Page 228 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 14] OXIDATION AND REDUCTION 217

(+2 →+4) =+2
| |
(g) HNO 3 + Pb(NO 3 ) 2 −→ Pb(NO 3 ) 4 + NO
| |
(+5 →+2) =−3
6 HNO 3 + 2 HNO 3 + 3Pb(NO 3 ) 2 −→ 3Pb(NO 3 ) 4 + 2NO + 4H 2 O
(for extra
nitrate ions)
8 HNO 3 + 3Pb(NO 3 ) 2 −→ 3Pb(NO 3 ) 4 + 2NO + 4H 2 O

(h)8 H + 2NO 3 + 3Sn 2+ −→ 2NO + 3Sn 4+ + 4H 2 O
+
−
2− + 2I + 4H −→ SO 2 + I 2 + 2H 2 O
+
−
(i)SO 4
( j)10 Cr 3+ + 6 MnO 4 + 11 H 2 O −→ 5Cr 2 O 7 2− + 6Mn 2+ + 22 H +
−
+ 2− 2− + 2Ag
(k)2 Ag + 2S 2 O 3 −→ S 4 O 6
14.34. Consider the following part of an equation:
NO 2 + MnO 4 −→ Mn 2+ +
−
(a) If one half-reaction is a reduction, what must the other half-reaction be? (b) To what oxidation state can the
nitrogen be changed? (c) Complete and balance the equation.
Ans. (a) Since one half-reaction is a reduction, the other half-reacttion must be an oxidation.
(b) The maximum oxidation state for nitrogen is +5, because nitrogen is in periodic group V. Since it starts
out in oxidation number +4, it must be oxidized to +5.
−
(c) NO 2 + MnO 4 −→ Mn 2+ + NO 3 −
−
5NO 2 + MnO 4 −→ Mn 2+ + 5NO 3 −
−
H 2 O + 5NO 2 + MnO 4 −→ Mn 2+ + 5NO 3 + 2H +
−
2−
14.35. What is the oxidation number of sulfur in S 2 O 8 , the peroxydisulfate ion?
Ans. If you calculate the oxidation number assuming that the oxygen atoms are normal oxide ions, you get an
answer of +7, which is greater than the maximum oxidation number for sulfur. That must mean that one of
the pairs of oxygen atoms is a peroxide, and thus the sulfur must be in its highest oxidation number, +6.
O O 2−
O S O O S O
O O

14.36. Which of the following reactions (indicated by unbalanced equations) occur in acid solution and which occur in
basic solution?
(a) HNO 3 + Cu −→ NO + Cu 2+
2−
(b) CrO 4 + Fe(OH) 2 −→ Cr(OH) 3 + Fe(OH) 3
(c) N 2 H 4 + I −→ NH 4 + I 2
+
−
Ans. (a) Acid solution (HNO 3 would not be present in base.) (b) Basic solution (The hydroxides would not be
+
present in acid.) (c) Acid solution (NH 3 rather than NH 4 would be present in base.)
14.37. Complete and balance the following equations:

−
(a) Cl + MnO 2 + H −→ Mn 2+ + H 2 O + Cl 2 (e) V 2+ + H 3 AsO 4 −→ HAsO 2 + VO 2+
+
−
−
−
(b) ClO −→ Cl + ClO 3 − ( f ) Hg 2 2+ + CN −→ C 2 N 2 + Hg
2− 2+ 3− − +
(c)Pb + PbO 2 + SO 4 −→ PbSO 4 + H 2 O (g)VO + AsO 4 −→ AsO 2 + VO 2
(d) I 3 + Co(CN) 6 4− −→ Co(CN) 6 3− + I −
−
−
Ans. (a)2 Cl + MnO 2 + 4H −→ Mn 2+ + 2H 2 O + Cl 2
+
−
−
(b) 3 ClO −→ 2Cl + ClO 3 −