1 / 7

A.. Electrolytic Cell

+. -. A.. Electrolytic Cell. Electrolytic cell. An electric current is used to force a chemical reaction to occur. 1. Description. Only one cell is needed. Cell contains a power source. e-. e-. Battery pulls electrons off of one electrode. And puts them on another electrode. e-. e-.

damon
Download Presentation

A.. Electrolytic Cell

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. + - A.. Electrolytic Cell Electrolytic cell An electric current is used to force a chemical reaction to occur 1. Description Only one cell is needed Cell contains a power source e- e- Battery pulls electrons off of one electrode And puts them on another electrode e- e- - e- e- +

  2. If we add melted NaCl (MOLTEN NaCl) to the cell - + - + Na+ and Cl- will be free to move Na+ + e- --> Na 2 Cl- --> Cl2 + 2e- oxidation reduction Na+ Cl- ANODE CATHODE Na+ move to the negative electrode and gains electrons This is reduction The negative electrode is the cathode Cl- move to the positive electrode and loses electrons This is oxidation The positive electrode is the anode

  3. To calculate the minimum voltage needed to run the electrolysis, determine the voltage of the reaction. 2 NaCl  2 Na + Cl2 Na+ + e-  Na 2 Cl-  Cl2 + 2 e- B. Problems with solutions When you try electrolytic cells with solutions, sometimes the water reacts before the ions. At the Anode, H2O can be oxidized 2 H2O  O2 + 4H+ + 4 e- At the cathode, water can be reduced 2 H2O + 2 e-  H2 + 2OH- This happens if a. cathode – metal is very strong Group 1 or heavy group 2 b. anode – if negative ion is polyatomic -2.71 V -1.36 V -4.07 V We need 4.07 V to run this!

  4. C. Faraday’s Laws 1. The same amount of electricity causes the same amount of chemical change Apply X coulombs of charge, and always get the same amount of product 2. One mole of electrons = 96,500 C of charge. D. Calculating with Faraday’s Laws Combine his ideas with the concept of current. Current = charge from e- passing per second I = q t I = current (amps) q = charge (coulombs) t = seconds

  5. To apply Faradays Laws to determine how much reactant or product is involved: • Use time and current to get total coulombs of charge • Convert C to moles of e-, using Faraday’s constant (96,500 C/mole of e-) • Use the balanced ½ reaction to set up a proportion to find the moles of reactant or porduct • Convert to moles • Example – A current of 2.34 A is delivered to an electrolytic cell for 85 minutes. How many grams of Au will be obtained from molten AuCl3? • Get C • I = q/t • 2.34 = q • (85)(60) • q = 1.19 x 104 C 2. Convert to moles 1.19 x 104 C x mole of e- 96,500 C 0.124 moles of e-

  6. 3. Au  Au+3 + 3 e- Ratio 1 3 Moles X 0.124 0.0412 moles 4. g = gmm x moles g = 197g/mole x 0.0412 g = 8.12 g • Example – How long must a 4.55 A current be used to cause 133.5 g of copper to be plated on an electrode from a CuCl2 solution? • Turn g into moles • g = gmm x moles • 2. Use half reaction to get moles of e- • Cu+2 + 2 e-  Cu • 2 1 • x 2.10 • X = 4.20 moles of e- 133.5 g = 63.55 g/ mole (x) X= 2.10 moles

  7. 3. Turn moles of e- into coulombs of charge 4.20 moles x 96,500 C/mole 4.05 x 105 C 4. Use current formula to get time I = q/t 4.55 = 4.05 x 105 t t = 8.91 x 104 sec Or 1.48 x 103 min Or 24.75 hours

More Related