Page 219 - Theory and Problems of BEGINNING CHEMISTRY

208 OXIDATION AND REDUCTION [CHAP. 14

4. In acid solution, balance the net charge with hydrogen ions, H .
+
5. Balance the hydrogen and oxygen atoms with water.
6. Multiply every item in one or both equations by small integers, if necessary, so that the number of electrons
is the same in each. The same small integer is used throughout each half-reaction, and is different from that
used in the other half-reaction. Then add the two half-reactions.
7. Cancel all species that appear on both sides of the equation. All the electrons must cancel out in this step,
and often some hydrogen ions and water molecules also cancel.
8. Check to see that atoms of all the elements are balanced and that the net charge is the same on both sides of
the equation.
Note that all the added atoms in steps 4 and 5 have the same oxidation number as the atoms already in the
equation. The atoms changing oxidation number have already been balanced in steps 1 and 2.

EXAMPLE 14.19. Balance the following equation by the ion-electron, half-reaction method:
−
+
Cr 2+ + H + NO 3 −→ Cr 3+ + NO + H 2 O
Ans. Step 1: Cr 2+ −→ Cr 3+ NO 3 −→ NO
−
Step 2: Cr already balanced. N already balanced.
Step 3: Cr 2+ −→ Cr 3+ + e − 3 e + NO 3 −→ NO
−
−
−
+
−
Step 4: Cr 2+ −→ Cr 3+ + e − 4H + 3 e + NO 3 −→ NO
+
−
−
Step 5: Cr 2+ −→ Cr 3+ + e − 4H + 3 e + NO 3 −→ NO + 2H 2 O
Step 6: Multiplying by 3 :
3Cr 2+ −→ 3Cr 3+ + 3 e −
Adding:
−
3Cr 2+ + 4H + 3 e + NO 3 −→ NO + 2H 2 O + 3Cr 3+ + 3 e −
−
+
Step 7: 3 Cr 2+ + 4H + NO 3 −→ NO + 2H 2 O + 3Cr 3+
−
+
Step 8: The equation is balanced. There are three Cr atoms, four H atoms, one N atom, and three O atoms on each
side. The net charge on each side is 9+.
EXAMPLE 14.20. Complete and balance the following equation in acid solution:
−
2− + Cl 2 −→ ClO 4 + Cr 3+
Cr 2 O 7
Ans. Step 1: Cr 2 O 7 2− −→ Cr 3+ Cl 2 −→ ClO 4 −
Step 2: Cr 2 O 7 2− −→ 2Cr 3+ Cl 2 −→ 2 ClO 4 −
Step 3: 2 atoms are reduced 3 units each 2 atoms are oxidized 7 units each
−
− 2− −→ 2Cr 3+ Cl 2 −→ 2 ClO 4 + 14 e −
6 e + Cr 2 O 7
−
−
+
Step 4: 14 H + 6 e + Cr 2 O 7 2− −→ 2Cr 3+ Cl 2 −→ 2 ClO 4 + 14 e + 16 H +
−
−
−
−
Step 5: 14 H + 6 e + Cr 2 O 7 2− −→ 2Cr 3+ + 7H 2 O 8 H 2 O + Cl 2 −→ 2 ClO 4 + 14 e + 16 H +
+
+ − 2− −→ 2Cr 3+ +7H 2 O] 3 [8 H 2 O+Cl 2 −→ 2 ClO 4 +14 e +16 H ]
+
−
−
Step 6: 7 [14 H + 6 e +Cr 2 O 7
3+
−
−
+ − 2− −→ 14Cr +49H 2 O 24H 2 O+3Cl 2 −→ 6ClO 4 +42e + 48H +
98H +42e +7Cr 2 O 7
−
−
+ − 2− −→14 Cr 3+ + 49 H 2 O + 6 ClO 4 + 42 e + 48 H +
Step 7: 24 H 2 O + 3Cl 2 + 98 H + 42 e + 7Cr 2 O 7
+ 2− −→14 Cr 3+ −
3Cl 2 + 50 H + 7Cr 2 O 7 + 25 H 2 O + 6 ClO 4
Step 8: The equation is balanced. There are 6 Cl atoms, 50 H atoms, 14 Cr atoms, and 49 O atoms on each side,
as well as a net 36+ charge on each side.
If a reaction is carried out in basic solution, the same process may be followed. After all the other steps have
+
−
been completed, any H present can be neutralized by adding OH ions to each side, creating water and excess
OH ions. The water created may be combined with any water already on that side or may cancel any water on
−
the other side.