(a) A Cr3+1aq2 solution is electrolyzed, using a current of 7.60 A. What mass of Cr(s) is plated out after 2.00 days?

Video transcript

Hello everyone. So in this video we're dealing with calculating for the mass of our gallium in a solid state. So we have a three electron reduction given to us right here in our equation we see that we're adding three electrons, so is being reduced by three electrons. Each mole of gallium requires three Faraday's. So our electric current, which is denoted as a, is equal to columns per second. Were given this value to be 8.20 Of course just using our units breakdown for this just be 8. columns per second. Alright, so for Faraday's constant then is equal to F equals columns Permal. And we also have our molar mass. So these are just kind of givens the molar mass of gallium is 69.7 - three g per mole. Now calculate for the mass of gallium is just a mental analysis. Alright? We're gonna go ahead and start with our three days. So we need to go from days and two seconds. I don't know the direct conversion. So I'll use units that I do know. So going from days to hours for every one day we have 24 hours For every one hour we have 60 minutes And for every one minute there are 60 seconds now incorporating my electric current. Now we have 8.20 columns for every one second and for Faraday's constant for every constant that we have here is going to be 96485 columns. And we said that we have a three electron production and every mole of our gallon request three Faraday's. So for every three Faraday's there is one mole of gallium And finally using the molar mass of our gallium. That's 69.7-3 g of gallium for every one mole of gallium. So we can see here that the units will cancel out beautifully. Days with days, hours, minutes, seconds, Cool. Ums Faraday's and last three moles. So we put all the numerical values into our calculator. We can see here that the numerical value that we get will be 512 units being grams of our gallium solid. And this right here is going to be my final answer for this problem. Thank you all so much for watching.