Page 227 - Theory and Problems of BEGINNING CHEMISTRY

216 OXIDATION AND REDUCTION [CHAP. 14

(c) Part a is the net ionic equation for part b. It is easier to balance part a. To balance part b, either amend the
balanced net ionic equation of part a, or do part b, starting from scratch. (There must always be at least
one type of ion represented in a balanced half-reaction equation.)

14.32. Balance the following equation by the oxidation number change method:

NH 3 + O 2 −→ NO + H 2 O
Ans. The O 2 is reduced to the −2 oxidation state; both products contain the reduction product.

(−3 →+2) =+5
| |
NH 3 + O 2 −→ NO + H 2 O
| |
2(0 →−2) =−4

We know that five O 2 molecules are required, but we still do not know from this calculation how many of
the oxygen atoms go to NO or H 2 O.

4NH 3 + 5O 2 −→ 4NO + H 2 O
Finish by inspection:
4NH 3 + 5O 2 −→ 4NO + 6H 2 O

14.33. Complete and balance the following equations:
(a) H 2 O 2 + Sn 2+ −→ Sn 4+ + H 2 O (g) HNO 3 + Pb(NO 3 ) 2 −→ Pb(NO 3 ) 4 + NO
−
+
(b) Br + BrO 3 −→ Br 2 + H 2 O (h)H + NO 3 + Sn 2+ −→ NO + Sn 4+
−
−
+
−
(c) Ce 4+ + SO 3 2− −→ Ce 3+ + SO 4 2− (i) SO 4 2− + I + H −→ SO 2 + I 2 + H 2 O
−
(d) Zn + OH −→ Zn(OH) 4 2− + H 2 ( j) Cr 3+ + MnO 4 −→ Cr 2 O 7 2− + Mn 2+
−
+
+
(e) Cu 2 O + H −→ Cu + Cu 2+ + H 2 O (k) Ag + S 2 O 3 2− −→ S 4 O 6 2− + Ag
( f )H 2 SO 4 (conc) + Zn −→ H 2 S + Zn 2+
+2
| |
H 2 O 2 + Sn 2+ −→ Sn 4+ + 2H 2 O
Ans. (a) | |
2(−1 →−2) =−2
+
2H + H 2 O 2 + Sn 2+ −→ Sn 4+ + 2H 2 O
+
−
−
(b)6 H + 5Br + BrO 3 −→ 3Br 2 + 3H 2 O
−
(c) e + Ce 4+ −→ Ce 3+
−
2− 2− + 2 e + 2H +
H 2 O + SO 3 −→ SO 4
2Ce 4+ + H 2 O + SO 3 2− −→ SO 4 2− + 2H + 2Ce 3+
+
− 2− −
(d) 4OH + Zn −→ Zn(OH) 4 + 2 e
− −
2 e + 2H 2 O −→ 2OH + H 2
− 2−
2H 2 O + 2OH + Zn −→ Zn(OH) 4 + H 2
(e)Cu 2 O + 2H −→ Cu + Cu 2+ + H 2 O
+
Cu is not stable in aqueous solution. Cu(I) is stable in solid compounds like Cu 2 O, but when that reacts
+
with an acid, the Cu disproportionates—reacts with itself—to produce a lower and a higher oxidation
+
state.
( f ) 8 e + 8H + H 2 SO 4 −→ H 2 S + 4H 2 O
+
−
Zn −→ Zn 2+ + 2 e −
4Zn + 8H + H 2 SO 4 −→ H 2 S + 4H 2 O + 4Zn 2+
+