Electrolysis Electroplating Practice Problems Electrochemistry

in this tutorial we're going to work on some quantitative problems associated with electroplatin and electrolysis now what mass of nickel metal can be plated on to the cathode from nickel sulfate solution using current of 8 amps for 15 minutes so how can we find the answer well first let's write the half reaction that occurs at the cathode so nickel is going to be reduced to nickel metal so this tells us that one mole of nickel can be plated onto the cathode for every two moles of electrons that pass through the solution now keep in mind one mole of electrons that flows through the circuit is equal to 96 485 coulombs and charge is equal to current multiplied by time so the unit for electric charge is the column and the unit for current is amps and the basic unit for time is seconds so one coulomb represents current of one amp that flows for one second so how can we use this information to calculate the mass of nickel that can be plated onto the cathode well first let's start with the time in minutes and let's convert minutes into seconds so there's 60 seconds in one minute and then let's multiply the time by the current which is 8 amps so just going to be 8 amps over 1. so we can cancel the unit minutes and so right now we have seconds times amps which is equivalent to column so in the next step we could say that one amp times one second is equal to one coulomb based on this conversion factor so now we can cancel the unit seconds and we could cancel the unit amps so now we have the charge in columns now we can convert from columns to moles of electrons 96 485 coulombs is equal to one mole of electrons now we know that two moles of electrons that passes through the wires represent one mole of nickel that's plated onto the cathode so we can write two moles of nickel mean two moles of electrons per one mole of nickel you can write nickel two plus or nickel it won't change your answer so these units will cancel and we can cancel moles of electrons now the last thing we need to do is multiply this fraction by the molar mass of nickel and so using the periodic table the molar mass of nickel is 58.69 grams per one mole of nickel and so we could cancel these units now let's get the answer so it's going to be 15 times 60 times eight divided by ninety six thousand four hundred eighty five divided by two and then times fifty eight point six nine so this will give you two point one nine grams of nickel metal so that's how you can calculate the mass of metal plated onto cathode if you're given the electric current and the time for which the electric current is active so therefore is the right answer in this problem number two how much current is needed to plate five grams of copper metal on the cathode from copper sulfate solution in two hours so how can we find the answer for this problem well let's write the reduction reaction so the copper two plus ion is going to pick up two electrons and be reduced to copper metal and so we can see that one mole of copper is associated with two moles of electrons now since we need to calculate the current let's start with the mass of copper so we have five grams of copper metal and we need to convert it to moles so the molar mass of copper is 63.55 so one mole of copper has mass of 63.55 grams and we know that there's two moles of electrons per one mole of copper so we could cancel the unit grams and moles of copper at this point so now that we have moles of electrons keep in mind one mole of electrons is equal to 96 485 coulombs so we can use that in the next fraction now one coulomb is equal to one amp times one second so we could cancel these units and also the unit coulomb now to get the current in amps we just need to divide by the number of seconds so let's convert two hours into seconds now one hour is equal to 60 minutes and there's 60 seconds in each minute so two hours is going to be two times sixty times sixty and so two hours is equivalent to seventy two hundred seconds so we're going to divide by 1700 seconds so the seconds will cancel and the only unit that we have is the unit for electric current which is in amps and so this will give us the answer so it's 5 divided by 63.55 times 2 times 96 0485 and then take that result divided by 7200. so the answer is 2.1 amps so that's the electric current that we need to apply for two hours if we want to play five grams of copper metal on the cathode so is the right answer in this problem now let's work on the third problem how long will it take to play 10 grams of iron metal from an iron two sulfate solution using an electric current of 5 amps so let's write reaction so the iron two plus ion will pick up two electrons to turn into iron metal at the cathode and believe the cell potential for this is like negative point four four volts so we can see that one mole of fe relates to two moles of electrons so what should we do if you're given the mass it might be best to start with the grams so i'm going to start with 10 grams of iron metal now let's convert it to moles so the molar mass of iron metal is 55.85 grams per mole so now we have moles of fe now let's convert it to moles of electrons so for every one mole of iron metal that is deposited on the cathode two moles of electrons flow through the wires of the circuit now let's convert this to columns based on faraday's constant one mole of electrons is 96 485 coulombs and one clue is one amp times one second now we could cancel moles of electrons and also we could cancel the unit coulombs so our goal is to get the time and look at the answer choices we need it in hours so we need to get rid of the unit amps so what we need to do is divide by the current which is 5 amps in this problem so we no longer have this unit so now let's convert seconds into hours so we know that there's 60 seconds per minute and there's 60 minutes per hour so now we could cancel the unit seconds and we could cancel the unit minutes so now this will give us the time in hours so it's 10 divided by 55.85 times 2 times 96 485 divided by 5 divided by 60 and then divided by 60 again so this is equal to 1.9 hours so that's how long it's going to take to plate 10 grams of iron metal using this current it's like 1.9195 hours it takes an electric current of 5.739 amps applied to solution of mso4 for one hour to plate seven grams of an unknown metal identify the unknown metal so recommend that you pause the video and try the problem so what do we need to do to identify the metal anytime you wish to identify substance the best way to do so is to calculate the molar mass of the substance and see which of the answers that it matches now to calculate the molar mass we need to take the mass in grams and divide it by the number of moles now we already have the mass of the unknown metal so that's half the battle somehow we need to use the current and the time to calculate the moles of substance that we have now what charge does the metal have well we know sulfate has negative two charge so therefore the metal has to have plus two charge so the reduction reaction for the metal is as follows so we know that one mole of electrons mean two moles of electrons rather corresponds to one mole of the metal so let's start with the time in hours so we have time of one hour and let's convert that to seconds so one hour is equal to 60 minutes and one minute is equal to 60 seconds and then let's multiply the time in seconds by the current in amps so this will help us to get the charge in coulombs amp times one second is equal to an electric charge of one column so the unit hours cancel we can cancel minutes we can cancel seconds and we can cancel amps so now we have the charging plumes now what's next what do you think we should do at this point now we need to convert coulombs to moles of electrons so we know that one mole of electrons is 96 485 coulombs and then we know that two moles of electrons correspond to one mole of the metal and so we could do away for unit coulombs and moles of electrons so that's how we can calculate the moles of metal that's going to be plated onto the cathode so it's going to be 60 times 60 times 5.739 divided by 96 485 and then divide that result by 2. so this is equal to 0.107065 moles of the metal so now let's calculate the molar mass so we have 7 grams of the metal divided by 0.107065 moles and so the molar mass is approximately 65.4 grams per mole so now all we need to do is match this molar mass of the substance the molar mass of copper is 63.55 so that can't be it the molar mass of nickel is 58.69 so that doesn't match it the molar mass of iron metal is 55.85 but the molar mass of zinc is 65.38 so that matches it and the molar mass of lead is 207.2 so that's definitely out so we can see that the unknown metal is zinc in this example and zinc does have two plus charge or plus two charge so zinc fits the description number five during the electrolysis of an aqueous solution of one molar sodium hydroxide hydrogen gas is produced at the cathode and oxygen gas is produced at the anode if current of 5 amps is applied to this solution for 45 minutes what volume of oxygen gas will be produced at the anode at stp so let's start with picture and so we need two electrodes and these electrodes have to be inner electrodes so we can use two carbon based graphite electrodes or platinum electrodes and we're going to attach battery to the cell so we're going to fill this beaker with water and we're going to have one molar sodium hydroxide solution in water now this is the positive terminal of the battery and on the left side we have the negative terminal and so electrons are going to flow towards the positive terminal and they're going to emanate away from the negative terminal now knowing this which electrode is the anode and which one is the cathode you need to know that electrons always flow from the anode to the cathode so on the right side we have the anode on the left side is the cathode now oxidation occurs at the anode and reduction occurs at the cathode so we need to know what reactions are occurring at the anode and what reactions are occurring at the cathode and so you need to look up the standard reduction potentials for water under basic conditions now reduction will always have the electrons on the left side oxidation will always have the electrons on the right side of the half reaction so at the anode hydroxide will be oxidized into oxygen gas and water and it's going to release four electrons per oxygen molecule the cell potential for this is negative 0.4 volts now at the cathode we're going to have different reaction water is going to acquire two electrons and it's going to turn into hydrogen gas plus hydroxide so at the cathode hydrogen gas will be produced and at the anode oxygen gas will be produced now our goal is to calculate the volume of oxygen gas that will be produced at stp for gases recall that one mole of gas occupies volume of 22.4 liters so if we could calculate the moles of o2 then we could use that to calculate the volume of o2 and looking at our answers it's all in milliliters so we need to convert liters to milliliters one liter is equal to thousand milliliters and in this half reaction notice the ratio between o2 and the number of electrons so the coefficient for o2 is one and for the number of electrons it's four so we can say that one mole of oxygen gas will be produced when four moles of electrons flow through the wires of the circuit so using what you know go ahead and calculate the volume of oxygen gas in milliliters that will be produced at the anode at stp standard temperature and pressure so let's start with the time in minutes so the current is being applied for 45 minutes and let's convert that to seconds one minute is equal to 60 seconds and then let's multiply that by the current which is five amps and then let's convert that to coulombs one coulomb is one amp times one second so we no longer have the unit minutes or seconds or amps now let's convert columns to moles of electrons so there's 96 485 coulombs per 1 mole of electrons now we have the ratio between the moles of oxygen and the moles of electrons for every 4 moles of electrons that passes through the battery or the wires of the circuit we could say that one mole of oxygen gas will be generated at the anode now oxygen is gas so we can convert that to liters using this conversion factor so one mole of oxygen is equivalent to 22.4 liters of oxygen and finally let's convert liters to milliliters one liter is thousand milliliters and so that's all we need to do in this example so now we have the final unit in milliliters so it's going to be 45 times 60 times 5 divided by 96 485 divided by 4 times 22.4 and then times thousand so the final answer is 783.54 milliliters of oxygen gas so you can round that and say it's about 784 milliliters which means is the right answer for the problem you