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Lecture13221
 

Lecture13221

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a supplemental resource for students

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    Lecture13221 Lecture13221 Presentation Transcript

    • Chemical Reactions: Oxidation-Reduction Lecture 13
    • Oxidation-reduction reactions everywhere.
    • Oxidation-reduction reactions:
    • Oxidation-reduction reaction is a chemical reaction in which one substance loses electrons while, at the same time, another substance gains electrons.
    • Electrons move from the reactant with less attraction for electrons to reactant with more attraction for electrons.
    • Electrons move in the formation of both ionic and covalent compounds.
    • Consider the following reactions:
      • 2Mg (s) + O 2 (g)  2MgO (s)
      • 2H 2(g) + Cl 2(g)  2HCl (g)
    • Electrons always move but it is a transfer of electron charge in case of ionic compounds and a shift of electron charge in case of covalent compounds.
    • Consider formation of magnesium oxide, MgO
    • Oxidation is the loss of electrons. Oxidation (electron loss by Mg): Mg   Mg +2 + 2e - 
    • Reduction is the gain of electrons. Reduction (electron gain by O 2 ): 1/2O 2 + 2e -   O -2
    • Oxidizing agent is that oxidizes. Oxygen gains electrons from magnesium, it is the oxidizing agent.
    • Reducing agent is that reduces. Magnesium loses electrons to oxygen, it is the reducing agent.
    • Give-and-take of electrons:
      • Oxidation : Mg   Mg +2 + 2e -
      • The reducing agent is oxidized.
      • Reduction : 1/2O 2 + 2e -   O -2
      • The oxidizing agent is reduced. 
    • Consider formation of hydrogen chloride, HCl
    • Oxidation is the loss of electrons. Oxidation (electron loss by H 2 ): 1/2H 2 - e -   H +
    • Reduction is the gain of electrons. Reduction (electron gain by Cl 2 ): 1/2Cl 2 + e -   Cl -
    • Oxidizing agent is that oxidizes. Chlorine gains electrons from hydrogen, it is the oxidizing agent.
    • Reducing agent is that reduces. Hydrogen loses electrons to chlorine, it is the reducing agent.
    • Give-and-take of electrons:
      • Oxidation : 1/2H 2 - e -   H +
      • The reducing agent is oxidized.
      • Reduction : 1/2Cl 2 + e -   Cl -
      • The oxidizing agent is reduced. 
    • Redox general chart
    • How do we know which atom loses electron charge and which atom gains it? Each atom is assigned an oxidation number (O.N.), or oxidation state.
    • Oxidation state is a hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic.
    • Different denotation:
      • Oxidation number – sign before the number: Mg +2 , Cr +3 , Cr +2 , Al +3 , S -2 , O -2
      • Ionic charge – sign after the number: Mg 2+ , Cr 3+ , Cr 2+ , Al 3+ , S 2- , O 2-
    • Memorize these
    • Memorize these
    • A sample problem on determining the oxidation number of an element.
    • Solve these
    • Memorize this
    • Memorize this
    • A sample problem on recognizing oxidizing and reducing agents.
    • To define a redox reaction, always monitor if oxidation numbers of the atoms change in the reaction, from the reactants to products.
    • An oxidation-reduction reaction only, not an “oxidation reaction” or a “reduction reaction”.
    • The number of electrons lost by the reducing agent equals the number of electrons gained by the oxidizing agent.
    • The oxidation number method
      • Assign oxidation numbers to all elements in the reaction.
      • From the changes in the oxidation numbers, identify the oxidizing and the reducing species.
      • Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes.
    • The oxidation number method
      • Multiply one or both of these numbers by appropriate factors to make the electrons lost equal the electrons gained, and use the factors as balancing coefficients.
      • Complete the balancing by inspection, adding states of matter.
    • A sample problem on balancing redox equations by the oxidation number method.
    • In any titration, one solution of known concentration is used to determine the concentration of another solution through a monitored reaction.
    • In a redox titration, a known concentration of oxidizing agent is used to determine an unknown concentration of reducing agent through a monitored redox reaction.
    • A sample problem on finding a concentration by a redox titration.
    • Whenever a reaction includes a free element as either reactant or product, it is a redox reaction.
    • Types of chemical reactions
      • Combination: X + Y  Z
      • Decomposition: Z  X + Y
      • Displacement: X + Y Z  X Z + Y
    • How two elements may react
      • Metal and nonmetal form an ionic compound:
      • 2 K (s) + Cl 2 (g)  2 K Cl (s)
      • 4 Al (s) + 3 O 2 (g)  2 Al 2 O 3 (s)
      • Two nonmetals form a covalent compound:
      • N 2 (g) + 3 H 2 (g)  2 N H 3 (g)
      • P 4 (s) + 6 Cl 2 (g)  4 P Cl 3 (l)
      • N 2 (g) + O 2 (g)  2 N O (g)
    • How an element and a compound may react
      • 2 N O (g) + O 2 (g)  2 N O 2 (g)
      • 2 S O 2(g) + O 2 (g)  2 S O 3 (g)
      • P Cl 3(l) + Cl 2 (g)  P Cl 5 (l)
      • 4 Fe (OH) 2(s) + 2H 2 O (l) + O 2 (g)  4 Fe ( O H) 3(s)
    • How a compound may decompose
      • Thermal decomposition:
      • 2K Cl O 3 (s)  2K Cl (s) + 3 O 2 (g)
      • 2 Hg O (s)  2 Hg (s) + O 2 (g)
      • ( N H 4 ) 2 Cr 2 O 7(s)  N 2 (g) + Cr 2 O 3(s) + 4H 2 O (g)
      • Electrolytic decomposition:
      • 2 H 2 O (l)  2 H 2 (g) + O 2 (g)
      • Mg Cl 2 (l)  Mg (l) + Cl 2 (g)
    • How one element may displace another
      • A metal displaces H 2 from water or acid:
      • 2 Al (s) + 6 H 2 O (g)  2 Al (OH) 3(s) + 3 H 2 (g)
      • 2 Na (s) + 2 H 2 O (l)  2 Na OH (aq) + H 2 (g)
      • Zn (s) + H 2 SO 4(aq)  Zn SO 4(aq) + H 2 (g)
      • Fe (s) + 2 H Cl (aq)  Fe Cl 2(aq) + H 2 (g)
      • Fe (s) + 2 H + (aq) + 2C l - (aq)  Fe 2+ (aq) + 2Cl - (aq) + H 2 (g)
      • Fe (s) + 2 H + (aq) +  Fe 2+ (aq) + H 2 (g)
    • How one element may displace another
      • A metal displaces another metal ion from solution:
      • Cu SO 4 + Zn  Cu + Zn SO 4
      • Cu 2+ + SO 2- 4 + Zn  Cu + Zn 2+ + SO 2- 4
      • Cu 2+ + Zn  Cu + Zn 2+
    • Why does copper displace silver and not vice versa?
    • Metal activity series
    • How one element may displace another
      • A halogen displaces another halogen:
      • 2K Br + Cl 2  2K Cl + Br 2
      • 2K + + 2Br - + Cl 2  2K + + 2 Cl - + Br 2
      • 2Br - + Cl 2  2 Cl - + Br 2
      • Halogen activity series:
      • F 2 > Cl 2 > Br 2 > I 2
    • A sample problem on identifying the type of redox reaction.
    • All the combustion reactions are redox because elemental oxygen is a reactant.
    • THE END