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Oxidation Numbers & Redox Reactions

Oxidation Numbers & Redox Reactions. How to Make Balancing Redox Reactions a Relatively Painless Process. What Do We Need to Consider?. Definitions Rules for Assigning Oxidation Numbers Balancing Redox Reactions. What are Oxidation Numbers?.

yolanda-hensley
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Oxidation Numbers & Redox Reactions

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  1. Oxidation Numbers & Redox Reactions How to Make Balancing Redox Reactions a Relatively Painless Process

  2. What Do We Need to Consider? • Definitions • Rules for Assigning Oxidation Numbers • Balancing Redox Reactions

  3. What are Oxidation Numbers? • The oxidation number of an element indicates the number of electrons lost, gained, or shared as a result of chemicalbonding. • The change in the oxidation state of a species lets you know if it has undergone oxidation or reduction.

  4. Oxidation can be defined as "an increase in oxidation number". • In other words, if a species starts out at one oxidation state and ends up at a higher oxidation state it has undergoneoxidation.

  5. Conversely, • Reduction can be defined as "a decrease in oxidation number". • Any species whose oxidation number is lowered during the course of a reaction has undergone reduction.

  6. Example: Na + Cl2 2NaCl • The Na starts out with an oxidation number of zero (0) and ends up having an oxidation number of 1+. • It has been oxidized from a sodium atom to a positive sodium ion.

  7. Example: Na + Cl2 2NaCl • The Cl2 also starts out with an oxidation number of zero (0), but it ends up with an oxidation number of 1-. • It, therefore, has been reduced from chlorine atoms to negative chloride ions.

  8. Oxidizing agents • The substance bringing about the oxidation of the sodium atoms is the chlorine, thus the chlorine is called an oxidizing agent. • In other words, the oxidizing agent is being reduced (undergoing reduction).

  9. Reducing agents • The substance bringing about the reduction of the chlorine is the sodium, thus the sodium is called a reducing agent. • Or in other words, the reducing agent is being oxidized (undergoing oxidation).

  10. Redox reactions • Oxidation is ALWAYS accompanied by reduction. • Reactions in which oxidation and reduction are occurring are usually called Redox reactions.

  11. Rules for Assigning Oxidation Numbers • There are several rules for assigning the oxidation number to an element. • Learning these rules will simplify the task ofdetermining the oxidation state of an element, and thus, whether it has undergone oxidation or reduction.

  12. Rules 1 and 2 • The oxidation number of an atom in the elemental state is zero. • Example: Cl2 and Al both are 0 • The oxidation number of a monatomic ion is equal to its charge. • Example: In the compound NaCl, the sodium has an oxidation number of 1+ and the chlorine is 1-.

  13. Rule 3 • The algebraic sum of the oxidation numbers in the formula of a compound is zero. • Example: the oxidation numbers in the NaCl above add up to 0

  14. Rule 4 • The oxidation number of hydrogen in a compound is 1+, except when hydrogen forms compounds called hydrides with active metals, and then it is 1-. • Examples: H is 1+ in H2O, but 1- in NaH (sodium hydride).

  15. Rule 5 • The oxidation number of oxygen in a compound is 2-, except in peroxides when it is 1-, and when combined withfluorine. Then it is 2+. • Example: In H2O the oxygen is 2-, in H2O2 it is 1-.

  16. Rule 6 • The algebraic sum of the oxidation numbers in the formula for a polyatomic ion is equal to the charge on that ion. • Example: in the sulfate ion, SO42-, the oxidation numbers of the sulfur and the oxygens add up to 2-. The oxygens are 2- each, and the sulfur is 6+.

  17. Application Problems: • What is the oxidation number of chromium in • Na2CrO4 • In Na2CrO4the Cr has an oxidation number of 6+ and it can be said to be in the 6+ oxidation state • Cr2O72- • In Cr2O72-the Cr also has an oxidation number of 6+ and it can be said to be in the 6+ oxidation state.

  18. Given the unbalanced equation below • Cr2O3(s) + Al(s) Cr(s) + Al2O3(s) • identify the oxidation state of each element • Cr3+ O2- Al0 Cr0 Al3+ O2- • identify the oxidizing agent • Cr3+ in the chromium(III) oxide • identify the reducing agent • Al(s)

  19. Balancing Redox Reactions • Redox reactions often can be most easily balanced by the following method usually called the half-reaction method. • Example: • Cr3+(aq) + Cl1-(aq) Cr(s) + Cl2(g)

  20. first • divide it into half-reactions; that is, the oxidation reaction and the reduction reaction. • Oxidation: Cl1-(aq) Cl2(g) + 2e- • Reduction: Cr3+(aq) + 3e- Cr(s)

  21. Next balance • each half-reaction with respect to atoms first, then with respect to electrons. • Oxidation: 2Cl1-(aq) Cl2(g) + 2e- • Reduction: Cr3+(aq) + 3e- Cr(s) • Oxidation: 3(2Cl1-(aq) Cl2(g) + 2e-) • Reduction: 2(Cr3+(aq) + 3e- Cr(s))

  22. Which gives • Oxidation: 6Cl1-(aq) 3Cl2(g) + 6e-) • Reduction: 2Cr3+(aq) + 6e- 2Cr(s)

  23. add • the two half-reactions together canceling the electrons which are now equal on each side of the arrow • Final Equation: • 2Cr3+(aq) + 6Cl1-(aq) 2Cr(s) + 3Cl2(g)

  24. Granted, • this example is a simple one, however, the same basic procedure is carried out for most redox reactions, with theaddition of a couple of other steps, depending on the reaction conditions.

  25. Example: • copper metal added to concentrated nitric acid • Cu(s) + HNO3(aq)  Cu(NO3)2(aq) + NO2(g)

  26. Half-reactions • Oxidation: Cu(s) Cu2+(aq) + 2e- • Reduction: NO31-(aq) + e- NO2(g)

  27. balance • each half-reaction with respect to atoms first, then with respect to electrons • Oxidation: Cu(s) Cu2+(aq) + 2e- • Reduction: 2NO31-(aq) + 2 e-2NO2(g) • Add H2O on the right to balance out the oxygen atoms. • Then add enough H+ on the left to balance the hydrogens in the H2O

  28. balance • Oxidation: Cu(s) Cu2+(aq) + 2e- • Reduction: 2NO31-(aq) + 4H+ + 2 e- 2NO2(g)+ 2H2O • Add the two equations together canceling out the 2e- on each side

  29. balance • Cu(s) + 2NO31-(aq) + 4H+ • Cu2+(aq) + 2NO2(g)+ 2H2O • You now have the net ionic equation that shows that copper reacts with nitric acid to produce copper(II) ions, nitrogen dioxide gas and water. The other two nitrate ions not shown in this form of the equation give copper(II) nitrate and are spectator ions.

  30. An example: the zinc-copper voltaic cell This cell produces 1.10 volts

  31. The cell works because the valence electrons of zinc have a higher energy than the valence electrons of copper. This gives zinc a greater tendency to give up electrons than copper. In the external circuit, electrons will always move from the positive anode to the negative cathode.

  32. This is an oxidation half-reaction.Oxidation always occurs at the anode. • Zinc is the anode: • A piece of Zn metal is immersed in ZnSO4 (aq) • Atoms of the zinc metal lose electrons and become • zinc ions in solution. • The half-reaction is represented by: Zn    Zn+2 + 2e- A shorthand for representing the half-cell reaction is: Zn|Zn+2

  33. This is a reduction half-reaction.Reduction always occurs at the cathode. • Copper is the cathode: • A piece of Cu metal is immersed in CuSO4 (aq) • Copper ions in solution gain electrons to add copper • atoms to the copper metal. • The half-reaction is represented by: Cu+2 + 2e-  Cu • A shorthand for representing the half-cell reaction is: Cu+2|Cu

  34. This cell operates because: Copper metal takes electrons away from zinc metal. These electrons move from the anode to the cathode through the external circuit. The newly formed Zn+2 go into solution. SO4-2 ions in solution move from the cathode to the anode through the salt bridge. The cell continues to operate as long as there is a potential energy difference between the half-cells. The shorthand for the zinc-copper cell is:Zn|Zn+2||Cu+2|Cu The anode (oxidation reaction) is always on your left. The cathode (reduction reaction) is always on your right. The two vertical lines between the half-cells represent the "salt bridge".

  35. Some practical applications of electrochemical cells: Dry cell - acid form This is the source of power for an ordinary flashlight. Most "flashlight batteries" produce 1.5 volts. The case of the cell is zinc metal acting as the anode. At the center of the cell is a stick of graphite for the cathode. The graphite stick is surrounded by a paste of MnO2 and NH4Cl.

  36. Half-reactions for this cell are: Anode - Zn (s)  Zn+2 + 2e - Cathode - 2NH4+ (aq) + 2MnO2 (s) + 2e -  Mn2O3 (s) + 2NH3 (aq) + H2O (l) Remember that the term "battery" refers to two or more cells connected together. The 9-volt transistor battery is a true battery. It contains six individual 1.5 volt cells connected in series to produce 9 volts.

  37. Dry cell - alkaline form: The NH4Cl in the "acid form" cell is replaced by KOH and the zinc is in powder form rather than a solid piece of metal. The graphite cathode is eliminated and acid corrosion of the container does not occur. The alkaline cell is more efficient and can be miniaturized to fit more varied applications.

  38. Your level of understanding of this material will determine your success with the electrochemistry concept. Study the information until you are comfortable with it. Write the answer for the following questions: • What causes metallic conduction? • How are metallic and electrolytic conduction different? • At which electrode does reduction occur during electrolysis? • What is the purpose of a generator? • What does voltage measure? • Are cations positive or negative? • Why can't the potential energy of a single half-cell be measured? • Oxidation always takes place at which electrode? • What would cause electrons to stop flowing in a cell? • What does a single vertical line represent in the shorthand for a voltaic cell? • What does a double vertical line represent in the shorthand for a voltaic cell? • How is the voltage of a voltaic cell determined? • What direction do electrons flow in the external circuit of a cell?

  39. Identify the oxidizing agent and reducing agent in each equation: • H2SO4 + 8HI   H2S + 4I2 + 4H2O • CaC2 + 2H2O   C2H2 + Ca(OH)2 • Au2S3 + 3H2  2Au + 3H2S • Zn + 2HCl   H2 + ZnCl2

  40. Fuel Cell Electrode DiagramThis picture contains active zones.Click and explore! Menu Prev Next A diagram of one electrode pair is shown opposite. Hydrogen and oxygen are supplied to the cell as molecular gases which are dissociated by catalyst material on the electrodes. At the Anode (negative electrode) hydrogen gas is dissociated into two protons (H+) and twoelectrons (e_). The protons can travel throughthe polymeric proton exchange membrane:PEM (a solid electrolyte) and combine with oxygen ions generated at the Cathode (positive electrode) to form water.

  41. TheFuel Cell Apparatus These pictures contains active zones. Click and explore!

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