Loading....
Coupon Accepted Successfully!

 

Balancing Simple Equations

A chemical equation is said to be balanced if all the atoms present in the reactants appear in the same numbers among the products. Here is an example.

 

Example 2-1

 Balance the following equation.

 
Solution

Start by balancing the oxygen atoms. There are two oxygen atoms on the reactant side and three oxygen atoms on the product side. To balance this, put 3 as the coefficient of oxygen on the reactant side. When we write '3 O2,' that means we have 6 oxygen atoms on the reactant side. To make the same number of oxygen atoms on the product side, let's put 2 as the coefficient of Fe2O3. Now the oxygen atoms seem to be balanced.

Let's take a look at Fe. Since the coefficient of Fe2O3 is 2, we have 4 atoms of Fe on the product side. We can balance this by writing 4 as the coefficient of Fe on the reactant side. So the balanced equation is as follows:

 


 

 

Problem 2-1

Balance the following equations:

 

 
Answers

image\24266.jpg
 

Oxidation Number

Electrons are exchanged during oxidation-reduction reactions. The behavior of atoms or ions in terms of the number of electrons transferred is expressed as the oxidation state (oxidation number). We can define oxidation number as the charge of an atom or ion, based on a set of standard rules. If the given species is an ion containing a single atom, then its oxidation state is its charge itself. Let's analyze this by looking at a few examples.
In NaCl, the oxidation state of sodium is +1 and the oxidation state of chlorine is –1. Generally, the elements at the top right corner of the periodic table are assigned negative oxidation numbers. Some of the elements on the right side of the periodic table can have positive or negative oxidation numbers depending upon the atom to which the given element is bonded to. The elements in the middle and the left portions of the periodic table have almost exclusively positive oxidation numbers.

 

Table 1-1
 
General guidelines for assigning oxidation numbers
  1. The elemental natural state oxidation number of any atom is zero. For example, the oxidation number of oxygen atom in O2 is zero.
  2. The sum of the oxidation numbers of the atoms in a compound should be zero.
  3. The sum of the oxidation numbers of the atoms in an ionic species (a species with a net charge) should equal the net charge of the ionic species.
  4. The oxidation number of a given ion containing a single atom is its charge itself.
 
Oxidation numbers of some common elements
  1. The common oxidation number of Group IA metals is +1. E.g., lithium, sodium, potassium.
  2. The common oxidation number of Group IIA metals is +2. E.g., beryllium, magnesium, calcium.
  3. The common oxidation number of Group IIIA is +3. E.g., aluminum, boron.
  4. The common oxidation number of Group IVA is +4. +2 is also seen in some compounds such as CO.
  5. The common oxidation numbers of Group V A are +5 and –3.
  6. The common oxidation number of Group VIA is –2.
  7. The common oxidation number of Group VIIA is –1.
  8. The common oxidation number of H is +1. In some metal hydrides, hydrogen shows an oxidation number of –1.

 

The above list of common oxidation numbers is not comprehensive. Nevertheless, it gives you a basic and essential picture about assigning oxidation numbers in common compounds and ionic species. Most elements can have multiple oxidation states, depending on the element or ionic species they are bonded to. You have to always follow the general guidelines in Table 1-1, and check whether the items listed are satisfied.
 
Now that we have learned the theory of assigning oxidation numbers, let's do an example to see how it works.
 

Example 2-2

What is the oxidation number of sulfur in sulfuric acid?

 

Solution

Sulfuric acid is H2SO4. The oxidation number of hydrogen is +1. But we have two hydrogens which add up to a charge of +2. Since the total charge of this molecule should be zero, we can say that the charge of sulfate ion is –2. We also know that the oxidation number of oxygen is –2. But there are four oxygens in a sulfate ion. So the charge adds to –8. Now let's solve this algebraically.


ONSulfur - Oxidation number of sulfur

ONOxygen - Oxidation number of oxygen

 

ONSulfur + 4 (ONOxygen) = –2

ONSulfur + 4 (–2) = –2

ONSulfur = +6

 

So the oxidation number of sulfur in sulfuric acid is +6.
 

 

Problem 2-2
Calculate the oxidation state of the element indicated in each of the following problems.
  1. What is the oxidation state of hydrogen in MgH2?
  2. What is the oxidation state of chlorine in ClO3– ?
  3. What is the oxidation state of oxygen in Na2O2?
  4. What is the oxidation state of nitrogen in NH3?
  5. What is the oxidation state of oxygen in O2?
  6. What is the oxidation state of bromine in HBrO2?
  7. What is the oxidation state of manganese in KMnO4?
Solution
  1. –1
  2. +5
  3. –1
  4. –3
  5. 0
  6. +3
  7. +7




Test Your Skills Now!
Take a Quiz now
Reviewer Name