An Investigation into Electrolysis - Copper Sulphate.

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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.

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.”

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Research:

I have researched other possible ways to work out a rough idea of what the results may be:

Charge (C) = Current (A) x Time (sec.)

  • Moles of Electrons or Faradays = Charge (C) / 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 (5x60)
  • Faradays = 60/96500
  • Moles of Copper = 0.0006217/2
  • Mass = 0.0003108 x 64
  • Mass = 0.0199 grams
  • If the Current is 0.2A and the time taken 10 minutes
  • Charge ...

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