Learn about Senior Chemistry, Redox 5, in this comprehensive video by bannanaiscool.
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So we’ve got a current of electrons that goes from the anode to the cathode, well if we know the current in terms of its amperage or how many coulombs of charges going over from anode to cathode per second and we actually can measure the time, we can do some fancy calculating in terms of how many moles of electrons are going. And moles of electrons can tell us how much is deposited at a cathode or broken down in anode. Here’s the formula you need in order to do that. The number of moles of electrons equals the current and that’s funny because I stands of current, but its measured in amperes that’s current but we don’t write amperes in our calculation we write coulombs per seconds—c/s. The time is always measured in seconds so that cancels out each of the coulombs and F is the Faraday’s constant in honor of Michael Faraday. He’s one of the great scientist of all time back in the 19th century. F- Faraday’s constant is 9.65 x104 coulombs of charge every time you have one mole of electrons. And with that now we can do some really cool calculations. To do this question, you’ve got to get first that Cu2So4 which is an ionic compound and break it down to copper ions, sulfate ions and water in solution. Now when you look on your chart you'll find that Cu2 is the strongest oxidizing agents. One of the equations here, water is actually the reducing agent. But since the question is saying how much copper can we plate, I only care about the copper half reaction. So find the metal you're looking for and this calculations and write the half reaction on the data booklet for that metal. Here’s the half reaction for copper ions turning in to solid copper. Now with that as the basis, take that number of moles of electrons equals IT/F and plug in your current and you'll notice that I wrote amps in here, but really that won’t cancel with coulombs here unless you write coulombs per second. It’s the best way to do it. Now you see when you do this math and look what I did, I took the 20 minutes and turned it into times 60 seconds over minutes I didn’t have to actually just worry about just putting in the seconds. I do all my conversions in the calculations itself. So coulombs is going to cancel coulombs here and you're going to get seconds cancelling seconds over here and minutes here and you're left with one over one over moles. But that’s moles of what? Number of moles of electrons. It’s the moles of electrons here. But how do you get the moles of copper? It’s a two to one ration. So look, here are the moles of electrons from there and now I cancel out the moles of electron and leave myself with moles of copper. But we don’t want moles of copper, so we put that in the next part of ratio and we want grams of copper that’s the molar mass of copper and then when you do this math you'll get 0.79 grams of copper that can form in this reaction.
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