Chemistry 103A; Sections 5, 6, 7, 8; Lecture 10, 13 Sep 00

Redox reactions, continued
 
 

Here are some examples which we worked or will work out in class:

HNO3 (N has oxidation number +5)

SO42-(S has oxidation number +6)

PO43-(P has oxidation number +5)

Cr2O72-

B2O3.


Now that we know about oxidation numbers we can define what we mean by oxidation and reduction.

Oxidation is an increase in oxidation number.

Oxidation is a loss of electrons from the atom oxidized.

Other tell-tale signs,

Gain of one or more O atoms

Loss of one or more H atoms
 

Reduction is a decrease in oxidation number.

Reduction always is a gain of electrons to the atom reduced.

Other tell-tale signs,

Loss of one or more O atoms.

Gain of one or more H atoms.


Balancing Redox Equations
 

We have already balanced a fair number of chemical reaction equation and some of these were (without telling you) redox reactions. For example, in the last homework set you balanced the following redox reactions:

2 Cr + 3 Cl2 ® 2 CrCl3

SiO2 + 2 C ® Si + 2 CO

3 Fe + 4 H2O ® Fe3O4 + 4H2

2 NaBH4 + H2SO4 ® B2H6 + 2 H2 + Na2SO4


Let's look at what is being oxidized and what is being reduced in each reaction. Let us also note the oxidation numbers involved.
 

Notice that in each reaction there is something that is oxidized and something else that is reduced.
 

That is our first general rule about redox reactions: If something is oxidized something else has to be reduced.

That is easy to understand. If oxidation is a loss of electrons, the electrons don't just disappear. Something else has to gain electrons.


These reactions were fairly easy to balance, but not all redox reactions are that easy to balance. I am going to show you a general method for balancing redox reactions that works for even the most complicated redox reactions.
 

This method of balancing redox reactions is called the half-reaction method, or sometimes the H2O, H+, e- , method. Here's how it works:

Identify the element that is being oxidized and the element that is being reduced. (There should be no more than one of each.)

Write a "half reaction" which contains only the compounds containing the element which is oxidized and another half reaction that contains only the element being reduced.

Balance each half reaction separately by balancing everything but O, H, and e- .

Balance O in each half reaction by adding H2O to the appropriate side of each half reaction. (Don't worry, if there isn't any water in the reaction the H2O's will cancel out.)

Balance H by adding H+ ions to the appropriate side. (Don't worry, if there is no H+ in the reaction they will cancel out.)

Balance the charge in each half reaction by adding electrons (e- ) as needed.

Now add these two half reactions in such a way that the electrons cancel out. That means that we may have to multiply one or both half reactions by a small whole number to make the number of electrons on each side cancel.

 
(If you find that there are some H+'s left in the reaction then you have balanced the reaction as if it were in an acid solution. If you know that the reaction occurred in a basic solution then add enough OH- ions to both sides of the balanced reactions so that the H+ ions can all be converted to H2O.) Here are some examples:

Let's start out easy. The following reaction was on your homework, but I have removed the spectator ions:

NaBH4(s) + H+(aq) ® B2H6(g) + H2(g) + Na+(aq). Here's one that's a little harder. You will learn, over the course of your chemistry career, that the permanganate ion, MnO4-, is a good oxidizing agent. That is, it is able to oxidize things - lots of things. In acid solution the manganese is reduced to Mn2+ ions. Here is a reaction that is often used in the quantitative analysis for calcium.

CaC2O4(s) + MnO4-(aq) + H+(aq) ® Mn2+(aq) + CO2 + H2O + Ca2+
 

In addition, you will learn that in basic solution manganese is reduced to MnO2. Let's try a reaction in basic solution.

MnO4- + S2-® MnO2 + SO42-
(We may end up with some H2O and OH- , but no H+.)