5 Steps to a 5 AP Chemistry

6. C—The balanced equation is:


7. C—It takes 4 mol of electrons (4 F) to change the
   platinum ions to platinum metal. The calculation
   would be: (0.80 F)(1 mol Pt/4 F) =0.20 mol Pt


8. D—For a substance to serve as a reducing agent,
   it must be capable of being oxidized. The man-
   ganese, in the MnO 4 −, is already in its highest
   oxidation state, so it could not be oxidized. All
   other answers contain a substance that may be
   oxidized.


9. D—The dichromate ion oxidizes the sulfide ion
   to elemental sulfur, as the sulfide ion reduces the
   dichromate ion to the chromium(III) ion.
   Chromium goes from +6 to +3,while sulfur goes
   from −2 to 0. The hydrogen remains at +l, so it
   is neither oxidized nor reduced.


10. C—The balanced chemical equation is:


11. C—Using the equation:

You should realize that log K = 4 gives a

K= 104. This will give a Kof about 10^4 (actually,

K=1.1 × 104 ).

12. A—The Sn^2 +, from SnCl 2 , reduces the man-
    ganese from +7 to +4. This makes SnCl 2 a reduc-
    ing agent. The tin is oxidized to Sn^4 +, so KMnO 4
    is an oxidizing agent. NaF did nothing, so it
    behaves as neither an oxidizing nor as a reducing
    agent.


13. C—The impure silver must be oxidized so it will
    go into solution. Oxidation occurs at the anode.
    Reduction is required to convert the silver ions to
    pure silver. Reduction occurs at the cathode. The
    cathode must be pure silver, otherwise it could be
    contaminated with the cathode material.


14. D—The MnO 4 −oxidizes the sulfide ion to ele-
    mental sulfur, while the sulfide ion reduces the
    permanganate ion to the manganese(II) ion.
    15. B—If the voltage was not equal to E°, then the
    cell was not standard. Standard cells have 1 M
    concentrations, and operate at 25°C with a partial
    pressure of each gas equal to 1 atm. No gases are
    involved in this reaction, so the cell must be oper-
    ating at a different temperature or a different
    concentration (or both).

16. A—

You can estimate the answer by replacing 96,500

with 100,000 and 200.6 with 200.

17. A—The gases produced are hydrogen (at the
    cathode) and chlorine (at the anode). Hydrogen
    is odorless, while chlorine has a distinctive odor.


18. C—The KIO 4 oxidizes the H 2 O 2. Thus, H 2 O 2 is
    the reducing agent. The iodine is reduced from
    +7 to +5, while the oxygen in the H 2 O 2 is oxi-
    dized from −1 to 0 (O 2 ).


19. E—The bromate ion, BrO 3 −, is gaining electrons,
    so it is being reduced. Reduction always occurs at
    the cathode.


20. B—Recall that 5.0 amp is 5.0 C/s. The calculation
    would be:


21. E—The half-reactions giving the overall reaction
    must be:

Thus, -0.76 +? =0.90, giving? =1.66 V. The

half-reaction under consideration is the reverse of

the one used in this combination, so the sign of

the calculated voltage must be reversed. Do not

make the mistake of multiplying the voltages

when the half-reactions were multiplied to equal-

ize the electrons.

32 [()^2 +−+→()]Zn aq e Zn s V

Ms M e

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076

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.

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23n() n^232 +++()aq→+ °= 23 M ()aq Zn ()aq E 090 .V

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432 HNO e NO HO+−−++→+ 32

SO 232 −−−+→++ 10 OH 2 SO^245 HO e 2 8 −

254  Step 4. Review the Knowledge You Need to Score High