A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for 2.00 min, 12.3 mL of water-saturated H21g2 is collected. The temperature of the system is 25.5 °C, and the atmospheric pressure is 768 torr. What is the magnitude of the current in amperes?

Video transcript

Welcome back everyone in this example, we're told that an amateur which measures electric current is created by a student using the electrolysis of water into hydrogen and oxygen gasses. If the student wants to collect 15.5 ml of water saturated hydrogen gas, how long should an electrical current of .670 amperes be run through the device? The temperature of the system is 27°C and the atmospheric pressure is 850 tour. So our first step because we have an electrical system is to write out redox reactions first for the formation of our water into hydrogen as given from the prompt where we want to recognize we need to balance out our atoms first. We have two moles of hydrogen and two moles of hydrogen on both sides here, but one mole of oxygen on the reacting side and recall that to balance oxygen, we use our hydroxide o h minus. And so we're going to add one mole of hydroxide to the product side, where we will now have one mole of oxygen on the product side. But now three moles of hydrogen, meaning to balance out hydrogen, we're going to place a coefficient of two in front of water. And we're also going to make our hydroxide have a coefficient of two, which is now going to give us a total of two oxygen's four waters on the product side and we would have two oxygen's and sorry for hydrogen on the product and reactant side. Now that our atoms are balanced, we look at net charge, we have a net charge of -2 on the product side and a net charge that is neutral on the reactant side. So to cancel out this or to rather gain a negative charge on the reactant side so that the charges are balanced. We're going to add two electrons to our reactant side. And because we added two electrons to our react inside, this tells us that our reaction is going to be a reduction, which we would recall occurs at the cathode of our electrode. And so now we want to go to our second redox reaction, which is for the conversion of liquid water to oxygen gas 02. So again beginning by bouncing out our atoms, we have one mole of oxygen on the reacting side and two on the product side. So we're going to go ahead and place the coefficient of two in front of our water on the reacting side, which will now give us four moles of hydrogen on the reactant side. And to balance hydrogen, we're going to use our hydride cat ion. And so we're going to go ahead and add a total of Four moles of hydrogen to our product side. Now giving us four moles of hydrogen and four moles of hydrogen on both sides of our reaction. So now that our atoms are balanced, we want to balance net charge. We have a net charge of plus four on the product side. And so the net charge of our react inside is zero. So we want to cancel out this plus four charge by adding electrons to our product side. So we're going to add four electrons to our product side. And because we added electrons to our product side this time, we would understand our reaction as an oxidation, which we recall occurs at the an ode of our electrode. And so now that we have these oxidation reactions written out, we want to get an overall reaction by adding these together. But before we can add them together, we need to make sure that the electrons are balanced. So we're going to take this entire first equation and multiply it by two. And so this is going to give us a total of four moles of water. We would have four electrons, we would have two moles of hydrogen gas and we would have four moles of hydroxide. And so canceling out what's similar in both equations. We can get rid of the four electrons we can get rid of when we recognize combining the four moles of hydroxide with the formals of hydride. This is going to form four moles of water together. And so we can cancel out these formals of water with the formals of water here And now, what we're left with is going to make up our overall equation. So we're left with two moles of liquid water, Yields two moles of hydrogen gas Plus one mole of oxygen gas. And so now we're going to go ahead and make note of our electrons that are transferred. So this is going to be based on hydrogen gas since we need to figure out The time our amateur is running to collect the 15.5 ml of water uh sorry of water saturated hydrogen gas that the student is trying to attain. So we're going to recognize that based on our hydrogen gas to form these four moles of hydrogen gas per our overall recreation we have and the transfer of electrons where we recall is represented by N where we have two moles of electrons transferred per mole of hydrogen gas. And again, that's because our formals of hydride need to gain four electrons total to form our hydride or hydrogen gas and we have two moles of hydrogen gas, so two moles of electrons. So hopefully that part is clear. Now let's move on to our next step which is to find are moles of hydrogen gas. And so for this step, we're going to begin by focusing on total pressure, which we recall is P. T. And we would find that by taking our pressure of our hydrogen gas in our system and adding that to our pressure of water. Now we don't know a pressure of hydrogen gas but we do know both total pressure and pressure of water. So we're going to say that our pressure of hydrogen gas is equal to our pressure total minus our pressure of water. And according to our prompt, we have a temperature of 27°C. And so we would recall that at 27°C, our pressure of water According to our chart in our textbooks of C temperature vs pressure and tour that has an equivalent of 26.7 tour. So this comes from our textbooks and we are given a total pressure in the prompt equal to 850 Tour. So plugging in what we know, we would say that our pressure of hydrogen gas is equal to 850 Tour subtracted from 26.7 Tour. And so taking this difference, we would have a total pressure of hydrogen gas equal to 823.3 Tour. But we want this to be an A. T. M. Because we're ultimately in our next step going to be using the ideal gas equation. So we're going to recall the conversion factor to go from To 1 80 M, which we recall has an equivalent of 760 tour. So canceling out our units of tour, we're left with A T. M. And this is going to yield a pressure for hydrogen gas equal to 1.833 A. T. M's. And so now we're going to move forward again and get to our ideal gas law, which we should recall is pressure times volume equal to the most of our gas times the gas constant R times temperature in kelvin. And we are given a temperature according to the prompt of 27 degrees Celsius, which we want to convert to kelvin. So we're going to add to 73 And this is going to give us our Kelvin temperature of 300 Kelvin. So now we have our kelvin temperature. Let's plug in what we know to solve for our moles of hydrogen. So we're going to reorganize this and say that sulfur moles of hydrogen, we would have our pressure times volume divided by r gas constant R times temperature. And so we would say that our pressure of hydrogen is equal to our pressure given from or what we calculated above as 1. ATMs. So 0833 ATM of H two Multiplied by the volume. Given in our prompt of 15.5 ml that the student wants to collect of hydrogen gas. So we're going to recall that we need to cancel out male leaders to get to leaders because our gas constant R uses units of leaders. And so we would recognize the conversion factor that 10 to the third power of our middle leaders is equivalent to one leader. This allows us to get rid of middle leaders were left of leaders in the numerator for volume. And then for our gas constant R. We should recall that it's equivalent to a value of 0. leaders times a T. M's divided by moles, times kelvin And now plugging in our Kelvin temperature. We have 300 Kelvin as we calculated above. So canceling out our units, we can get rid of ATMs with ATMs in the denominator, we can get rid of Kelvin with Kelvin in the denominator, we can get rid of leaders with leaders and what we are left with is just units of moles, which is what we want. And this is going to give us our moles of hydrogen gas equal to a value of 6.8 to 07 times 10 to the negative fourth power moles. And now that we have our moles of hydrogen gas, we want to figure out the time that it takes to collect our volume of hydrogen gas that the student wants to tame the 15.5 mL uh using that and meter machine. So beginning with the molds that we calculated, we have six point and sorry, this is going to be the next step the time the meter takes to collect 15.5 ml of hydrogen. So we're going to recall that. We just calculated 6.8-07 times 10 to the negative fourth power moles of hydrogen. We want our final unit to be time. So let's get rid of moles by plugging in from our redox step, the moles transferred per mole of hydrogen gas. So we're going to plug in that we have a total of two moles of electrons transferred per mole of hydrogen gas. And so this will allow us to cancel out our moles of hydrogen. Now we want to cancel out our moles of electrons by recalling Faraday's constant, which is 96,485 columns, which is equal to one mole of electrons. So this now allows us to cancel our moles of electrons and moving on in our conversion, we want to get rid of columns. So we're going to recall that we have one Kalume equal to one ampere second and now canceling out columns. We want to get rid of one ampere second. And so we're going to recognize based on this whole unit here that we are going to cancel amp amps actually by placing it in the denominator and placing that unit one in the numerator. So we would just say we have one over amp. This allows us to get rid of the amp term as well as the Unit of one. We just multiply by each other and that's just one. So now we just need to get rid of the second unit. So we're going to recall that we have 60 seconds equal to one minute, which leaves us with our final unit of time in minutes after we cancel out seconds. And so now that we have our final unit of time and minutes. This is going to yield our final answer of three 27 minutes as the amount of time it takes the amateur to collect 15.5 mL of the hydrogen gas. And this is our final answer to the correct amount of sig figs, three sig figs. So I hope everything I went through is clear. If you have any questions, leave them down below, and I will see everyone in the next practice video.