1 / 13

Electrochemistry Part V: The Electrolytic Cell

Electrochemistry Part V: The Electrolytic Cell. Jespersen Chap. 20 Sec 7 & 8. Dr. C. Yau Fall 2013. Electrolytic Cell (Sec. 20.7). We had been considering 2 half-reactions, one of which is the "drive" for the forward reaction.

reese-lynch
Download Presentation

Electrochemistry Part V: The 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. Electrochemistry Part V: The Electrolytic Cell Jespersen Chap. 20 Sec 7 & 8 Dr. C. Yau Fall 2013

  2. Electrolytic Cell (Sec. 20.7) We had been considering 2 half-reactions, one of which is the "drive" for the forward reaction. In the electrolytic cell (or electrolysis cell), the driving force is the electric current. "Lysis" means to break apart. "Electrolysis" means to break with an electric current. We know 2Na + Cl22NaCl is spontaneous. We can do the reverse reaction by applying an electric current. 2NaCl  2Na + Cl2 What are the physical states? Usually the electrolytic cell has only one chamber.

  3. Easiest to be reduced Easiest to be oxidized

  4. That’s what nonmetals do in redox rxns: X X- (reduction) Easiest to be reduced Periodic Table Easiest to be oxidized That’s what metals do (oxidation): M M+

  5. Electrolysis in Aqueous Solutions Can we "break apart" potassium sulfate? It is not easy to melt an ionic compound. What if we did this in an aqueous solution? Water can be oxidized as well as reduced. Why? How? H2O  H2 reduction at the … H2O  O2 oxidation at the… Overall: 2 H2O (l)  2H2(g) + O2(g) This cannot take place without an electrolyte present (acting as a salt bridge).

  6. Example 20.12 p. 957 Electrolysis is planned for an aq soln that contains a mixture of 0.50 M ZnSO4 and 0.50 M NiSO4. On the basis of standard reduction potentials, what products are expected to be observed at the electrodes? What is the expected net cell reaction? First determine which species are present. Next determine which might be reduced, which might be oxidized. (Remember those to be reduced are on the left side of Table 19.1, those to be oxidized are on the right side. Which is reduced the easiest on the table? Which is oxidized the easiest?)

  7. Example 20.12 p. 957 (continued) Ans. Ni2+ Ni2+ + 2e-  Ni Eo = - 0.25 V Zn2+ + 2e-  Zn Eo = - 0.76 V 2H2O + 2e-  H2 + 2OH-Eo = - 0.83 V Which is most easily reduced? Now examine what is most easily oxidized. (Examine what is on the right side of the table.) S2O82- + 2e-  2SO42- Eo = +2.01 V O2 + 4H+ + 4e-  2H2O Eo =+1.23 V What is the net cell rxn? What is the Eocell? Ans. H2O

  8. Example 20.12 p. 957 (continued) Ni2+ + 2e-  Ni Eo = - 0.25 V O2+ 4H+ + 4e-  2H2O Eo=+1.23 V 2H2O O2 + 4H+ + 4e-Eo =-1.23 V What is the net cell rxn? 2Ni2+ + 4e-  2Ni Eo = - 0.25 V 2H2O O2 + 4H+ + 4e-Eo = -1.23 V 2Ni2+ + 2H2O  2Ni + O2 + 4H+Eo = -1.48 V Do PractExer 23, 24 p. 958 How can it be negative?

  9. Electrolysis Stoichiometry Type of question to answer: • How many g of metal are deposited in an electrolytic cell given the current and time? • How long does it take to deposit a certain g of metal given the current?

  10. Basics you need to know... • Charge = electric current x time Units: (coulomb) (ampere) (seconds) C A s • 1 C = 1 A.s • Faraday’s constant: 1 F = 96500 C mol-1 • This means 96500 C = 1 mol electrons • 3 sig. fig. • No need to memorize constant but must know it’s there.

  11. Example 20.13 How many grams of Cu are deposited on the cathode of an electrolytic cell if an electric current of 2.00 A is run through a soln of CuSO4 for a period of 20.0 min? Gather everything you need to know... • Charge delivered in 20.0 min? • Relationship between charge and g Cu? Ans. 0.790 g

  12. Example 20.14 Electrolysis provides a useful way to deposit a thin metallic coating on a electrically conducting surface. This technique is called electroplating. How much time would it take, in minutes to deposit 0.500 g of metallic nickel on a metal object using a current of 3.00 A? The nickel is reduced from the +2 oxidation state. Ans. 9.13 min Practice Exer. 20.25, 20.26, 20.27, 20.28

More Related