welcome electrochemistry is one of the toughest topics on the ap chem exam you want to ace the exam so today we are going to tackle electrochemistry and get you ready so let's get started um i'm going to share my screen with you and we will do it so today we're going to talk about representations as they relate to electrochemistry like reactions or equations cell diagrams and particle diagrams we're going to talk about types and properties of electrochemical cells like voltaic versus electrolytic and standard versus non-standard we're going to look at all the calculations and quantities that relate to electrochemistry like cell potential free energy equilibrium constants faraday's law and others and we're going to talk about other topics too and kind of mix them in especially thermodynamics which is also part of unit 9 the the last unit of the ap chemistry course and one of the more challenging ones so today we're just going to kind of take an overview we're going to look at free response questions but you may need more and if you do the unit 9 ap daily videos will be really helpful for you they go into a lot of detail on all the topics and we are going to focus mostly on free response questions but before we get to the free response questions which are very dense we're going to warm up by looking at some review so first of all let's just do a quick comparison of voltaic cells and electrolytic cells so these are the two different kinds of electrochemical cells and they have some things in common but they also have a lot of differences and which type of cell you are working with is going to really affect how you solve problems and what kind of questions you will see about it so an electrolytic cell uses a battery or power source a voltaic cell typically has two components or two compartments whereas an electrolytic cell usually has one so that's kind of a visual indicator that you can see oxidation occurs at the anode in both so that's very important to remember oxidation at the anode and reduction at the cathode in both types of cells thermodynamically favorable is voltaic cells also called galvanic cells so they are spontaneous is the old terminology we use those reactions happen without energy input negative cell potential is an electrolytic cell so if e cell is negative you know you're talking about an electrolytic cell if you have a voltaic cell you've got to have a positive e cell and a negative free energy change that of course indicates that you have a thermodynamically favorable process we already said that's a voltaic cell so negative free energy changes for voltaic cells and equilibrium constants less than one well that corresponds to a process that is um not thermodynamically favorable and so at least in the forward direction and so that would be an electrolytic cell okay so that's kind of a quick contrast compare and uh an easy way to just kind of keep all of these quantities straight and how they relate to each other is just to kind of memorize these this relationship that you see there in orange right now which is a positive e cell means a negative delta g or free energy which means that you've got a k that is bigger than one and then the the opposite is also true you got a negative e cell that means a positive delta g and a k value that's less than one so those are helpful relationships to remember all right so let's look at a very common type of diagram that you might see on a free response question a cell kind of image or cell diagram so first of all we look at this what kind of electrochemical cell is it is it voltaic or is it electrolytic well there's a couple of clues that we can see here first of all there's no battery there's no power source here there's nothing that is supplying energy to it which is needed in an electrolytic cell also we see two compartments and that as i said is typically found in a voltaic cell so we can tell right away that this is voltaic which side of the cell is the anode well remember anode means oxidation oxidation produces or liberates electrons so on the left side of this cell we can see that electrons are leaving right so electrons are being produced on the left side and so that indicates oxidation which means the anode so that's an important definition to remember okay so what else can we extract from this particular picture let's take a look at that um what will happen to the concentration of magnesium ions mg2 plus as the cell operates well magnesium is being magnesium metal is being oxidized we decided at in the magnesium half cell on the left if magnesium is oxidized as the cell operates that means we're making magnesium ions they are a product so the concentration of magnesium ions is going to increase as the cell operates describe the motion of ions in the salt bridge which is in the middle there where we see kno3 as the cell operates so an easy way to remember this is cations go to the cathode and anions go to the anode and understanding why is about charge balance inside the cell so nitrate ions are anions and they're going to flow to the anode which is on the left side in this particular diagram it's not always on the left side it could be on the right but here we decided magnesium is being oxidized and so the nitrate ions are going to go in that direction and then that is to compensate for the negative charge that's lost as the electrons leave and go through the wire so negative charges are leaving through the wire so we need negative charges to come back through the salt bridge to equalize those charges and then potassium ions our cations are going to flow to the right and into the cathode so cations go to the cathode and that's because electrons are coming in to the cathode side through the wire and we're adding negative charge we need to essentially cancel that out for charge balance so we've got to add positive ions through the salt bridge in order to equalize the charge and complete the circuit essentially so no salt bridge the cell does not operate if the salt bridge sometimes there's questions about like what happens if you remove the salt bridge if the salt bridge isn't there in a bullet excel it's not happening okay no no cell potential no voltage okay so now this is another type of representation you see all the time in electrochemistry problems and that is a half reaction with a standard cell potential and we have two of them here and we're going to use them to try to write a balanced net ionic equation for the reaction that occurs in a galvanic cell so this is the thermodynamically favorable reaction involving these two half reactions well one of them has to be the oxidation you can't have two reductions happening right now it's written they're both written as reductions but one thing is going to be oxidized one thing is going to be reduced so which one is which well the more positive e-reduction so the value um or e-naught reduction the standard reduction potential is going to be the one that is reduction it stays the other one is the oxidation so the less positive or more negative value is the oxidation and that one you're gonna flip so you reverse the oxidation half reaction and in this case that's the aluminum so we flip it around make the reactants the products and vice versa then we need to multiply the half reactions as needed to make the electrons be equal in the two half reactions because when we write a balanced net ionic equation electrons are not reactants or products they have to so they have to cancel out when we add them together so in this case we need to multiply the silver half reaction by three to make it have three electrons and that way it'll cancel out the aluminum the electrons that go with the aluminum reaction all right so we do that now we add together these manipulated equations the reversed aluminum reaction the multiplied silver reaction and the electrons will drop out and we end up with this as our balanced net ionic equation three silver ions plus aluminum produces three silver plus one aluminum ion so that's the process you go through to take half reactions and make a balanced overall net ionic equation all right and then next question what is the value of the standard cell potential e naught cell so for this we use the relationship e naught equals e naught cathode minus e naught anode and the cathode of course is the reduction and we decided that that was the silver and we subtract the anode notice here the anode standard reduction potential is a negative number so we're subtracting a negative and of course that means we're adding mathematically speaking so we end up with 2.46 volts it is a positive number which we expect in a galvanic cell but it better be positive or we made a mistake somewhere so do not use coefficients when you're calculating um e standard potentials we don't multiply by the 3 for the silver reaction we just use the values as they are and you want to be sure to watch your plus and minus signs as i said sometimes you're subtracting a negative and it's really easy to just make a silly mistake there all right this might be one of the trickiest things in electrochemistry so rank the cell potentials e cell of the following non-standard cells so notice there's no like little degree sign no not right so that means that it's not under standard conditions these are non-standard cells standard conditions is one molar concentration of all ions or all aqueous species 298 kelvin and or 25 degrees celsius and one atmosphere of pressure if you have gases around under non-standard conditions you have to think about q the reaction quotient for the process and i'm going to walk you through how you use q to figure out what a non-standard cell potential is compared to standard cell potential or other non-standard cell potentials so here we're being asked to rank the cell potentials for three non-standard cells and you'll see that they have different concentrations of the two species two aqueous species and so the balanced chemical equation we already wrote so that's here and then the q expression of course is equivalent to the k expression we take the product concentration and divide by the reactant concentration raised to the power of its stoichiometric coefficient which is three for silver ions and we don't include solids so the aluminum metal and the silver metal are not part of our q expression so that's q all right so in a voltaic cell k we said is greater than one because it's thermodynamically favorable in the forward direction so under standard conditions all concentrations are one molar so q is one if you plug ones into q you end up with one so what does this mean well q is less than k in a standard cell and so the reaction moves in the forward direction and comparing cell potentials requires comparing q values for those cells a smaller q means that q is further away from from k and so the smaller q is the larger the e cell will be because the the greater the potential essentially driving the reaction toward equilibrium so first we got to calculate our q values using the expression so for cell a where we have 0.5 molar of both species we end up with a value of 4 for q for cell b we end up with a value of 2 and for cell c we end up with 0.59 now we said the smaller q is the bigger e cell is in a voltaic cell um which is what we're talking about here and so that means that a which has the biggest value of q is going to have the smallest potential b is in the middle and then c has the smallest value of q so it's going to have the biggest potential now this is a lot of different concepts to link together here and so i like i said i think this is probably one of the tougher topics in electrochem so i really recommend um the ap daily 9.9 to take a deep dive into non-standard cells and thinking about this more practice all right so we're almost ready to do some free response on electrochem so let's take a quick look at this image which is taken from the ap chemistry equation and formula sheet um and constant sheet that you'll see on the exam which is so helpful so it really is your friend um many equations related to this unit are on the formula sheet and also all other units um so this equation delta g naught equals minus n f e naught is a one that you use to connect thermodynamics to electrochemistry so it's very helpful here we have faraday's law of electrolysis um and current equals charge divided by time and we also have the nernst equation which is is a tool for helping us to understand on standard cell potentials and calculate those um and this is just one small portion as i said of the formula sheet all right so let's do some practice if you haven't already grabbed it um the handout for this session is found here at this link and qr code so we're going to do these free response questions um they're all based on actual frqs that have been given in previous years but i changed them up a little bit by changing quantities and amounts and adding and taking away things to make sure that they're fresh and new so you haven't seen them before and also just to focus on the things that are probably going to be the hardest on the on the exam all right so let's go um the following questions refer to the electrochemical cell diagram and the standard reduction potentials below all right so we have this diagram and hopefully right away you're noticing okay there's no battery here there's no power source and i've got two compartments so this is a voltaic cell um we have two uh half reactions and the standard reduction potential values given so calculate the standard cell potential for the thermodynamically favorable reaction that occurs in the cell well we know we need to um take the cathode minus the anode so how do we do that well we the cathode is going to be the more positive value here the 0.8 and the anode is going to be the less positive or more negative so because we have to come up with a positive number it's a thermodynamically favorable reaction and when we do that we end up with 1.56 volts so write a balanced net ionic equation for the reaction in part a i actually think it's easier to write the balanced equation first and then calculate the standard cell potential but that's not how the question was asked on the ap exam there's nothing to stop you if you notice part b before you do part a from from flipping them around all right so we have the zinc half reaction and we have the silver half reaction that we're given we need to reverse the zinc because that's our oxidation and then we need to multiply the silver reaction by two because again we need to equalize those electrons so they drop out when we add them together and make our overall balanced net ionic equation for this cell so when we do that here's what we come up with one zinc solid plus two silver ions aqueous produces one zinc aqueous plus two silver solid and i had a few questions um from previous sessions feedback like do i need to write the phase symbols the aqs and the s's on the exam you don't they are helpful though just to help you remember things like oh well this species which is aqueous goes in the q expression and this species which is solid needs to doesn't need to go in the q expression so if you write them down it's it's not going to hurt you it could help you but you won't be you won't lose points for not having your phases if you don't put charges though then all bets are off you're going to lose the point for that so you've got to remember to put charges for ions it's really really important okay um in the diagram below label the anode and the cathode on the dotted lines provided and indicate in the boxes below the half cells the concentration of silver nitrate and the concentration of zinc nitrate that are needed to generate e naught okay so we already sort of figured this out that silver is being reduced and so you just have to remember reduction cathode and um zinc is being oxidized and so therefore that needs to be the anode so this is just a diagram labeling it's a kind of a common question type on the exam and then the standard conditions one molar so you just put a one under there and there you go you earn those points um yeah that's really all there is to say it's just things to remember in the expanded view of the center portion of the potassium nitrate salt bridge shown in the diagram below draw and label a particle view of what occurs in the salt bridge as the cell begins to operate omit solvent molecules and use arrows to show the movement of particles okay so first of all let's write down what our particles are so it's silver nitrate and we need to write down the um there's going to be potassium ions and there's i'm sorry it's not silver nitrates potassium nitrate so there's potassium ions and there are nitrate ions and so we have k plus we have no3 minus you got to put those charges in there all right we know that cations move to the cathode and in this particular diagram the cathode is on the left side and so we need to draw an arrow showing that the potassium moves to the left and conversely the anion the anions the nitrate is going to move to the right so we need an arrow showing to the right and that was pretty much all you needed to do there to earn those two points okay so now we're getting into the tricky stuff this is the non-standard cells and trying to figure out how non-standard conditions affect the cell potential using that q versus k reasoning all right so the following questions involve possible alterations to the cell will the cell potential e cell increase decrease or remain unchanged if a g plus concentration is doubled justify your answer all right first thing i got to tell you is do not use equilibrium or le chatelier reasoning in order to try to explain non-standard electrochemical cells because voltaic cells or electrochemical cells in general are not at equilibrium during operation and you will not earn credit for this type of argument um they're very explicit about that so you need to use a q versus k comparison in this situation i've seen many responses where students kind of went into the equilibrium line and they just don't earn credit so in this particular reaction we know by looking at the balanced equation that's up at the top in blue that the q expression is zinc ions divided by silver ions concentration squared so if we double the silver ion concentration we are doubling the denominator when the denominator gets bigger then what's going to happen well doubling the silver concentration will cause q to decrease when q goes down that means that the cell is essentially further from equilibrium so decreasing q relative to its value in a standard cell moves it further from k and that causes the cell potential to go up so the answer here is it will increase so when q goes down in a voltaic cell you're getting further away from k generally speaking so next non-standard condition or next next alteration i should say to the cell will the cell potential e cell increase decrease or remain unchanged if the initial mass of the silver electrode is doubled justify your answer so this is kind of a trick because the silver electrode is not part of the q expression the silver electrode is a solid it doesn't appear in the q expression so it has no effect on the cell potential if you change the electrode mass and that question has been asked several times so it remains unchanged e cell will not change if you change electrode mass all right part f will the cell potential increase decrease or remain the same as the cell operates justify your answer all right so what happens as the cell operates well the reactant is consumed and the product is increases because it's being made right so the reactant concentration goes down the product concentration goes up so if the product concentration the zinc gets bigger and the silver concentration the reactant goes down the denominator is getting smaller here and the numerator is getting bigger when that happens what happens to the value of the quantity well q is going to get bigger and if q gets bigger it moves closer to k and if it moves closer to k the cell potential goes down so i know that's a lot of mental gymnastics but the more you do of these the more um it makes sense so zinc increases silver decreases therefore q increases and increasing q relative to its original value in the cell moves it closer to k therefore the value of the cell potential goes down and actually as the cell operates the cell potential continually decreases until it reaches zero at which point the cell is in equilibrium and there's no more cell potential and the cell is what we'd say like dead it's all used up so anyway the answer here is decrease all right so what should you take away from this first problem on electrochemical cells and voltaic cell diagram well in voltaic cells more positive reduction potential is the cathode or reduction in the cell don't multiply by coefficients when you calculate e cell electrons and particles both must be balanced in the cell overall cell equation e naught cell is only accurate under standard conditions which is one molar for all solutions cations move to the cathode anions move to the anode that nice like alliteration should help you remember it use q versus k to compare cell potentials and voltaic cell cells potentials decrease as they operate so let's try another this is going to be a different kind of problem no diagram here a cell produces 110 grams of aluminum via electrolysis as represented in the equation below so we've got 2 al2o3 plus 3 carbon 3c produces 4al plus 3co2 determine the number of moles of electrons that must be transferred in the cell to produce the 110 grams of aluminum all right so this is a stoichiometry problem so we need to figure out how many moles of aluminum is 110 grams and then how many moles of electrons what's the mole ratio of electrons to aluminum in order to do that so the answer to the mole ratio question is three to one there's three electrons per atom of aluminum that is reduced per ion that is reduced to an atom so first step i'm going to convert 110 grams to moles using the molar mass of aluminum and then i'm going to use that mole ratio to convert from moles of aluminum to moles of electrons so correct answer here 12.2 moles notice that i have three significant figures in that answer because 110 grams with that decimal point has three sig figs so that's where i'm starting part b the process required 139 minutes using a constant current determine the current used in amperes so this is a dimensional analysis problem meaning using units is really really important and will help guide you through the steps of calculations that you need to do and that's generally true of these electrolysis problems so first of all you need times to be in seconds because amperes the unit of current is coulombs which is a unit of charge per second so you've got to be in seconds in order to do this and so that's the first step next thing is you're going to use faraday's constant 96 485 coulombs per mole of electrons so that's how much charge required for one mole of electrons to change into coulombs or unit of charge so we convert into coulombs you don't need to remember that number 96485 because it's on the equation sheet um and then we need amperes as i said is coulombs divided by seconds or coulombs per second so we do that math and we end up with a current of 141 amperes amp can be used as an abbreviation or capital a is also used as an abbreviation and as i said faraday's constant is this fancy cursive f and that is found the numbers found on the equation sheet okay um so part c a student hypothesizes that oxygen is oxidized in the reaction do you agree or disagree justify your answer in terms of oxidation numbers so we need to find the oxidation numbers for oxygen in order to answer this question and justify in the way that we're being directed to so the first thing is to find the oxidation number for oxygen that's something you need to remember if you don't remember how to do that you want to review but a basically oxygen is minus two in almost all compounds um and so in this case on both sides of the equation the oxygen is found in a compound and so therefore its oxidation number is going to be -2 both in the reactants and the products so the oxidation number of oxygen does not change during this process and therefore oxygen is neither oxidized or reduced when something is oxidized its oxidation number gets more positive and or less negative and the opposite is true for reduction um so this would be an explanation that aligns with the prompt and you want to be sure that you state your choice i always leave a little space above my explanation if i'm writing that first to just state my choice here it said disagree or agree so at the end go back and write your choice at the top just to be super clear about exactly what it is that you're saying all right calculate the volume of co2 measured at standard temperature and pressure that is produced in the process so here comes a gas law question and it's very common in a free response question about some that is mostly about some other topic that there'll be a gas law thing thrown in there so you want to be ready for that for this we're going to use the ideal gas law to help us answer this question about volume first thing we need to do is convert from moles of aluminum which we calculated in part a into moles of carbon dioxide which is the species we're being asked about using the mole ratio of three to four from the balanced chemical equation that gives us 3.058 moles of carbon dioxide then we're going to use the ideal gas law and rearrange to solve for um volume which is the requested quantity and when we do that we get an answer 68.5 liters notice again that this is three sig figs which is what we were given in this problem and it's kind of the starting the benchmark for all of the calculations that we do um and both pv equals nrt the equation and the values for stp that 273 kelvin and one atmosphere are on the equation seat sheet so if you forget those you can look there um and it'll help you out all right part e for the electrolytic cell to operate the al2o3 must be in a liquid state rather than in solid state explain why this is true so in order for something to conduct electricity charged particles have to be free to move ionic solids don't conduct electricity because the particles although they have charges the anions and cations they're locked in place in a rigid crystal lattice in the liquid state of course those particles have been freed from the lattice the lattice is broken up and they can move so they're mobile and that is really the uh the key to conduction of electricity is mobile particles mobile charge particles so the key idea that earned the point here is that solids no no mobile ions liquids have mobile ions and the exact wording that was used wasn't terribly important okay is the entropy change for this process positive or negative justify your answer so here we're being asked to apply the thermodynamic concept of entropy to the balanced equation that's given and the first thing we want to think about is moles of gas because um more gases are far more dispersed than liquids or solids um and that means that they have much greater entropy so any process that produces more moles of gas is going to have a positive entropy change and we do see that here there's zero moles of gas in the reactants and three moles of gas in the products the three moles of carbon dioxide gases are more dispersed and so they have greater entropy so you notice how i'm constructing my argument so i i refer to this specific scenario here then i explain using the concept that's requested which is entropy and then i connect the two things um and therefore the products have more entropy so the change in entropy from reactants to products is positive another piece another way that you could construct this argument is to refer to delta n gas where n gas is the moles of gas so delta n gas in this case is three because three in the products minus zero in the reactants is three all right and don't forget to put your choice the entropy the entropy change is positive for this process um okay now we are ready to think about the next part of the equation when al solid is placed in a concentrated solution of potassium hydroxide at 25 degrees celsius the reaction represented here occurs so we have 2al plus 2oh minus plus 6h2o produces 2 aloh4 minus ions and 3h2 okay so using the table of standard reduction potentials shown above calculate the following e naught in volts so standard cell potential for the formation of aloh4 minus aqueous and h2 at 25 degrees celsius so you'll notice here that the two things that are being formed the aloh4 minus ion and the h2 if we go up to the half reactions in the table ball they're found in the top reaction but as it's written in the standard reduction table those two things are are shown as um reactants or the al oh4 minus is a reactant and it needs to be a product so we're going to flip this around and another way to think about this is that we're told at the beginning of this prompt that aluminum solid is a reactant so we need the aluminum solid to be on the reactant side of the half reaction in order to interpret this correctly so we reverse this top reaction because al is a reactant in this scenario and when we do that then it becomes clear that the top reaction is the anode that's the oxidation and the bottom reaction is the cathode and then we can substitute into our e naught um equation and find the standard cell potential so that gives us minus 0.8383 minus a negative 2.35 volts and so we end up with 1.52 volts as our standard cell potential for this reaction and notice that it's a positive number which we expect because this reaction is happening um it's thermodynamically favorable it's happening kind of on its own of its own accord when these two things are mixed according to the description all right delta g naught in kilojoules per mole of reaction for the formation of aloh4 minus and h2 so we need to now use that equation that relates the free energy delta g naught to the standard cell potential e naught so first though we need to figure out how many electrons we are going to be using in this reaction so we look here and we can see that two and three are the numbers of electrons in the half reactions and we need um the least common multiple of those two numbers in order to have the number of electrons that are being transferred in the overall reaction so in order for them to be equalized essentially so n is 6 because the least common multiple of 2 and 3 is 6. and so n is 6 in this equation faraday's constant f is a constant it's on the equation sheet and e naught we just calculated in part g i so we're going to plug in all of those things to our equation and we're going to remember that there's this negative sign here and when we do that we end up with a value of negative eight point eight zero times ten to the fifth joules per mole of reaction at or negative eight point eight zero times times ten squared kilojoules per mole of reaction you could also write it without scientific notation as long as you remembered to put the decimal point um after because there are three sig figs in this number again um that has pretty much been determined by the given values um this 1.52 volts here that we calculated already in the previous part has three sig figs all right um let's move on aluminum metal can be recycled from scrap metal by melting the metal to evaporate impurities calculate the amount of heat needed to purify one mole of aluminum originally at zero degrees celsius by melting it the melting point of aluminum is 660 degrees celsius the heat capacity of aluminum is 0.90 joules per gram per degree celsius and the heat of fusion of aluminum is 0.397 kilo kilojoules per gram so watch your units this problem uses both moles and grams and joules and kilojoules in the given information so you've got to be sure that you're converting appropriately so that you can use the information that's given to calculate the quantities all right so this occurs in two steps in terms of what happens and how we calculate it heating the aluminum up to reach its melting point is one calculation that uses one formula and then melting it at its melting point how much energy does that take that's a separate calculation a separate step and a different equation that we're going to use so for heating something within one phase we use q equals m c delta t um which is something you've probably seen in calorimetry problems and other places well one mole we're going to convert this to grams using the molar mass of aluminum because the specific heat is given in units of joules per gram degrees celsius and then 660 degrees celsius is our delta t because we were heating from zero up to 660 the melting point so then we need to multiply all of these numbers and we end up with 16 kilojoules or 16 000 joules notice that these numbers have two sig figs because 0.90 the specific heat we were given has only two sig figs all right now we need to do the calculation for the second step which is once we get the aluminum up to its melting point temperature of 660 degrees celsius now we actually need to melt it and to do that we use this relationship which is the heat of fusion times the mass that of aluminum that was melted so we have the heat of fusion 0.397 kilojoules per gram and we have the number of grams which is one mole or 26.98 grams and that gives us 10.7 kilojoules so now to answer to give our final answer to this question we need to add those two numbers together because that's the total amount of energy that is needed to heat the aluminum up to the melting point temperature and then melt it when we do that we end up with 27 kilojoules again keeping our sig figs correct here we can't keep the tenths place because we only have the ones place in our first number 16 kilojoules so we round and we get 27 kilojoules for the answer here um and just remember that heating and melting are two different steps so always if you're talking about a temperature change within a phase or a phase change those are going to require two different calculations to figure out the amount of energy involved okay the equation for the overall process of extracting aluminum from al2o3 is shown below which requires less energy recycling existing aluminum or extracting al from al2o3 justify your answer with the calculation okay so this is is a fairly straightforward problem but there is one little trick to it so this delta h equals kilojoules per mole of reaction so that means it's 1675 kilojoules for this reaction as it is written not per mole of aluminum because this reaction involves two aluminum we see that coefficient of two so that's really important to remember because we need to use the mole ratio one mole one reaction worth is two moles of aluminum and when we do that we end up with 837.5 kilojoules per mole of aluminum and so clearly 27 kilojoules is much much less than 837.5 kilojoules so recycling requires less energy all right so we want to be sure that we put our claim at the top after you do the work and you're sure what your answer is then you you can write your claim you can also write it first if you know but just make sure that you you state it really clearly at the beginning so the person scoring your exam sees what your answer is and there's no ambiguity all right so what should you take away from this long long problem well first of all a metal ions charge determines the moles of electron per mole of metal uh use dimensional analysis for faraday's law problems so units are really helpful use oxidation numbers and how they change or don't change to determine the species oxidized and reduced the ideal gas law stp and r the ideal gas constant are all found on the equation sheet mobile charges are needed for conductivity when delta n gas the number of moles of gas uh the change is positive then delta s the entropy change is positive and vice versa watch your units and signs it's always important in chemistry but especially in electrochemistry and thermodynamics so so important um the standard cell potential can be calculated from the standard reduction potentials of the cathode and anode using this equation balance the electrons in half reactions to find n um when you're using the delta g naught equals negative n times f times delta e naught equation heat transfer within one phase is q equals m c delta t energy required for heating and phase changes requires separate calculations and mole reaction and mole of a substance are not equivalent all right so here is a summary of a list of all of the ap daily live review sessions today we did the seventh one electrochemistry in ap chemistry i hope you'll watch them all i really think that they are useful in preparing for the exam here's two qr codes and links that you'll find helpful the one on the left is the feedback form and i would love for you to give me some feedback on this session tell me what was good what wasn't too fast too slow too hard too easy i encourage you to keep practicing preparation is really the key to being successful on this exam um yeah and here is the feedback form and the handouts and i really hope to see you next time so thank you so much you