An Investigation into Electrolysis - Copper Sulphate Introduction Decomposition caused by electricity is called electrolysis. The electrical energy causes a chemical change. When a salt is dissolved in water, its ions become free to move so the solution can be ?electrolyzed.? The products of the electrolysis depend on the chemical solution, its strength and the type of electrode. The cathode is negatively charged and therefore attracts to it positive ions. E.g. hydrogen. The more reactive substance stays in the solution whereas the less reactive is released and appears as a gas (hydrogen) or a coating of metal. At the anode (positively charged) the negative charges are given up; for example chlorine or oxygen ions become elements. Electrolysis is used to produce gases or purify metals. plan Aim: To find out if the current being transferred through copper sulphate affects the amount of copper transferred.
Prediction: I predict that as the amount of electrical current is passed through the copper sulphate solution more copper will be transferred. I also predict that the rise in electrical current and the rise in copper transferred will be related, and follow a trend. e.g. Text Box: Copper Transferred[image] I believe that as the current doubles so will the amount of copper transferred. E.g. 1amp = 0.1 gram of copper transferred ? 2amps = 0.2 grams of copper transferred. I think this will happen because it is logical that when something is doubled the affected will be doubled as well. Also to back up my theory is Faraday?s Law: Faraday?s First Law of electrolysis states that: ?The mass of any element deposited during electrolysis is directly proportional to the number of coulombs of electricity passed" Faraday?s Second Law of electrolysis states that: ?The mass of an element deposited by one Faraday of electricity is equal to the atomic mass in grams of the element divided by the number of electrons required to discharge one ion of the element." Research: I have researched other possible ways to work out a rough idea of what the results may be: Charge ? = Current (A) x Time (sec.) ? Moles of Electrons or Faradays = Charge ? / 96500 ? Moles of Copper = Moles of electrons or Faradays / ratio=2 ? Mass = moles x ram ? If the Current is 0.2A and the time taken 5 minutes ? Charge = 0.2 x (5×60) ? Faradays = 60/96500 ? Moles of Copper = 0.0006217/2 ? Mass = 0.0003108 × 64 ? Mass = 0.0199 grams ? If the Current is 0.2A and the time taken 10 minutes ? Charge = 0.2 x (10 × 60) ? Faradays = 120/96500 ? Moles Copper = 0.0012435/2 ? Mass = 0.0006217 × 64 ? Mass = 0.0398 grams ? If the current is 0.2A and the time taken 15 minutes ? Charge = 0.2 x (15 × 60) ? Faradays = 180/96500 ? Moles Copper = 0.0018652/2 ? Mass = 0.0009326 × 64 ? Mass = 0.0597 grams ? If the current is 0.2A and the time taken 20 minutes ? Charge = 0.2 x (20 × 60) ? Faradays = 240/96500 ? Moles Copper = 0.002487/2 ? Mass = 0.0012435 × 64 ? Mass = 0.0759 grams ? If the current is 0.2A and the time taken 25 minutes ? Charge = 0.2 x (25 × 60) ? Faradays = 300/96500 ? Moles Copper = 0.0031088/2 ? Mass = 0.0015544 × 64 ? Mass = 0.0995 grams Apparatus: Ammeter, power pack, 2x copper plates, copper sulphate, beaker, electrical wires, scales. Measurements: I am going to measure, by weighing, the weight of both the copper anode before and after each change in current. I will weigh the electrode to the nearest 100th of a gram to give me accurate results on which to base my conclusion. I am measuring the weight of one electrode only as I believe it would take to long, and would be a pointless exercise to weigh both. I would expect the weight of the anode to decrease with increased current. The change in mass of copper indicates reaction rate so that the more copper deposited at the cathode, the faster the reaction rate. I will need to ensure that the anode is washed, dried and re-weighed after each test to ensure a fair test. Method: 1. Set up the apparatus - 2.
Put copper sulphate solution into the beaker.
Weigh the anodes original weight and make note of this.
Clip the copper plates to the sides of the beaker, making sure
they are in the copper sulphate solution.
Connect the power pack and ammeter to the copper plates,
creating a circuit.