CK-12-Chemistry Intermediate

(Marvins-Underground-K-12) #1

http://www.ck12.org Chapter 22. Oxidation-Reduction Reactions


one type of solution or the other. The oxidation of Fe^2 +by Cr 2 O 72 −does not readily occur in basic solution, but it
was balanced that way to demonstrate the method.


In summary, the choice of which balancing method to use depends on the kind of reaction. The oxidation-number
method works best if the oxidized and reduced species appear only once on each side of the equation and if no
acids or bases are present. The half-reaction method is more versatile and works well for reactions involving ions in
aqueous solution.


Lesson Summary



  • In the oxidation-number-change method for balancing a redox reaction, the increase in oxidation number by
    one atom is compared to the decrease in oxidation number of another atom. The total number of electrons lost
    in the oxidation is then made equal to the total number of electrons gained in the reduction.

  • To balance a redox reaction by the half-reaction method, write separate half-reactions for the oxidation and
    reduction processes. Balance by atoms, adding in water and hydrogen ions if necessary. Put in electrons to
    balance by charge, equalize the electrons in the two reactions, and add them back together. For basic solutions,
    add hydroxide ions.


Lesson Review Questions


Reviewing Concepts



  1. When balancing an equation by the oxidation-number-change method, what must be true about the total
    change in oxidation number of the oxidized and reduced species?

  2. When balancing an equation by the half-reaction method, what must be true about the number of electrons
    lost in the oxidation compared to the number of electrons gained in the reduction?

  3. Which method works well for molecular substances?

  4. Which method works well for reactions between ions in aqueous solution?


Problems



  1. Balance the reactions below using the oxidation-number-change method.
    a. WO 3 (s) + H 2 (g)→W(s) + H 2 O(g)
    b. SbCl 5 (aq) + KI(aq)→SbCl 3 (aq) + KCl(aq) +I 2 (s)
    c. Bi 2 S 3 (s) + HNO 3 (aq)→Bi(NO 3 ) 3 (aq) + NO(g) + S(s) + H 2 O(l)
    d. Cl 2 (g) + KOH(aq)→KClO 3 (aq) + KCl(aq) + H 2 O(l)

  2. Balance the reactions below using the half-reaction method. All are in acidic solution.
    a. MnO 4 −(aq) + Cl−(aq)→Mn^2 +(aq) + Cl 2 (g)
    b. H 2 O 2 (l) + Fe^2 +(aq)→Fe^3 +(aq) + H 2 O(l)
    c. Cr 2 O 72 −(aq) + C 2 O 42 −(aq)→Cr^3 +(aq) + CO 2 (g)

  3. Balance the reactions below using the half-reaction method. All are in basic solution.
    a. CN−(aq) + MnO 4 −(aq)→CNO−(aq) + MnO 2 (s)
    b. Bi(OH) 3 (s) + SnO 22 −(aq)→SnO 32 −(aq) + Bi(s)
    c. Br 2 (l)→BrO 3 −(aq) + Br−(aq)

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