1 / 11

Electrolytic Cells & Electrolysis

Electrolytic Cells & Electrolysis. Up until now we have only examined oxidation-reduction situations which produced an electric current & voltage. It is interesting to note that the opposite is also possible – an electric current can be used to produce a chemical change. Electrolysis

hamish-barnett
Télécharger la présentation

Electrolytic Cells & Electrolysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Electrolytic Cells & Electrolysis Up until now we have only examined oxidation-reduction situations which produced an electric current & voltage. It is interesting to note that the opposite is also possible – an electric current can be used to produce a chemical change. Electrolysis • refers to a chemical change which is produced or caused by an electric current • such a reaction is said to occur in an “electrolytic cell” Example: Electrolysis of molten NaCl

  2. Predict the reactions which occur: • Cathode Reaction (reduction): • Na+ pick up electrons & “plate out” or remain “stuck” to the electrode as a layer of Na metal. • Reaction: Na+ + e Na(s) Anode Reaction (oxidation): • to provide a supply of electrons which flow back to the battery, Cl- ions give up electrons & become Cl2 molecules which bubble away • Reaction: 2Cl-Cl2(g) + 2e • The longer the current flows, the greater the amount of Na & Cl2 which is produced.

  3. Such electrolysis reactions were first studied by an English scientist, Michael Faraday, in the 1830’s. The main observation is given by: Faraday’s Law of Electrolysis • The mass of substance produced at an electrode is directly related to the amount of current which passes through the circuit.

  4. Example – Electroplating Apparatus: copper plating The object to be copper plated is placed into a solution of Cu2+ ions. The object is connected to the negative pole of the battery, a Cu strip is connected to the positive pole of the voltage source.

  5. Observations: Spoon: (negative pole) • becomes covered with a layer of copper atoms • They are apparently produced by the Cu2+ ions in the solution picking up electrons according to the reaction: Cu2+ + 2 e  Cu(s) Cu strip: • Dissolves, forming Cu2+ ions in solution & producing electrons which return to the battery The longer that current is passed through the set-up, the greater the amount of copper produced at the negative pole (as predicted by Faraday’s Law)

  6. Quantitative Relationships in Electrolysis An equation like: Cu+2 + 2e Cu(s) tells us that 2 moles of electrons (or 2 moles of electric charge) will produce 1 mole of Cu metal. (63.5 g) To make this relationship more practical, we need to be able to relate the mass of Cu produced to the amount of electric charge flowing, measured in everyday units of amperes.

  7. The following are the needed factors in the relationship: Coulomb (C) • the unit used in science to specify the quantity of electric charge (q) • 1 coulomb is the amount of electric charge which flows or is transferred when a current of 1 ampere flows for 1 second Mathematically: Charge(coulombs) = current(amperes) x time(seconds) q= It Coulombs of electric charge are related to “number of electrons” through another unit called the FARADAY (sometimes called Farad). This is the amount of electric charge carried by (or associated with) one mole of electrons. It works out that 1F = 96 500 C.

  8. By use of these relationships we can now relate the amount of substance produced by electrolysis to the current in amperes and the length of time the current flows (in seconds). Note: to find the number of grams deposited Amps  coulombs = amps x time  faradays (moles of electrons 1F=96 500C)  moles of the substance  grams of substance

  9. Sample Problems: Molten CaCl2 is being decomposed by electrolysis. At the negative pole, Ca(s) is being produced according to the equation: Ca2+ + 2e  Ca(s) If a current of 300 amperes flows for 550 s, find: • # of coulombs of charge that flow • # of faradays of charge • # of moles of electrons that flowed • moles of Ca produced • grams of Ca produced

  10. Example 2: In the electrolysis of molten NaCl, if 0.69 g of Na is formed at one pole, how many grams of chlorine gas will be formed at the other pole?

  11. Example 3: How many minutes would it take to deposit 1.26 g of solid copper from a copper (II) chloride solution using 8 A of current?

More Related