balancing_eg.html

Balancing Redox Reactions Using the Half Reaction Method

Many redox reactions occur in aqueous solutions or suspensions. In this medium most of the reactants and products exist as charged species (ions) and their reaction is often affected by the pH of the medium. The following provides examples of how these equations may be balanced systematically. The method that is used is called the ion-electron or "half-reaction" method.

Example 1 -- Balancing Redox Reactions Which Occur in Acidic Solution

Organic compounds, called alcohols, are readily oxidized by acidic solutions of dichromate ions. The following reaction, written in net ionic form, records this change. The oxidation states of each atom in each compound is listed in order to identify the species that are oxidized and reduced, respectively.

An examination of the oxidation states, indicates that carbon is being oxidized, and chromium, is being reduced. To balance the equation, use the following steps:

First, divide the equation into two halves; an oxidation half-reaction and reduction half- reaction by grouping appropriate species.
```
(red.)   (Cr₂O₇)^-2   ---->   Cr⁺³
 
(ox.)    C₂H₆O	 ---->   C₂H₄O	
```
Second, if necessary, balance both equations by inspection. In doing this ignore any oxygen and hydrogen atoms in the formula units. In other words, balance the non-hydrogen and non-oxygen atoms only. By following this guideline in the example above, only the reduction half-reaction needs to be balanced by placing the coefficient, 2 , in front of Cr⁺³ as shown below.
```
(red.)    (Cr₂O₇)^-2  ---->  2 Cr⁺³
 
(ox.)	C₂H₆O     ---->  C₂H₄O  
```
(as there are equal numbers of carbon atoms on both sides of this equation, skip this step for this half-reaction. Remember, in this step, one concentrates on balancing only non-hydrogen and non-oxygen atoms)
The third step involves balancing oxygen atoms. To do this, one must use water (H₂O) molecules. Use 1 molecule of water for each oxygen atom that needs to be balanced. Add the appropriate number of water molecules to that side of the equation required to balance the oxygen atoms as shown below.
```
(red.)   (Cr₂O₇)^-2   ---->  2 Cr⁺³  +  7 H₂O
 
(ox.)    C₂H₆O      ---->  C₂H₄O 
```
(as there are equal numbers of oxygen atoms, skip this step for this half-reaction)
The fourth step involves balancing the hydrogen atoms. To do this one must use hydrogen ions (H⁺). Use one (1) H⁺ ion for every hydrogen atom that needs to balanced. Add the appropriate number of hydrogen ions to that side of the equation required to balance the hydrogen atoms as shown below
```
(red.)   14 H⁺ + (Cr₂O₇)^-2   --->   2 Cr⁺³ + 7 H₂O
```
(as there are 14 hydrogen atoms in 7 water molecules, 14 H⁺ ions must be added to the opposite side of the equation)
```
(ox.)    C₂H₆O ---> C₂H₄O + 2 H⁺
```
(2 hydrogen ions must be added to the "product" side ofthe equation to obtain a balance)
The fifth step involves the balancing of positive and negative charges. This is done by adding electrons (e-). Each electron has a charge equal to (-1). To determine the number of electrons required, find the net charge of each side the equation.

The electrons must always be added to that side which has the greater positive charge as shown below.

note: the net charge on each side of the equation does not have to equal zero.

The same step is repeated for the oxidation half-reaction.

As there is a net chargae of +2 on the product side, two electrons must be added to that side of the equation as shown below.

At this point the two half-reactions appear as:
```
(red)   6e^-  +  14 H⁺  +  (Cr₂O₇)^-2  ------->  2 Cr⁺³ + 7 H₂O
 
(ox)    C₂H₆O  ------>  C₂H₄O  +  2 H⁺  +  2e^-
```
The reduction half-reaction requires 6 e-, while the oxidation half-reaction produces 2 e-.
The sixth step involves multiplying each half-reaction by the smallest whole number that is required to equalize the number of electrons gained by reduction with the number of electrons produced by oxidation. Using this guideline, the oxidation half reaction must be multiplied by "3" to give the 6 electrons required by the reduction half-reaction.
```
(ox.)   3 C₂H₆O ---> 3 C₂H₄O + 6 H⁺ + 6e^-
```
The seventh and last step involves adding the two half reactions and reducing to the smallest whole number by cancelling species which on both sides of the arrow.
```
6e^-  +  14 H⁺  +   (Cr₂O₇)^-2  ----->   2 Cr⁺³ + 7 H₂O
                    3 C₂H₆O   ----->   3 C₂H₄O + 6 H⁺ + 6e^-
```
adding the two half-reactions above gives the following:

6e^-+ 14H⁺ + (Cr₂O₇)^-2 + 3C₂H₆O ---> 2Cr⁺³ + 7H₂O + 3C₂H₄O + 6H⁺ + 6e^-

Note that the above equation can be further simplified by subtracting out 6 e- and 6 H⁺ ions from both sides of the equation to give the final equation.

Note: the equation above is completely balanced in terms of having an equal number of atoms as well as charges.

Example 2 - Balancing Redox Reactions in Basic Solutions

The active ingredient in bleach is the hypochlorite (OCl^-) ion. This ion is a powerful oxidizing agent which oxidizes many substances under basic conditions. A typical reaction is its behavior with iodide (I^-) ions as shown below in net ionic form.

I^- (aq) + OCl^-(aq) ------> I₂ + Cl^- + H₂O

Balancing redox equations in basic solutions is identical to that of acidic solutions except for the last few steps as shown below.

First, divide the equation into two halves; an oxidation half-reaction and reduction-reaction by grouping appropriate species.
```
(ox)    I^-   ----> I₂
 
(red)   OCl^- ----> Cl^- + H₂O
```
Second, if needed, balance both equations, by inspection ignoring any oxygen and hydrogen atoms. (The non-hydrogen and non-oxygen atoms are already balanced,hence skip this step)
Third, balance the oxygen atoms using water molecules . (The hydrogen and oxygen atoms are already balanced; hence, skip this step also.
Fourth, balance any hydrogen atoms by using an (H+) for each hydrogen atom
```
(ox)    2 I^- ----> I₂
```
(as no hydrogens are present, skip this step for this half-reaction)
```
(red)   2 H⁺ + OCl^- -----> Cl^- + H₂O
```
(two hydrogen ions must be added to balance the hydrogens in the water molecule).
Fifth, use electrons (e-) to equalize the net charge on both sides of the equation. Note; each electron (e-) represents a charge of (-1).
Sixth, equalize the number of electrons lost with the number of electrons gained by multiplying by an appropriate small whole number.
```
(ox)   2 I^-  ---->  I₂  +  2e^-
(red)  2e^-  +  2 H⁺  +  OCl^-  ---->  Cl^-  +  H₂O
```
(as the number of electrons lost equals the number of electrons gained, skip this step)
Add the two equations, as shown below.
```
2 e^-  +  2 I^-  +  2 H⁺  +  OCl^-  ----> I₂  +  Cl^-  +  H₂O  +  2e^-
```
and subtract "like" terms from both sides of the equation. Subtracting "2e-" from both sides of the equation gives the net equation:
To indicate the fact that the reaction takes place in a basic solution, one must now add one (OH-) unit for every (H+) present in the equation. The OH- ions must be added to both sides of the equation as shown below.
```
2 OH^- + 2 I^- + 2 H⁺ + OCl^-  ----->  I₂ + Cl^- + H₂O + 2 OH^-
```
Then, on that side of the equation which contains both (OH^-) and (H⁺) ions, combine them to form H₂O. Note, combining the 2 OH^- with the 2 H⁺ ions above gives 2 HOH or 2 H₂O molecules as written below.
```
2 H₂O + 2 I^- + OCl^-  ---->  I₂ + Cl^- + H₂O + 2 OH^-
```
Simplify the equation by subtracting out water molecules, to obtain the final, balanced equation.

Note that both the atoms and charges are equal on both sides of the equation, and the presence of hydroxide ions (OH^-) indicates that the reaction occurs in basic solution.