Page 217 - Theory and Problems of BEGINNING CHEMISTRY

206 OXIDATION AND REDUCTION [CHAP. 14

One of the most important uses of oxidation numbers is in balancing redox (oxidation-reduction) equations.
These equations can get very complicated, and a systematic method of balancing them is essential. There are
many such methods, however, and each textbook seems to use its own. There are many similarities among the
methods, and the following discussion will help no matter what method your instructor and your textbook use.
It is of critical importance in this section to keep in mind the difference between oxidation number and
charge. You balance charge one way and changes in oxidation number another way.
Therearetwoessentiallydifferentmethodstobalanceredoxreactions—theoxidationnumberchangemethod
and the ion-electron method. The first is perhaps easier, and the second is somewhat more useful, especially for
electrochemical reactions (Sec. 14.6).

Oxidation Number Change
The total of the oxidation numbers gained in a reaction must equal the total of the oxidation numbers lost,
since the numbers of electrons “gained” and “lost” must be equal. Therefore, we can balance the species in which
the elements that are oxidized and reduced appear by the changes in oxidation numbers. We use the numbers
of atoms of each of these elements that will give us equal numbers of electrons gained and lost. If necessary,
first balance the number of atoms of the element oxidized and/or the number of atoms of the element reduced.
Finally, we balance the rest of the species by inspection, as we did in Chap. 8.

EXAMPLE 14.15. Balance the following equation:
? HCl + ? HNO 3 + ? CrCl 2 −→? CrCl 3 + ?NO + ?H 2 O

Ans. By inspecting the oxidation states of all the elements, we find that the Cr goes from +2to +3 and the N goes from
+5to +2. They are the only elements undergoing change in oxidation number.
(+2 −→ +3) = 1
| |
? HCl + ? HNO 3 + ? CrCl 2 −→? CrCl 3 + ?NO + ?H 2 O
| |
(+5 −→ +2) =−3
To balance the oxidation numbers lost and gained, we need three CrCl 2 and three CrCl 3 for each N atom reduced:
? HCl + 1 HNO 3 + 3 CrCl 2 −→ 3 CrCl 3 + 1NO + ?H 2 O
It is now easy to balance the HCl by balancing the Cl atoms and to balance the H 2 O by balancing the O atoms:
3 HCl + 1 HNO 3 + 3 CrCl 2 −→ 3 CrCl 3 + 1NO + 2H 2 O

or 3 HCl + HNO 3 + 3 CrCl 2 −→ 3 CrCl 3 + NO + 2H 2 O
We check to see that there are now four H atoms on each side, and the equation is balanced.

EXAMPLE 14.16. Balance the following equation:
HCl + K 2 CrO 4 + H 2 C 2 O 4 −→ CO 2 + CrCl 3 + KCl + H 2 O
Ans. Here, Cr is reduced and C is oxidized. Before attempting to balance the oxidization numbers gained and lost, we
have to balance the number of carbon atoms. Before the carbon atoms are balanced, we don’t know whether to work
with one carbon (as in the CO 2 ) or two (as in the H 2 C 2 O 4 ).
HCl + K 2 CrO 4 + H 2 C 2 O 4 −→ 2CO 2 + CrCl 3 + KCl + H 2 O
Now proceed as before:
(+6 −→ +3) =−3
| |
HCl + K 2 CrO 4 + H 2 C 2 O 4 −→ 2CO 2 + CrCl 3 + KCl + H 2 O
| |
2(+3 −→ +4) =+2
HCl + 2K 2 CrO 4 + 3H 2 C 2 O 4 −→ 6CO 2 + 2 CrCl 3 + KCl + H 2 O