoms, so that we have eight hydrogen atoms on the right side, so we have to add eight on the left side, so now let's consider the charges, the total charge on the right side is positive two and on the left side we have eight times one, which is eight plus negative one, so it will be plus seven for the total charge on the left side, so seven minus two is five, so what we need to add five electrons to the side. with the highest charge and that's the left side so this is the reduction half reaction since the electrons are on the right side I mean on the left side excuse me so now what we have to do is making sure that the number of electrons is equal to the least common multiple between 5 and 3 is 15.

So to get to 15, we need to multiply the first equation by five and the second equation by three, so for the first equation it will be five fe e produces five fe, three more ions plus 15. electrons and for the second equation it will be 15 electrons plus 24 h more ions and then three permanganate ions and that will become three manganese ions, two more and twelve water molecules , so now what we can do is add these two reactions, so let's go I put that here, so on the left side we will have 5 fe plus 24 h more and three permanganate ions, the number of electrons will cancel out and in the right side there will be five fe, three more ions, three manganese ions and 12 water molecules, so this will be the equilibrium reaction in acidic conditions, so in basic conditions what we need to do is add 24 hydroxide ions to both sides to balance it in basic conditions and so these two will cancel and they will become 24 molecules of water, so it will be 5 molecules of faith plus 24 h2 and everything else will be the same, but on the right side we are going to add 24 ions hydroxyl.

Now there is one more step that we have to take care of and that is the water molecule, so we have to subtract both sides by the smaller of the two numbers, so that there will only be h2o on one side of the reaction, so these two will cancel out, so now what we have What will be left over will be 5 fe3 plus 12 h2o molecules and three permanganate ions and that will become five fe3 plus ions and three manganese ions two plus and 24 hydroxide ions on the right side, so now this reaction is balanced this is how we can balance it in acidic and basic conditions so let's say we have an electrochemical cell so we have an energy source and we only have one beaker instead of two and let's say the electrons they flow from the left side to the right side, so with this information and let's say the solution contains copper sulfate and hydrochloric acid with this information, how can you determine what reactions are occurring at the anode and at the cathode?

You can also use the standard reduction potential and if you don't have it, With you, these are the important reactions that you need to consider so that hydrogen can pick up two electrons and can become hydrogen gas and the cell potential is about zero volts. This is the standard reduction potential for copper and is now approximately 0.34 volts. The standard reduction potential for oxygen in acidic conditions is about 1.23 volts, we also have the reduction potential for chlorine and it will be converted to chloride with a cell potential of 1.36 volts and also sulfate can be reduced to sulfur dioxide in acidic conditions and sulfur dioxide is polar so it can dissolve in water and the cell potential for this part uh will be 0.17 volts so with these reduction potentials what will occur at the anode and which one will happen at the cathode, so go ahead and take a minute break. video and see if you can figure this out so let's get started, first let's identify the anode and the cathode.

We know that oxidation occurs at the anode. The reduction occurs at the cathode, so we know that electrons also flow from the anode to the cathode, so the anode. is on the left side and for this example the cathode is on the right side, the cathode doesn't always have to be on the right side, sometimes it can be on the left side, just keep in mind that electrons flow from the anode to the cathode, so what species? Do we have the solution right now? In the solution we have h plus ions, we have chloride, we have copper two plus and we also have sulfate, so which of these four species is most likely to be reduced?

So if it's going to be reduced, that means. which is gaining electrons hydrogen ions are capable of gaining electrons the same thing happens with copper sulfate it can gain electrons but the chloride does not want to gain electrons so the chloride will not be reduced because if it gains electrons it will be like cl minus two and That It is not possible, so of these three species that we highlight h plus cu 2 plus and sulfate, which of them is most likely to be reduced, then that is when we must look at the cell potential to know which one has the highest or most positive cell. potential will be copper 2 plus wants to gain 2 electrons and the potential of the cell is greater than h more or less2, that is, more or less4 sulfate, therefore, this will be the reaction that will occur at the cathode, this is the average reduction reaction. now all of these are reduction potentials now we know that oxidation occurs at the anode so at the anode we need to reverse one of these reactions so that h2 cannot occur at the angle because we don't have hydrogen gas in the solution so This can't be this, we have to look at the left side now that we have water because it dissolves in water, so the water could be oxidized to oxygen gas, it could go to the left, so we have to look that we do have chloride the chloride could be oxidized to chlorine gas for oxidation, electrons are lost, so if we reverse it it can be an oxidation reaction.

We don't have sulfur dioxide, so we can't use the third reaction, so we won't be able to either. use the fifth reaction, we could use the third, you could say the third, not the fifth, so it will be water or chloride that we can oxidize, so if we reverse those two reactions, which one is better if we reverse h2o with o2? will be negative 1.23 and if we reverse the chloride half reaction, it will actually be negative 1.36 because these values ​​are relatively close to each other. Both can occur, but it will be easier for water to oxidize to oxygen than to chloride.

Because it takes less energy to oxidize water into oxygen gas, why can we say that to oxidize chloride to chlorine we need at least 1.36 volts, but to oxidize water to oxygen we only need 1.23 volts? That is why it is easier to oxidize water into gaseous oxygen. but both will happen so the copper at the ethical cathode will be reduced a or b will be reduced to metallic copper and at the anode the oxygen or rather water will be oxidized into gaseous oxygen plus hydrogen ions and give up four electrons and the potential of the cell. because that is negative 1.23, so these two reactions will be the predominant reactions that occur at the cathode and the anode;

However, other reactions can occur so that the chloride can be oxidized into chlorine gas because the cell potential is not far from that of water. Now the sulfate can also be reduced at the cathode because the cell potential is quite close to that of copper, but it will be easier to reduce the copper 2 plus instead of the sulfate ion, as you can see the cell potential is not very far. but if you just want the best reactions it will be h2o, h2o will be oxidized at the anode and copper 2 plus will be reduced at the cathode.

Those reactions are the most favorable that will occur at those two electrodes. The others can occur. but just on the lower performance they're just less likely to happen but they can still happen so that's it for this video thanks for watching and have a great day.