Page 226 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 14] OXIDATION AND REDUCTION 215

Supplementary Problems

14.26. Explain why we were able to use the charges on the monatomic cations in names in Chap. 6 instead of the required
oxidation numbers.
Ans. For monatomic ions, the charge is equal to the oxidation number.

14.27. Determine the oxidation number of oxygen in (a)Na 2 O 2 ,(b) RbO 2 , and (c)OF 2 .
1
Ans. (a) −1(b) − 1 (+1 + 2x = 0, therefore x =− )(c) +2
2 2
14.28. Since the reactions in Problem 14.23 are the same, and the Daniell cell produces electric energy, what kind of energy
does the direct reaction produce?
Ans. It produces more heat energy than the electrochemical reaction produces.
14.29. Complete and balance the following redox equations:

(a) MnO 4 + Hg 2 2+ −→ Hg 2+ + Mn 2+ (d)P 4 + OH −→ PH 3 + HPO 3 2−
−
−
−
(b)I + H 2 O 2 −→ I 2 + H 2 O (e)H 2 C 2 O 4 + MnO 4 −→ Mn 2+ + CO 2
−
− − −
(c)Br 2 + OH −→ Br + BrO 3 ( f )Cr(OH) 2 + H 2 O 2 −→ Cr(OH) 3
−
+
Ans. (a)16 H + 2 MnO 4 + 5Hg 2 2+ −→ 10 Hg 8 2+ + 2Mn 2+ + 8H 2 O
−
+
(b) 2 H + 2I + H 2 O 2 −→ I 2 + 2H 2 O
−
−
−
(c) 3 Br 2 + 6OH −→ 5Br + BrO 3 + 3H 2 O
−
(d) 2 H 2 O + P 4 + 4OH −→ 2PH 3 + 2 HPO 3 2−
−
(e)6 H + 5H 2 C 2 O 4 + 2 MnO 4 −→ 2Mn 2+ + 10 CO 2 + 8H 2 O
+
( f ) 2 Cr(OH) 2 + H 2 O 2 −→ 2Cr(OH) 3
14.30. Which of the equations in Problem 14.29 represent reactions in basic solution? How can you tell?
Ans. Reactions (c), (d), and ( f ) occur in basic solution. The presence of OH ions shows immediately that the
−
solution is basic. (The presence of NH 3 also indicates basic solution; in acid solution, this base would react
+
to form NH 4 .) The other reactions are not in base; H is present. Also, if they were in base, the metal ions
+
would react to form insoluble hydroxides and the simple ions would not be present.
14.31. Complete and balance the following equations:
+
+
(a)Zn + H + NO 3 −→ NH 4 + Zn 2+ (b)Zn + HNO 3 −→ NH 4 NO 3 + Zn(NO 3 ) 2
−
(c) How are these equations related? Which is easier to balance?
−
Ans. (a) NO 3 −→ NH 4 + Zn −→ Zn 2+ + 2 e −
− − +
8 e + NO 3 −→ NH 4
+ − − +
10 H + 8 e + NO 3 −→ NH 4
−
+
+
10 H + 8 e + NO 3 −→ NH 4 + 3H 2 O 4 Zn −→ 4Zn 2+ + 8 e −
−
+
+
−
10 H + 4Zn + NO 3 −→ NH 4 + 3H 2 O + 4Zn 2+
(b) Either add 9 NO 3 to each side of the equation in part a:
−
10 HNO 3 + 4Zn −→ NH 4 NO 3 + 3H 2 O + 4Zn(NO 3 ) 2
or
2 HNO 3 −→ NH 4 NO 3 Zn −→ Zn(NO 3 ) 2 + 2 e −
− 2 HNO 3 + Zn −→ Zn(NO 3 ) 2 + 2 e −
8 e + 2 HNO 3 −→ NH 4 NO 3
+ − 2 HNO 3 + Zn −→ Zn(NO 3 ) 2 + 2 e + 2H +
−
8H + 8 e + 2 HNO 3 −→ NH 4 NO 3
−
+
8H + 8 e + 2 HNO 3 −→ NH 4 NO 3 + 3H 2 O 8 HNO 3 + 4Zn −→ 4Zn(NO 3 ) 2 + 8 e + 8H +
−
10 HNO 3 + 4Zn −→ 4Zn(NO 3 ) 2 + NH 4 NO 3 + 3H 2 O