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Electro-Chemistry

Electro-Chemistry. Electrochemistry. Electrochemistry is the study of the relationships between electrical energy and chemical reactions It’s the study of how chemical energy is changed to electrical energy through the exchange (flow) of electrons. Oxidation-Reduction Reactions.

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Electro-Chemistry

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  1. Electro-Chemistry

  2. Electrochemistry • Electrochemistry is the study of the relationships between electrical energy and chemical reactions • It’s the study of how chemical energy is changed to electrical energy through the exchange (flow) of electrons

  3. Oxidation-Reduction Reactions Oxidation-reduction reactions are chemical reactions involving the exchange of electrons between two substances. • During an oxidation reaction, there is a loss of electrons. • For example, the oxidation of Fe(s) to Fe+2(aq) is accompanied by the loss of two electrons. Fe(s)  Fe+2(aq) + 2 e -

  4. Oxidation-Reduction Reactions • During a reduction reaction, there is a gain of electrons. • Example: reduction of Cu+2(aq) to Cu(s) is accompanied by the gain of two electrons. Cu+2(aq) + 2 e -  Cu(s) • Short-cut: • Oxidation Is Loss (OIL) of electrons (electrons on the right-hand side of the equation) • Reduction Is Gain (RIG) of electrons (electrons on the left-hand side of the equation) • Remember OIL-RIG.

  5. Redox Reactions • Oxidation and reduction occur together. • Hence, they are called redox reactions. • In a redox reaction, one substance gains electrons (reduction), while the other substance loses electrons (oxidation). Cu+2(aq) + Fe(s)  Fe+2(aq) + Cu(s) • In this reaction, Cu+2(aq) is reduced to Cu (s) and Fe(s) is simultaneously oxidized to Fe+2(aq).

  6. Redox Reactions • The substance being reduced is called the oxidizing agent. • The substance being oxidized is called the reducing agent.

  7. Oxidation Number • The oxidation number is a number which tells us how oxidized or reduced a given element of a given substance is. • The higher the oxidation number is, the more oxidized the element is.

  8. Assigning Oxidation Numbers • Rule # 1: The oxidation number for an element in the elemental state is 0. • the oxidation numbers for Al in Al(s) is 0. • Rule # 2: The oxidation number for oxygen in most oxygen compounds is equal to - 2. • Rule # 3: Group IA compounds, have an oxidation number of +1 • Na in NaCl(s) or in Na2SO4(s) has an oxidation number of +1 • Rule # 4: Group IIA compounds, the oxidation number of the metal is +2. • Ca in CaCO3 and in Ca(NO3)2 has an oxidation number of +2

  9. Assigning Oxidation Numbers • Rule # 5: In all HALOGEN-containing compounds, the oxidation number for the halogen is -1. • The oxidation number of F in NaF, CaF2 and AlF3 is always -1. • In NaCl, Cl has an oxidation number of -1. • Rule # 6: In a compound, the oxidation number of hydrogen is +1 if H is bonded to a nonmetal. • H in NH3, in CH4, in H2O and in HCN has the same oxidation number of +1. • Rule # 7: In a compound, the oxidation number of hydrogen is -1 if H is bonded to a metal. (Note: in this case, H behaves like an anion and is called hydride) • H in NaH (sodium hydride), in CaH2 (calcium hydride) has the oxidation number of -1.

  10. Assigning Oxidation Numbers • Rule # 8: The sum of the oxidation numbers of all elements in a compound is equal to 0 (the charge of the compound). • For example in NO2 Ox.# of N + 2 x Ox # O = 0. • Rule # 9: The sum of the oxidation numbers of all elements in a polyatomic ion is equal to the charge of the polyatomic ion. • For example in CO3-2, Ox # C + 3 x Ox # O = -2.

  11. Assigning Oxidation # Examples 1. Na2SO4: • Oxidation number of Na = “Na” = +1 • Oxidation number of O = “O” = -2 • Oxidation number of S: = “S” to be calculated • 2 “Na” + “S” + 4 “O” = 0 • 2 x (+1) + “S” + 4 x (-2) = 0 • “S” – 6 = 0 • “S” = +6

  12. Assigning Oxidation # Examples • Ox # of Pt in K2PtCl4 • Ox # of Mn in KMnO4 • Ox # of Pb in PbSO4 • Ox # of Pb in PbS2 • Ox # of each C in C2H3O2-

  13. Redox Reactions Identifying: • What is Oxidized (the Reducing Agent) • What is Reduced (the Oxidizing Agent) … in a Redox Reaction How’s It Done?

  14. Redox Reactions 1. Identify the substance oxidized, substance reduced, oxidizing agent, reducing agent, and write oxidation and reduction half reactions 2H2 + O2→ 2H2O Oxidized Reduced (RA) (OA) 2H20 → 4H+1 + 4e- Oxidation Half reaction O20 + 4e- → 2O-2 Reduction Half Reaction

  15. Redox Reactions 2. Identify the substance oxidized, substance reduced, oxidizing agent, reducing agent, and write oxidation and reduction half reactions Mg + Zn+2→ Mg+2 + Zn Oxidized Reduced (RA) (OA) Mg → Mg+2 + 2e- Oxidation Half reaction Zn+2 + 2e- → ZnReduction Half Reaction

  16. Redox Reactions 3. Identify the substance oxidized, substance reduced, oxidizing agent, reducing agent, and write oxidation and reduction half reactions Sn+ SnO2 + 2H2SO4→ 2SnSO4 + 2H2O Oxidized Reduced (RA) (OA) Sn Sn+2 + 2e-Oxidation Half reaction Sn+4 + 2e- Sn+2Reduction Half Reaction

  17. Redox Reactions Redox Reactions Worksheet

  18. Electrochemical Cells Let’s Look at the Handouts as an Intro to Electrochemical Cells

  19. - +2 Fe à Fe + 2 e Oxidation E° = + 0.44 V (s) (aq) anode - + Ag + e à Ag Reduction E° = + 0.80 V (aq) (s) cathode Calculating Standard Cell Potentials Example: Calculate the standard cell potential for: Fe (s)  Fe+2 (aq)  Ag+ (aq)  Ag (s) Fe+2 + 2e- Fe(s) Eo = -0.44V Ag+1 + 1e- Ag(s) Eo = +0.80V The Ag reduction has a higher E0 (higher on the list), so it is the reduction … the Fe reaction is turned around to an oxidation:

  20. - +2 Fe Fe + 2 e Oxida tion E° = + 0.44 V à (s) (aq) anode - + 2 (Ag + e Ag ) Reduction E° = + 0.80 V à x (aq) (s) cathode - +2 Fe à Fe + 2 e Oxidation E° = + 0.44 V (s) (aq) anode - + 2 Ag + 2 e à 2 Ag Reduction E° = + 0.80 V (aq) (s) cathode Calculating Standard Cell Potentials • In order to write the overall redox reaction, multiply the reduction half-reaction by 2, but NOT the potential value. Thus, Hence, Overall Redox reaction: • Fe (s) + 2 Ag+ (aq)  Fe+2 (aq) + 2 Ag (s)

  21. Calculating Standard Cell Potentials • Since E°cell is positive, the cell operates spontaneously. • Reaction will take place only if E°cell is positive E° = E° E° + cell red ox E° = (+ 0.80 V) + (+0.44 V) cell E° = + 1.24 V cell

  22. Calculating Standard Cell Potentials Example: Calculate the standard cell potential for: Al (s)  Al+3 (aq)  Hg+2 (aq)  Hg (s) Al+3 + 3e- Al(s) Eo = -1.66V Hg+2 + 2e- Hg(s) Eo = +0.85V The Hg reduction has a higher E0 (higher on the list), so it is the reduction … the Al reaction is turned around to an oxidation: Al(s)  Al+3 + 3e-Eo = +1.66V Hg+2 + 2e- Hg(s) Eo = +0.85V

  23. Calculating Standard Cell Potentials The Hg reduction has a higher E0 (higher on the list), so it is the reduction … the Al reaction is turned around to an oxidation: 2 x (Al(s)  Al+3 + 3e-) Eo = +1.66V (Hg+2 + 2e- Hg(s)) Eo = +0.85V 3 x 2Al(s) + 3Hg+2  2Al+3 + 3Hg Eo Cell: Eo red + Eoox Eo Cell = 0.85V + 1.66V = 2.51V

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