1 / 18

Oxidation and Reduction

Oxidation and Reduction. Chapters 20 & 21. Oxidation vs Reduction. Oxidation= A substance loses electrons Reduction A substance gains electrons 2Al (s) 0 + 3CuCl 2(aq) → 2AlCl 3(aq) + 3Cu (s) 0 Al (s) 0 → Al (aq) +3 aluminum oxno increases Cu (aq) +2 → Cu (s) 0 copper oxno decreases.

tuyet
Télécharger la présentation

Oxidation and Reduction

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. Oxidation and Reduction Chapters 20 & 21

  2. Oxidation vs Reduction • Oxidation= A substance loses electrons • Reduction A substance gains electrons • 2Al(s)0 + 3CuCl2(aq)→ 2AlCl3(aq) + 3Cu(s)0 • Al(s)0→Al(aq)+3aluminum oxno increases • Cu(aq)+2→Cu(s)0 copper oxno decreases

  3. What’s really happening… • 2Al(s)0 + 3CuCl2(aq)→ 2AlCl3(aq) + 3Cu(s)0 • 2Al(s)0→ 2Al(aq)+3+ 6e- Al oxno increases • 3Cu(aq)+2 + 6e- →Cu(s)0 Cu oxno decreases • These are called half reactions. Notice the number of electrons lost is the same as the number gained.

  4. What???? • Oxidation= increase in oxygen atoms, increase in oxno, loss in electrons • Reduction= loss of oxygen atoms, decrease in oxno, gain in electrons • Remember: • “L.E.O. says G.E.R.” • Loss of electrons oxidation • Gain electrons reduction

  5. Agents • The oxidizing agent causes oxidation of another substance. Example: copper • Cu(aq)+2→Cu(s)0 • The reducing agent causes reduction of another substance. Example: aluminum • Al(s)0→Al(aq)+3

  6. Activity Series of Metals (p. 668)

  7. When the the reaction happens, electrons move from Al to Cu • 2Al(s)0→ 2Al(aq)+3+ 6e- • 3Cu(aq)+2 + 6e- →Cu(s)0 • This electron flow can be measured as voltage! • We will see how later.

  8. Types of Redox Reactions • Direct Combination: • S + O2→ SO2 • Decomposition: • HgO → 2Hg + O2 • Single Replacement: • Cu(s) + 2AgNO3(aq)→ Cu(NO3)2(aq) + 2Ag(s) • Cu(s) + 2Ag+(aq)→ Cu+(aq) + 2Ag(s) (net ionic) • But: Cu(s) + ZnCl2(aq)→ NR • Cu(s) + Zn+2(aq)→ No reaction (due to relative reactivity rank)

  9. Balancing Redox Equations • Some equations are are difficult to balance by inspection or trial and error that worked up until now. • The fundamental principle is that the number of electrons lost in the oxidation process must equal the number of electrons gained in the reduction process.

  10. Electrochemical cells • Use redox reactions to either produce or use electricity.

  11. Voltaic Cells • In late 1700’s Italian physician Luigi Galvani twitched frog legs by connecting two metals. Italian scientist Alessandro Volta concluded the two metals in the presence of water produce electricity.

  12. Voltaic Cells • Zn(s) + CuSO4(aq)→ ZnSO4(aq) + Cu(s) • Zn(s) → Zn+2(aq) + 2e- oxidation • Cu+2(aq) + 2e- → Cu(s) reduction • Half Cell- Zn (anode) • Pushes e- to Cu • (cathode)

  13. Voltaic Cell • Electrons move spontaneously from the anode (-) to the cathode (+) • The salt bridge allows • Electrons to move freely • Without mixing solutions.

  14. Cell Potential • Ability to move e- through a wire from one electrode to another is the electrical or cell potential. It is measured in volts (v) • For Example: A Zn-Cu cell with 1 M solutions produces 1.10 volts. • Here is how it works:

  15. Standard Reduction Potentials (p. 693)

  16. Standard Electrode Potentials • Ecell = Eoxidation + Ereduction • Ecell0= sum of the oxidation potential (Eox0) plus reduction potential (Ered0) • The standard state conditions are noted with the 0. • E0 are determined by measuring half cell potential differences. • Zn(s) → Zn+2(aq) + 2e- E0 ox = + 0.76 V • Zn+2(aq) + 2e- → Zn(s) E0 red = - 0.76 V

  17. Calculating Cell Potentials • Zn(s) → Zn+2(aq) + 2e- E0 ox = + 0.76 V • Cu+2(aq) 2e- → Cu(s) E0 red = + 0.34 V • Total Voltage (Ecell) = + 1.10 Volts • Practice Problems #1 and 2 on P. 696.

  18. That’s it for this • Electrifying lecture!

More Related