how to balance equations using half reaction method??
in short steps.
(anyone???its urgent!!)
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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.
Cr2O72- + C2H6O Cr3+ + C2H4O
dichromate ethanol chromium(III) acetaldehyde
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; one will be an oxidation half-reaction and the other a reduction half- reaction, by grouping appropriate species. The nature of each will become evident in subsequent steps.
Cr2O72- Cr3+
C2H6O C2H4O
Second, if necessary, balance all elements except oxygen and hydrogen in both equations by inspection. In other words, balance the non-hydrogen and non-oxygen atoms only. By following this guideline in the example above, only the chromium reaction needs to be balanced by placing the coefficient, 2 , in front of Cr+3 as shown below.
Cr2O72- 2 Cr3+
C2H6O C2H4O
The third step involves balancing oxygen atoms. To do this, add 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.
Cr2O72- 2 Cr3+ + 7 H2O
C2H6O C2H4O
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
14 H+ + Cr2O72- 2 Cr3+ + 7 H2O
C2H6O C2H4O + 2 H+
The fifth step involves the balancing 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.
14 H+ + Cr2O72- 2 Cr3+ + 7 H2O
+12 +6
C2H6O C2H4O + 2 H+
0 +2
The electrons must always be added to that side which has the greater positive charge as shown below.
6 e- + 14 H+ + Cr2O72- 2 Cr3+ + 7 H2O
+6 +6
C2H6O C2H4O + 2 H+ + 2 e-
0 0
note: the net charge on each side of the equation does not have to equal zero.
The chromium reaction can now be identified as the reduction half-reaction and the ethanol/acetaldehyde as the oxidation half-reaction. 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.
6 e- + 14 H+ + Cr2O72- 2 Cr3+ + 7 H2O
3 x (C2H6O C2H4O + 2 H+ + 2 e-)
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.
6 e- + 14 H+ + Cr2O72- + 3 C2H6O 2 Cr3+ + 7 H2O + 3 C2H4O +
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.
Cr2O72- + C2H6O Cr3+ + C2H4O
dichromate ethanol chromium(III) acetaldehyde
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; one will be an oxidation half-reaction and the other a reduction half- reaction, by grouping appropriate species. The nature of each will become evident in subsequent steps.
Cr2O72- Cr3+
C2H6O C2H4O
Second, if necessary, balance all elements except oxygen and hydrogen in both equations by inspection. In other words, balance the non-hydrogen and non-oxygen atoms only. By following this guideline in the example above, only the chromium reaction needs to be balanced by placing the coefficient, 2 , in front of Cr+3 as shown below.
Cr2O72- 2 Cr3+
C2H6O C2H4O
The third step involves balancing oxygen atoms. To do this, add 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.
Cr2O72- 2 Cr3+ + 7 H2O
C2H6O C2H4O
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
14 H+ + Cr2O72- 2 Cr3+ + 7 H2O
C2H6O C2H4O + 2 H+
The fifth step involves the balancing 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.
14 H+ + Cr2O72- 2 Cr3+ + 7 H2O
+12 +6
C2H6O C2H4O + 2 H+
0 +2
The electrons must always be added to that side which has the greater positive charge as shown below.
6 e- + 14 H+ + Cr2O72- 2 Cr3+ + 7 H2O
+6 +6
C2H6O C2H4O + 2 H+ + 2 e-
0 0
note: the net charge on each side of the equation does not have to equal zero.
The chromium reaction can now be identified as the reduction half-reaction and the ethanol/acetaldehyde as the oxidation half-reaction. 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.
6 e- + 14 H+ + Cr2O72- 2 Cr3+ + 7 H2O
3 x (C2H6O C2H4O + 2 H+ + 2 e-)
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.
6 e- + 14 H+ + Cr2O72- + 3 C2H6O 2 Cr3+ + 7 H2O + 3 C2H4O +
debankursaha2001:
yup, thanx
wlcm
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