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Oxidation Reduction Reactions

Oxidation Reduction Reactions. Redox Reactions. Historically examples included corrosion, combustion and smelting. Now oxidation-reduction ( redox ) has a much broader meaning.

shelby-conley
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Oxidation Reduction Reactions

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  1. Oxidation Reduction Reactions Redox Reactions

  2. Historically examples included corrosion, combustion and smelting.

  3. Now oxidation-reduction (redox) has a much broader meaning.

  4. A simple redox demonstration involves placing pure copper wire in aqueous silver nitrate. The wire soon looks fuzzy from the formation of tiny silver crystals, The solution is turning blue.

  5. Cu (s) + 2 AgNO3(aq) Cu(NO3)2 (aq)+ 2Ag(s) • Cu(s) + 2Ag+(aq) +2NO3-(aq)  Cu2+(aq) + 2NO3- (aq) + 2Ag(s) • Removing the spectator ions we get our net ionic equation: Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag (s) Copper began as a neutral atom with no charge but changes into an ion with a 2+ charge. This happens when it loses 2 electrons. Cu (s) Cu2+ (aq) + 2 e- Copper was oxidized because it lost electrons. Silver went from an ion Ag+ to a neutral atom Ag. The only way this can happen is to gain electrons. It has been reduced.

  6. Oxidation – the loss of electronsReduction – the gaining of electrons LEO the Lion Says GER LEO Loss of Electrons is Oxidation GER Gain of Electrons is Reduction

  7. Oxidation Numbers • Oxidation numbers are our system for keeping track of what gains and what loses electrons. • An oxidation number is a positive or negative number assigned to an atom in a molecule or ion that reflects a partial gain or loss of electrons. • Main Rules: 1. The oxidation number of a pure element (not an ion) is zero (0). 2. The oxidation number of a monatomic ion (by itself or in an ionic compound) is equal to its charge.

  8. NaCl Na+ charge +1 Ox. No. +1Cl- charge -1 Ox. No. -1 3. The oxidation number of hydrogen is almost always +1 when it is in a compound. HCl H +1 Cl -1 H2S H +1 S -2 4. The oxidation number of oxygen is almost always -2 when it is in a compound. Two exceptions: peroxides O is -1 oxygen with fluorine O is +2.

  9. 5. The sum of the oxidation number in a compound is zero. Example: Mn2O7 O Is -2 (rule 4) -2 x 7 = -14 total The sum of the oxidation numbers must be zero. The total of the oxidation numbers of Mn must be +14. +14/2 = +7. The oxidation number of Mn must be +7.

  10. 6. The sum of the oxidation number of a polyatomic ion is equal to the charge on that ion. Example: Cr2O72- Oxygen - -2 x 7 = -14 The sum of the oxidation numbers must be -2 (instead of zero) as that is the overall charge on the ion. -14 + 12 = -2 +12/2 = +6 Cr The oxidation number always refers to each individual atom in the compound not the total for that element.

  11. Recognizing Redox Reactions • Oxidation cannot occur without reduction • Reducing Agent: substance that is oxidized • Oxidizing Agent: substance that is reduced • The oxidizing agent is usually the entire compound that the element that is being reduced is in, not just the element that is being reduced. Agents are always on the reactants side.

  12. Example • Sodium metal reacts with chlorine gas: Na + Cl2 2 NaCl 0 0 +1 -1 Na 0 to +1 lost 1 e oxidized (reducing agent) Cl 0 to -1 gained 1 e reduced (oxidizing agent) This is a redox reaction.

  13. Example 2 • Zn + HNO3  Zn(NO3)2 + NO2 + H2O Assign oxidation #’s Is this a redox reaction? What is the oxidizing agent? Remember that an increase in the oxidation number means oxidation Decrease in the oxidation number means reduction.

  14. Practice Set 2 Recognizing Redox Reactions • 4 HCl + O 2  2 H2O + 2 Cl2

  15. Oxidation – Reduction In a Shake Zn + CuSO4 • Observations:

  16. Balancing Redox Equations Using Oxidation Numbers • Cu(s) + Ag+ (aq)  Cu 2+(aq) + Ag (s) • Looking at the number of atoms the net ionic equation appears balanced. The charges however ARE NOT. in ox no final change in e’ change in e balance for electrons Coefficient Total Cu 0 2+ 2 lost 2 x 1 = 2 Ag +1 0 1 gained 1 x 2 = 2 Cu (s) + 2 Ag + Cu 2+(aq) + 2 Ag (s)

  17. Balancing Redox Reactions SnCl2 + HgCl2 SnCl4 + HgCl Initial Final Change Coeff Total e Sn +2 +4 2 x 1 2 Cl -1 -1 none Hg +2 +1 1 x2 2 SnCl2 + 2 HgCl2  SnCl4 + 2HgCl

  18. Example 2 MnO41- + Fe2+ + H1+  Mn2+ + Fe3+ + H2O element in ox final ox change Mn +7 +2 5 x 1 = 5 Fe +2 +3 1 x 5 = 5 MnO41- + 5 Fe2+ + H1+  Mn2+ + 5 Fe3+ + H2O But the H and O are not balanced: MnO41- + 5 Fe2+ + 8H1+  Mn2+ + 5Fe3+ + 4H2O

  19. Example 3 - A Diatomic Gas as Reactant NH3 + O2 NO2 + H2O in. ox final change no. atoms no. e’ balance N -3 +4 7 7 x 4 = 28 O 0 -2 2 2 4 x 7 = 28 Which O does the coefficients go in front of?????? Do the N first, O in diatomic second, H last. 4 NH3 + 7 O2 4 NO2 + 6 H2O

  20. Balancing Redox Using Half reactions • Using this method we will break an equation into the oxidation reaction and the reduction reaction. • These separate equations are referred to as half-reactions because the two halves cannot occur with out the other. (or two halves make a whole) • The spectator ions are removed from the equations. • Each half reaction is balanced separately. Electrons are added to balance the charges. • The electrons lost must equal the electrons gained. • Everything is put back together including the spectator ions.

  21. Example 1 – no spectator ions Mg (s) + Cl2(g) MgCl2(s) What is oxidized? What is reduced? Mg is oxidized 0 to +2 Cl is reduced 0 to -1 Mg  Mg+2 Cl2 2 Cl- Mg  Mg2+ + 2 e’ Cl2 + 2 e’  2 Cl- electrons are equal so it is already balanced. Mg (s) + Cl2(g) MgCl2(s)

  22. Example 2 Cu (s) + AgNO3(aq)  Cu(NO3)2 + Ag (s) Cu  Cu2+ + 2e’ Ag+ + 1 e’  Ag The electrons do not balance. Cu  Cu2+ + 2e’ 2Ag+ + 2 e’  2Ag Cu + 2Ag+  Cu2+ + 2Ag Return spectator ions: Cu (s) + 2 AgNO3(aq)  Cu(NO3)2 + 2Ag (s)

  23. Example 3: Reaction in an Acid Solution MnO4- + Fe2+ + H+  Mn2+ + Fe3+ + H2O Fe oxidized Fe +2 to +3 Mn reduced + 7 to +2. Fe2+ Fe3+ + 1 e’ Mn7+ + 5 e’  Mn2+ 5 Fe2+ 5 Fe3+ + 5 e’ MnO4- + 5 Fe + H+  Mn2+ + 5 Fe3+ + H2O (The hydrogen and the oxygen must be included in the half reaction and balanced.). MnO4- + 5 Fe2+ + 8 H+  Mn2+ + 5 Fe3+ + 4 H2O

  24. Electrochemistry • During redox reactions, electrons pass (flow) from one substance to another. • Electrochemistry is the branch of chemistry that deals with the conversion of chemical energy to electrical energy. • Electrochemical Cells – spontaneous chemical reactions convert chemical energy into electrical energy. Batteries are an example. • Electrolytic Cells – electrical energy is used to cause non-spontaneous chemical reactions to occur. Rechargeable batteries and electroplating are examples.

  25. Electrochemical CellsAlso called voltaic cells or galvanic cells • The electrons that are released by the oxidation half reaction are passed along to the reduction reaction. • An external circuit needs to be created. Reactions will occur without one but electricity will not be created. Zn(s) + Cu2+ (aq)  Zn2+ (aq) + Cu(s) zinc is oxidized (reducing agent) copper is reduced (oxidizing agent) An electrolytic solution (conducts electricity due to the presence of ions) is needed.

  26. Each beaker is a half cell. The metal strips become the electrodes.

  27. The electrodes are connected by a wire. This becomes the external circuit. A salt bridge contains an electrolytic solution. This allows the electrons to flow And becomes the internal circuit.

  28. Anode = Oxidation An Ox - post of the cell source of electrons Reduction = Cathode Red Cat + post of the cell Consumes electrons

  29. Electrons travel from the Zn anode to the Cu cathode through the wire of the external circuit. • At the anode Zn2+ ions go into solution. • The excess positive charges attract the negative NO3- ions from the salt bridge. Electrochemical Cell Simulation

  30. External Circuit – Electrons flow from anode to cathode • Internal Circuit- anions to anode; cations to cathode • An entire complete cell is comprised of: 2 half-cells (electrodes in their solution) internal circuit (salt bridge and the half cells) external circuit (the wire connecting the two electrodes) • When the cell reaches equilibrium, the voltage will be zero.

  31. Activity Series Revisited • What determines which element is oxidized and which is reduced? • Metals like to lose electrons so they tend to be oxidized. Metals are arranged according to the activity series. • In our cell example Zn is the anode (oxidation) while Cu is the cathode (reduction). • Where are these elements on the activity series? • Zinc is quite a ways higher on the series and therefore more easily oxidized.

  32. Standard Reduction Potentials For Half Reactions • This table allows us to determine the voltage of the electrochemical cells. • All values on the table are determined based on a half-reaction with a hydrogen half-cell. • 2H+(aq) + 2e’  H2(g) Eo = 0.00 v • Means: 25 oC, 100 kPa, 1 mol.L-1 Positive values on this table mean that they are better at competing for electrons and will be reduced. The hydrogen will be oxidized.

  33. E° Cu2+(aq) + 2e- → Cu(s)  0.34 V H2 (g)  → 2H+(aq) + 2e- + 0.00 V ___________________________ _____________ Cu2+(aq) + H2 (g) → 2H+(aq) + Cu(s)  0.34 V In the Table of Standard Reduction Potentials that zinc has a negative E° indicating that it is not as good at competing for electrons as hydrogen. Zn2+(aq) + 2e- → Zn(s)    E° = -0.76 V Therefore if zinc and hydrogen are paired together in an electrochemical cell, the hydrogen would be reduced (gain the electrons) and zinc would be oxidized (losing electrons). To determine the net redox reaction as well as the voltage of the electrochemical cell we reverse the zinc equation (write it in oxidation form), and also reverse it's sign before adding the equations and E° together: E° Zn(s)  → Zn2+(aq) + 2e- 0.76 V 2H+(aq) + 2e- → H2 (g)  + 0.00 V ______________________ ____________ Zn(s)+ 2H+(aq) → Zn2+(aq) + H2 (g)0.76 V

  34. Calculating Voltages of Electrochemical Cells • We can now use the table to calculate the voltage of our zinc-copper cell as well as use the table to explain why zinc is the anode (oxidized) and copper is the cathode (reduced). • Locate the Cu/Cu2+ half-reaction. • Cu: Cu2+ (aq) + 2e’ Cu(s) Eo = 0.34 V Zn: Zn2+ (aq) + 2e’ Zn(s)Eo = -0.76 V The copper is larger than zinc so it will be reduced. Zinc will be oxidized. Zn  Zn2+ + 2 e’ 0.076 V Cu2+ + 2 e’  Cu 0.034 V Cu2+ (aq) + Zn(s) Zn2+(aq) + Cu (s) 1.10 V

  35. Always reverse the half reaction that will result in a positive value for Eowhen the equations are added together. • A positive Eo value means that the reaction is spontaneous and electrochemical cells always involve a spontaneous chemical reaction.

  36. Batteries • Batteries are electrochemical cells used to generate power. Types of Batteries: • Dry Cells (Primary batteries) non rechargeable; electrolytes are paste not liquid; used for flashlights, radios; toys etc. Consists of a zinc case (anode) , a graphite rod (cathode) and an electrolytic paste

  37. 2. Secondary Batteries Rechargeable An example is a car battery (lead-acid).

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