1 / 15

Oxidation/Reduction Reactions

Oxidation/Reduction Reactions. REDOX REACTONS!. All chemical reactions fall into two categories those that are redox and those that are not redox ! Redox reactions involve the transfer of electrons. What is oxidation and reduction?. Oxidation

daw
Download Presentation

Oxidation/Reduction Reactions

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/Reduction Reactions • REDOX REACTONS! • All chemical reactions fall into two categories those that are redox and those that are not redox! • Redox reactions involve the transfer of electrons.

  2. What is oxidation and reduction? • Oxidation • Early chemists thought of oxidation as the combining of an element with oxygen, as in combustion. • Now we refer to oxidation as the loss of electrons. • Reduction • Early chemists thought of reduction as the loss of oxygen from a compound. • Now we refer to reduction as the gain of electrons.

  3. “LEO the lion says GER” • Oxidation • 4Fe(s) + 3O2(g)  2Fe2O3(s) - combustion or rusting • Fe  Fe3+ + 3e- - loss of electrons • Reduction • 2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g) -loss of oxygen • Fe3+ + 3e- Fe -gain of electrons

  4. Oxidation and Reduction must occur simultaneously: • Example: 2AgNO3(aq) + Cu(s)  2Ag(s) + Cu(NO3)2(aq) • Net Ionic: 2Ag+ + Cu  2Ag + Cu2+ • Nitrates ions are spectator ions! • Half Reactions: • 2Ag+ + 2e- 2Ag Reduction • Cu  Cu2+ + 2e- Oxidation • These reactions occur simultaneously and the electrons lost must equal the electrons gained.

  5. Reducing – Oxidizing Agents • Reducing Agent • The substance that loses electrons. • Cu is oxidized in the previous reaction but it is the REDUCING AGENT. • Oxidizing Agent • The substance that gains electrons. • Ag+ is reduced in the previous reaction but it is the OXIDIZING AGENT.

  6. Oxidation Numbers: • Elements always have an oxidation number of zero. • K -K is 0 • N2 -N is 0 • Oxygen is always -2 unless it is in a peroxide and then it is -1. • HNO3 - O is -2 • H2O2 - O is -1 • The charge on a monatomic ion is the oxidation number. • Fe3+ is +3 • Br1- is -1 • Hydrogen is always +1 unless it is in a metal hydride and then it is -1. • HNO3 - H is +1 • CaH2 - H is -1

  7. Determining Oxidation Numbers • K2CrO4 • O is -2 • K is +1 • Cr must be +6 • Cr2O3 • O is -2 • Cr must be +3 • Note that elements can have more than one oxidation number. • The sum of oxidation numbers in a neutral compound must be zero. • The sum of oxidation numbers in a polyatomic ions must equal the charge on the ion. • NO31- • O is -2 • N is +5

  8. Balancing Redox Reactions: • Example: • Sn + Ag+ Sn2+ + Ag • 1. Assign oxidation numbers. • 0 +1 +2 0 • Sn + Ag+ Sn2+ + Ag

  9. 2. Oxidation occurs when the oxidation number increases. Reduction occurs when the oxidation number decreases. Write the two half reactions. • Oxidation: Sn Sn2+ • Reduction: Ag+ Ag • 3. Use electrons to balance the charges in the half reactions. In oxidation the electrons appear on the right. In reduction the electrons appear on the left.

  10. Oxidation: SnSn2+ + 2e- • Reduction: Ag+ + 1e-  Ag • 4. If the number of electrons transferred is not equal multiple by a whole number so that the number of electrons lost equals the number gained. • Oxidation: Sn Sn2+ + 2e- • Reduction: (Ag+ + 1e- Ag) x2

  11. 5. Add the half reactions: • Oxidation: SnSn2+ + 2e- • Reduction: 2Ag+ + 2e-  2Ag • Net Balanced Redox Reaction: • Sn + 2 Ag+ Sn2+ + 2Ag

  12. Redox reactions are usually too complex to use trial and error method. • Balance the following example that is in an acidic solution (assume the presence of H2O and H+): • HNO3 + Fe2+ Fe3+ + NO2 • 1. Assign oxidation numbers. • +1+5-2 +2 +3 +4 -2 • HNO3 + Fe2+ Fe3+ + NO2

  13. 2. Write the half reactions: • Oxidation: Fe 2+ Fe3+ • Reduction: HNO3 NO2 • 3. Balance the half reactions. Use water to balance the oxygen (a), then hydrogen ions to balance the hydrogen(b), then electrons to balance the charges(c). • a) • Oxidation: Fe 2+ Fe3+ • Reduction: HNO3NO2 + H2O

  14. b) • Oxidation: Fe 2+ Fe3+ • Reduction: HNO3 + H+ NO2 + H2O • c) • Oxidation: Fe 2+ Fe3+ + 1e- • Reduction: HNO3 + H++ 1e- NO2 + H2O

  15. 4. Multiply the oxidation and reduction equations by whole numbers so that the number of electrons transferred is equal. • Oxidation: Fe 2+ Fe3+ + 1e- • Reduction: HNO3 + H+ + 1e- NO2 + H2O • 5. Add the reactions so that electrons cancel. • Net: Fe 2++ HNO3 + H+  Fe3+ + NO2 + H2O • 6. Check the equation is balance by charge and atom.

More Related