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Redox Reactions

Redox Reactions. DP Chemistry II External Assessment. Redox Reactions/Half Reactions. Oxidation can’t take place without Reduction Separating the 2 reactions can help determine what is being : Oxidized Reduced Oxidizing reagent Reducing reagent Half Reactions help do this:.

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Redox Reactions

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  1. Redox Reactions DP Chemistry II External Assessment

  2. Redox Reactions/Half Reactions • Oxidation can’t take place without Reduction • Separating the 2 reactions can help determine what is being : • Oxidized • Reduced • Oxidizing reagent • Reducing reagent • Half Reactions help do this:

  3. Half Reactions • Examples of Half reactions • 2 H+ + 2 e- ® H2MnO4- + 5 e- + 8 H+ ® Mn2+ + 4H2OZn ® Zn2+ + 2 e-Cu ® Cu2+ + 2 e- • A half reaction does not occur by itself, at least two such reactions must be coupled so that the electron released by one reactant is accepted by another in order to complete the reaction. • Thus, oxidation and reduction reactions must take place simultaneously in a system, and this type of reactions is called oxidation reduction reaction or simply redox reaction. For example,

  4. Half reactions • Zn + Cu2+ ® Zn2+ + Cu is such a redox reaction, • Zn being oxidized, and Cu2+ being reduced. • Redoxreactions take place in battery operations. • Half-reaction equations are useful for balancing redox reaction equations. • http://www.science.uwaterloo.ca/~cchieh/cact/c123/battery.html

  5. Half Reactions • Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) • This reaction may be written as two half-reactions and adding the two half reactions gives the overall equation representing a chemical process: • Zn Zn2+(aq) + 2 e-+) • Cu2+(aq) + 2 e-Cu(s) • ----------------------------------------- • Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)

  6. Balancing Redox reactions • A redox reaction may be balanced by first writing two half-reactions, and then canceling the electrons by adding them algebraically. • Identify the key element that undergoes an oxidation state change. • Balance the number of atoms of the key element on both sides. • Add the appropriate number of electrons to compensate for the change of oxidation state. • Add H+ (in acid medium), or OH- (in basic medium), to balance the charge on both sides of the half-reactions; and H2O, if necessary, to balance the equations.

  7. Balancing Redox reaction • H2O2 + I- -> I2 + H2O • Balance the two half reactions for the reaction in an acid solution • I- is oxidized (oxidation state increases from -1 to 0).O (oxygen) is reduced (oxidation state decreases from -1 to -2). • The two half-reactions with balanced number of key atoms are:2 I- ® I2; <--- (oxidation)H2O2 ® 2 H2O <--- (reduction of oxygen) • Add electrons to compensate for the changes of oxidation state 2 I- ® I2 + 2 e- (oxidation)H2O2 + 2 e- ® 2 H2O (reduction) • Obviously, H+ should be added to the reduction half-reaction, and the balanced equations are: 2 I- ® I2; <--- (oxidation)H2O2 + 2H+ ® 2 H2O (reduction)

  8. Balancing Redox reactions • ClO2 + OH-ClO2- + ClO3- • In this reaction, Cl from ClO2 is both oxidized and reduced. • The two half-reactions are: ClO2 ® ClO2-; (reduction)ClO2 ® ClO3-; (oxidation) • Add electrons to compensate for the oxidation changes: ClO2 + e- ® ClO2-; (reduced, 4 -> 3 for Cl)ClO2 ® ClO3- + e-; (oxidized, 4 -> 5) • Add H+, OH-, or H2O to balance both equations results in ClO2 + e- ® ClO2-ClO2 + 2 OH- ® ClO3- + e- + H2ONow add the two half reactions together to give the overall reaction: 2 ClO2 + 2 OH- ® ClO2- + ClO3- + H2O

  9. S(=S)O32- + I2I- + S4O62- • The S atoms are oxidized. For convenience, the best way is to assume the average oxidation state for both S atoms. S oxidized (oxidation state (+2) -> (+10/4 or +2.5)I reduced ( 0 -> -1) • Write the half-reactions and balance the key elements: 2 S2O32- ® S4O62-I2 ® 2 I- • Add electron to compensate the oxidation state changes: 2 S2O32- ® S4O62- + 2 e-I2 + 2 e- ® 2 I- • Since the half-reactions are balanced with respect to charges and number of atoms, no further work is required. Just add the two equations and get a balanced equation. 2 S2O32- + I2 ® 2 I- + S4O62-

  10. Reactivity • More reactive metals are stronger reducing agents • Not all oxidizing and reducing agents are of equal strength. • Some are stronger than others by loosing or gaining electrons • Metals tend to lose electronsand form positive ions, so they act as reducing agents. (stop and think about this) • More reactive metals lose their electrons readily and the are considered strong reducing agents • We can measure this by seeing if one metal is able to reduce the ions of another metal in solution.

  11. Review voltaic cells

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