Page 224 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 14] OXIDATION AND REDUCTION 213

14.10. What oxyacid of nitrogen can be prepared by adding water to N 2 O 5 ? Hint: Both compounds have nitrogen
in the same oxidation state.
Ans. HNO 3

H 2 O + N 2 O 5 −→ 2 HNO 3

PERIODIC RELATIONSHIPS OF OXIDATION NUMBERS
14.11. Predict the formulas of two compounds of each of the following pairs of elements: (a) S and O, (b)Cl
and O, (c) P and S, (d) P and F, (e) I and F, and ( f ) S and F.
Ans. The first element in each part is given in its highest oxidation state and in an oxidation state 2 less than the
highest. The second element is in its minimum oxidation state. (a)SO 3 and SO 2 (b)Cl 2 O 7 and Cl 2 O 5
(c)P 2 S 5 and P 2 S 3 (d)PF 5 and PF 3 (e)IF 7 and IF 5 ( f )SF 6 and SF 4
14.12. Predict the formulas of four fluorides of iodine.

Ans. IF 7 ,IF 5 ,IF 3 and IF. The oxidation states of iodine in these compounds correspond to the maximum oxidation
state for a group VII element and to states 2, 4, and 6 lower. (See Fig. 14-1.)

14.13. Write the formulas for two monatomic ions for each of the following metals: (a) Co, (b) Tl, (c) Sn, and
(d) Cu.

Ans. (a)Co 3+ and Co 2+ (the oxidation states of transition metals very in steps of one.) (b)Tl 3+ and Tl (the
+
maximum oxidation state of a group III element and the state 2 less than the maximum.) (c)Sn 4+ and Sn 2+
(the maximum oxidation state of a group IV element and the state 2 less than the maximum.) (d)Cu and
+
Cu 2+ (the maximum oxidation state for the coinage metals is greater than the group number.)

OXIDATION NUMBERS IN INORGANIC NOMENCLATURE
14.14. Name NO 2 and N 2 O 4 , using the Stock system. Explain why the older system using prefixes is still useful.
Ans. Both compounds have nitrogen in the +4 oxidation state, so if we call NO 2 nitrogen(IV) oxide, what do we
call N 2 O 4 ? We actually use the older system for N 2 O 4 —dinitrogen tetroxide.

BALANCING OXIDATION-REDUCTION EQUATIONS
14.15. Why is it possible to add H and/or H 2 O to an equation for a reaction carried out in aqueous acid solution
+
when none visibly appears or disappears?
Ans. The H 2 O and H are present in excess in the solution. Therefore, they can react or be produced without the
+
change being noticed.
14.16. How many electrons are involved in a reaction of one atom with a change of oxidation number from
(a) +2to −3 and (b) +5to −2?
Ans. (a)2 − (−3) = 5 5 electrons are involved (b)5 − (−2) = 7 7 electrons are involved

14.17. Identify (a) the oxidizing agent, (b) the reducing agent, (c) the element oxidized, and (d) the element
reduced in the following reaction:
+ 2− 2+ 3+ 3+
8H + CrO 4 + 3Co −→ Cr + 3Co + 4H 2 O
Ans. (a) CrO 4 2− (b)Co 2+ (c)Co (d)Cr.
An element in the reducing agent is oxidized; an element in the oxidizing agent is reduced.

14.18. Balance the equation for the reduction of HNO 3 to NH 4 NO 3 by Mn by the oxidation number change
method. Add other compounds as needed.