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How Does Changing The Amount of Electricity Effect How Much Copper is Deposited During Electrolysis?

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How Does Changing The Amount of Electricity Effect How Much Copper is Deposited During Electrolysis? Plan I am going to investigate how changing the current, effects how much copper is deposited during electrolysis. The equipment I will use is as follows: - A DC power pack An ammeter 125cm� of Copper Sulphate (CuSO4) solution A rheostat A stopwatch A thermometer Some wires 2 Copper sheets (for the anode and cathode) A beaker of water A beaker of propanone Piece of card with two slots about 4cm apart Sandpaper Electric Balance Crocodile Clips I will set up the equipment as follows: - I will cut two pieces of copper both 8cm long and 2.5cm wide. I will then sand both pieces of copper with the sand paper, so that they have a rougher edge, then I will dip it into the beaker of water then in the beaker of propanone. Then I will let it dry in the air. Doing this will ensure the anode isn't greasy, and has a rough edge, so that the copper will be able to stick to it. ...read more.


I will then dip each one in the beaker of water followed by the beaker of propanone. I will allow them to dry, and then measure their masses again. I expect that the cathode should get heavier, by the same amount of grams as the anode got lighter. I will record my results in a table like this: - Time (s) Current (A) Initial mass of cathode(g) Final mass of cathode(g) Mass of copper deposited(g) Initial mass of anode(g) Final mass of anode(g) Temperature (?C) When I am doing the experiment, I will make sure both electrodes are submerged in the same amount of liquid as the reaction only takes place on the part of the electrodes in the liquid. Another reason why I will use the cardboard is to keep the electrodes from touching each other. If they meet, a short circuit will be formed, so in order for the electrons to follow, I need to keep the electrodes apart. The reason I am using a copper anode is so that the concentration of the copper ions is kept the same. ...read more.


2 mol e ? 63.5g Cu, so if I had a quarter of the electrons, then, 0.5 mol e ?63.5g/4 Cu. from reading Chemistry by Hunt and Sykes, I know that the faraday constant is 96,500 C/mol, this is the amount of electricity carried by one mol of electrons. So, I know that I need 2*the faraday constant ?1 mol of copper atoms = 2*96500 C ? 1mol Cu =193000C ? 64g Cu If I used 193C, I would get: - 193 ? 64g/1000 Cu. if I did this, I would need to use 1A for 193secs. Also from reading Hunt and Sykes, Chemistry, I know that 1A = 1C/s. so if I use 1A and 480secs the amount of electricity used will be: - 480s*0.1C/s =48 C So the number of mol of electrons used is 48C/ 96500C/mol = 0.0005mol (to 4d.p.) So 2 mol of electrons are needed to give 1 mol of copper, so 0.0005 mol of electrons will give 0.00025 mol of copper. The mass of 0.00025 mol of copper is: - 0.00025 mol*64g/mol = 0.016g So I will expect 0.016g of copper to be deposited at the cathode, and an equal amount to be dissolved at the anode. ?? ?? ?? ?? Nicola Goodwin S11 ...read more.

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