Page 218 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 14] OXIDATION AND REDUCTION 207

Balance H 2 O from the number of O atoms, KCl by the number of K atoms, and finally HCl by the number of
Cl or H atoms. Check.
10 HCl + 2K 2 CrO 4 + 3H 2 C 2 O 4 −→ 6CO 2 + 2 CrCl 3 + 4 KCl + 8H 2 O

It is often easier to balance redox equations in net ionic form than in overall form.

EXAMPLE 14.17. Balance:
+ 2− 3+
H + CrO 4 + H 2 C 2 O 4 −→ CO 2 + Cr + H 2 O
Ans. First balance the C atoms. Then balance the elements changing oxidation state. Then balance the rest of the atoms
by inspection.
+ 2− 3+
10 H + 2 CrO 4 + 3H 2 C 2 O 4 −→ 6CO 2 + 2Cr + 8H 2 O
The net charge (6+) and the numbers of atoms of each element are equal on the two sides.

Ion-Electron, Half-Reaction Method
In the ion-electron method of balancing redox equations, an equation for the oxidation half-reaction and
one for the reduction half-reaction are written and balanced separately. Only when each of these is complete and
balanced are the two combined into one complete equation for the reaction as a whole. It is worthwhile to balance
the half-reactions separately since the two half-reactions can be carried out in separate vessels if they are suitably
connected electrically. (See Sec. 14.6.) In general, net ionic equations are used in this process; certainly some
ions are required in each half-reaction. In the equations for the two half-reactions, electrons appear explicitly; in
the equation for the complete reaction—the combination of the two half-reactions—no electrons are included.
During the balancing of redox equations by the ion-electron method, species may be added to one side of
the equation or the other. These species are present in the solution, and their inclusion in the equation indicates
that they also react. Water is a prime example. When it reacts or is produced, the chemist is not likely to
notice that there is less or more water present. In general, all the formulas for the atoms that change oxidation
number will be given to you in equations to balance. Sometimes, the formula of a compound or ion of oxygen
will be omitted, but rarely will the formula for a compound or ion of some other element not be provided. If a
formula is not given, see if you can figure out what that formula might be by considering the possible oxidation
states (Sec. 14.3).

EXAMPLE 14.18. Guess a formula for the oxygen-containing product in the following redox reaction:
−
H 2 O 2 + I −→ I 2 +
Ans. Since the iodine is oxidized (from −1 to 0), the oxygen must be reduced. It starts out in the −1 oxidation state; it
must be reduced to the −2 oxidation state. Water or OH is the probable product.
−
−
H 2 O 2 + 2I −→ I 2 + 2OH −
There are many methods of balancing redox equations by the half-reaction method. One such method is
presented here. You should do steps 1 through 5 for one half-reaction and then those same steps for the other
half-reaction before proceeding to the rest of the steps.
1. Identify the element(s) oxidized and the element(s) reduced. Start a separate half-reaction for each of these.
2. Balance these elements.
3. Balance the change in oxidation number by adding electrons to the side with the higher total of oxidation
numbers. That is, add electrons on the left for a reduction half-reaction and on the right for an oxidation
half-reaction. One way to remember on which side to add the electrons is the following mnemonic:

Loss of Electrons is Oxidation (LEO).
Gain of Electrons is Reduction (GER). “LEO the Lion says ‘GER’.”