Chemical Equations, Oxidation States
and Balancing of Equations

The common chemical equation could be A + B --> C + D. This is chemical A + chemical B, the two reacting chemicals will go to products C + D etc.

Oxidation

The term “oxidation” originally meant a reaction in which oxygen combines chemically with another substance, but its usage has long been broadened to include any reaction in which electrons are transferred.

Oxidation and reduction always occur simultaneously (redox reactions), and the substance which gains electrons is termed the oxidizing agent. For example, cupric ion is the oxidizing agent in the reaction: Fe (metal) + Cu++ --> Fe++ + Cu (metal); here, two electrons (negative charges) are transferred from the iron atom to the copper atom; thus the iron becomes positively charged (is oxidised) by loss of two electrons while the copper receives the two electrons and becomes neutral (is reduced).

Electrons may also be displaced within the molecule without being completely transferred away from it. Such partial loss of electrons likewise constitutes oxidation in its broader sense and leads to the application of the term to a large number of processes which at first sight might not be considered to be oxidation’s. Reaction of a hydrocarbon with a halogen, for example, CH₄ + 2 Cl --> CH₃Cl + HCl, involves partial oxidation of the methane; halogen addition to a double bond is regarded as an oxidation.

Dehydrogenation is also a form of oxidation, when two hydrogen atoms, each having one electron, a removed from a hydrogen-containing organic compound by a catalytic reaction with air or oxygen, as in oxidation of alcohol’s to aldehyde’s.

Oxidation Number

The number of electrons that must be added to or subtracted from an atom in a combined state to convert it to the elemental form; i.e., in barium chloride ( BaCl₂) the oxidation number of barium is +2 and of chlorine is -1. Many elements can exist in more than one oxidation state.

Now, let us look at some common ions. An ion is the reactive state of the chemical, and is dependant on it’s place within the periodic table.

Have a look at the “periodic table of the elements”. It is arranged in columns of elements, there are 18 columns. You can see column one, H, Li, Na, K etc. These all become ions as H⁺, Lⁱ⁺, K⁺, etc. The next column, column 2, Be, Mg, Ca etc. become ions Be²⁺, Mg²⁺, Ca²⁺, etc. Column 18, He, Ne, Ar, Kr are inert gases. Column 17, F, Cl, Br, I, ionise to a negative F^-, Cl^-, Br^-, I^-, etc. What you need to memorise is the table of common ions, both positive ions and negative ions.

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Table of Common Ions

Positive Ions

Negative Ions

Positive ions will react with negative ions, and vice versa. This is the start of our chemical reactions. For example:

Na+ + OH- --> NaOH (sodium hydroxide)

Na+ + Cl- --> NaCl (salt)

3H⁺ + PO₄^3- --> H₃PO₄ (phosphoric acid)

2Na⁺ + S₂O₃^2- --> Na₂S₂O₃

You will see from these examples, that if an ion of one (+), reacts with an ion of one (-) then the equation is balanced. However, an ion like PO₄^3- (phosphate will require an ion of 3+ or an ion of one (+) (but needs three of these) to neutralise the 3- charge on the phosphate. So, what you are doing is balancing the charges (+) or (-) to make them zero, or cancel each other out.

For example, aluminium exists in its ionic state as Al³⁺, it will react with many negatively charged ions, examples: Cl^-, OH^-, SO₄^2-, PO₄^3-.

Let us do these examples, and balance them.

Al³⁺ + Cl^- --> AlCl (incorrect)

Al³⁺ + 3Cl^- --> AlCl₃ (correct)

How did we work this out?

Al³⁺ has three positives (3+)

Cl- has one negative (-)

It will require 3 negative charges to cancel out the 3 positive charges on the aluminium ( Al³⁺).

When the left hand side of the equation is written, to balance the number of chlorine’s (Cl-) required, the number 3 is placed in front of the ion concerned, in this case Cl-, becomes 3Cl^-.

On the right hand side of the equation, where the ions have become a compound (a chemical compound), the number is transferred to after the relevant ion, Cl₃.

Another example:

Al³⁺ + SO₄^2- --> AlSO₄ (incorrect)

2Al³⁺ + 3SO₄^2- --> Al₂(SO₄)₃ (correct)

Let me give you an easy way of balancing:

Al is 3+

SO4 is 2-

Simply transpose the number of positives (or negatives) for each ion, to the other ion, by placing this value of one ion, in front of the other ion. That is, Al³⁺ the 3 goes in front of the SO₄^2- as 3SO₄^2-, and SO₄^2-, the 2 goes in front of the Al³⁺ to become 2Al³⁺. Then on the right hand side of the equation, this same number (now in front of each ion on the left side of the equation), is placed after each “ion” entity.

Let us again look at:

Al³⁺ + SO₄^2- --> AlSO₄ (incorrect)

Al³⁺ + SO₄^2- --> Al₂(SO₄)₃ (correct)

Put the three from the Al in front of the SO₄^2- and the 2 from the SO₄^2- in front of the Al³⁺. Equation becomes:

2Al³⁺ + 3SO₄^2- --> Al₂(SO₄)₃. You simply place the valency of one ion, as a whole number, in front of the other ion, and vice versa.

Remember to encase the SO₄ in brackets. Why? Because we are dealing with the sulphate ion, SO₄^2-, and it is this ion that is 2- charged (not just the O₄), so we have to ensure that the “ion” is bracketed. Now to check, the 2 times 3⁺ = 6⁺, and 3 times 2^- = 6^-. We have equal amounts of positive ions, and equal amounts of negative ions.

Another example:

NaOH + HCl --> ?

Na is Na⁺, OH is OH^-, so this gave us NaOH. Originally the one positive canceled the one negative

HCl is H⁺ + Cl ^-, this gave us HCl.

Reaction is going to be the Na⁺ reacting with a negatively charged ion. This will have to be the chlorine, Cl^-, because at the moment the Na⁺ is tied to the OH^-. So: Na⁺ + Cl^- --> NaCl

The H+ from the HCl will react with a negative (-) ion this will be the OH^- from the NaOH. So: H⁺ + OH^- --> H₂O (water).

The complete reaction can be written:

NaOH + HCl --> NaCl + H₂O. We have equal amounts of all atoms each side of the equation, so the equation is balanced.

or

Na⁺OH^- + H⁺Cl^- --> Na⁺Cl^- + H⁺OH^-

Something More Difficult:

Mg(OH)₂ + H₃PO₄ --> ? (equation on left not balanced)

Mg²⁺ 2OH^- + 3H⁺PO₄^3- --> ? (equation on left not balanced), so let us rewrite the equation in ionic form.

The Mg²⁺ needs to react with a negatively charged ion, this will be the PO₄^3-,
so: 3Mg²⁺ + 2PO₄^3- --> Mg₃(PO₄)₂

(Remember the swapping of the positive or negative charges on the ions in the left side of the equation, and placing it in front of each ion, and then placing this number after each ion on the right side of the equation)

What is left is the H⁺ from the H₃PO₄ and this will react with a negative ion, we only have the OH^- from the Mg(OH)₂ left for it to react with.

6H⁺ + 6OH^- --> 6H₂O

Where did I get the 6 from?  W hen I balanced the Mg²⁺ with the PO₄^3-, the equation became 3Mg²⁺ + 2PO₄^3- --> Mg₃(PO₄)₂

Therefore, I must have required 3Mg(OH)₂ to begin with, and 2H₃PO₄, ( because we originally had (OH)₂ attached to the Mg, and H₃ attached to the PO₄. I therefore have 2H₃ reacting with 3(OH)₂.  We have to write this, on the left side of the equation, as 6H⁺ + 6OH^- because we need it in ionic form.  The equation becomes:

6H⁺ + 6OH^- --> 6H₂O

The full equation is now balanced and is:

3Mg(OH)₂ + 2H₃PO₄ --> Mg₃(PO₄)₂ + 6H₂O

I have purposely split the equation into segments of reactions. This is showing you which ions are reacting with each other. Once you get the idea of equations you will not need this step.

The balancing of equations is simple.  You need to learn the valency of the common ions (see tables).  The rest is pure mathematics, you are balancing valency charges, positives versus negatives.  You have to have the same number of negatives, or positives, each side of the equation, and the same number of ions or atoms each side of the equation.

If one ion, example Al³⁺, (3 positive charges) reacts with another ion, example OH^- (one negative ion) then we require 2 more negatively charged ions (in this case OH^-) to counteract the 3 positive charges the Al³⁺ contains.

Take my earlier hint, place the 3 from the Al³⁺ in front of the OH^-, now reads 3OH^-, place the 1 from the hydroxyl OH^- in front of the Al³⁺, now stays the same, Al³⁺ (the 1 is never written in chemistry equations).

Al³⁺ + 3OH^- --> Al(OH)₃

The 3 is simply written in front of the OH^-, a recognised ion, there are no brackets placed around the OH^-. On the right hand side of the equation, all numbers in front of each ion on the left hand side of the equation are placed after each same ion on the right side of the equation. Brackets are used in the right side of the equation because the result is a compound. Brackets are also used for compounds (reactants) in the left side of equations, as in 3Mg(OH)₂ + 2H₃PO₄ --> ?

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The following page shows one of many emails I have received basically asking the same question, so I have posted the answer here... More Equations