Skills to develop
A half reaction is a reduction or an oxidation reaction, such as
2 H+ + 2 e- --> H2
Oxidation and reduction reactions are called redox reactions. The half reaction equations described earlier are useful in balancing redox reactions.
Theoretically, an oxidation-reduction reaction takes place even when the reactants are well separated in space, as long as the flow of electrons and ions are facilitated by electrical connections (salt bridge and wire). A simple redox reaction of the type:
Zn(s) + Cu2+(aq) --> Cu(s) + Zn2+(aq)
may be carried out using a galvanic cell.
This reaction may be written as two half-reactions and adding the two half reactions gives the overall equation representing a chemical process:
Zn --> Zn2+(aq) + 2 e-
+) Cu2+(aq) + 2 e- --> Cu(s)
-----------------------------------------
Zn(s) + Cu2+(aq) --> Cu(s) + Zn2+(aq)
In this case, the electrons travel from the Zn electrode to the Cu electrode via the wire connecting the two electrodes. The ions travel in a solution or through a salt bridge to balance the charge in the electrolyte solutions.
You have already seen the demonstration of this reaction in the past two weeks, and in the module called Battery.
A Redox reaction may be balanced by first writing two half-reactions, and then canceling the electrons by adding them algebraically. You will learn to balance half-reaction equations in this tutorial.
Some examples are given to illustrate how we use half reactions to describe and balance some reduction and oxidation (redox) reactions.
You may think that the two sulfur atoms in the formula are identical, but they are different. You have to understand the chemistry of these ions and then start to investigate their chemical reaction. The structure of HS2O3- may be compared to that of HSO4-:
S O- O O-
\\ / \\ /
S S
// \ // \
O OH O OH
Thus, one of the two S atoms has an oxidation state of -2, and we represent
this S atom by (=S) to indicate that it is attached to another S atom
by a double bond (=).
The chemistry of the above reaction is complicated, but you don't have to worry about that at this time. You may use the above method even if you do not know the structure of these species.
If your answer is...I'm lost!
Consider...
The oxidation state for S changed.
If your answer is...I
Consider...
The oxidation state of I goes from 0 to -1. It is therefore reduced.
If your answer is...S
Excellent...
S goes from -2 in H2S to 0 in S, an element.
If your answer is...I'm lost!
Consider...
The oxidation state of S goes from -2 in H2S to 0 in S, an element.
It is thus oxidized. What is reduced?
If your answer is...H
Consider...
The oxidation state for H does not change.
If your answer is...I
Excellent...
The oxidation state of I goes from 0 to -1. It is therefore reduced.
If your answer is...I'm lost!
Consider...
Zn is oxidized, NO3- is reduced.
If your answer is...N
Excellent...
The oxidation state of N goes from +5 to -3.
If your answer is...I'm lost!
Consider...
Enter NO3- or NH4+
If your answer is...NH4+
Consider...
You are probably thinking that NH4+ carries a postive charge,
and NO3- carries a negative charge. But you should add electrons to
compensate the oxidation state change.
If your answer is...NO3-
Excellent...
NO3- + 8 e --> NH4+.
If your answer is...I'm lost!
Consider...
Try a 2 or a 10.
If your answer is...2
Consider...
The 8 electrons carry 8 negative charges.
If your answer is...10
Excellent...
10 H+ + NO3- + 8 e --> NH4+ .
Each sides has a net +1 charge.
If your answer is...I'm lost!
Consider...
Try 3, H2O, or 3 H2O.
If your answer is...H2O
Consider...
Try '3 H2O'.
If your answer is...3
Consider...
Type the chemical formula carefully, 3 H2O.
If your answer is...3 H2O
Excellent...
If your answer is...I'm lost!
Consider...
Try -6.
If your answer is...-6
Excellent...
It changes from +5 to -1, a decrease of 6.
If your answer is...I'm lost!
Consider...
As in As2S3 has an oxidation state of +3.
As in H2As4- has an oxidation state of +5.
If your answer is...+2
Excellent...
(+3) -> (+5)
A "+" means increase in oxidation.
If your answer is...I'm lost!
Consider...
The oxidation state for S is -2 in As2S3,
+6 in SO42-.
If your answer is...S
Excellent...
Excellent! The oxidation state of S changes from -2 to +6, an increase of 8.
If your answer is...I'm lost!
Consider...
There are only two key elements, As, and S. As has been balanced for you.
If your answer is...S
Excellent...
Since there are 3 S on the left, and only one on the right.
If your answer is...I'm lost!
Consider...
Consider change of oxidation states:
As +3 -> +5
S -2 -> +6
Enter 'left' or 'right'.
If your answer is...right
Excellent...
Good work! Since both As and S are oxidized, both give off electrons.
If your answer is...I'm lost!
Consider...
As +3 -> +5 (2 e)*2
S -2 -> +6 (8 e)*3
Total number of electrons: (2*2 + 3*8) e = 28 e
If your answer is...28
Excellent...
If your answer is...I'm lost!
Consider...
You should balance the charge on the right using H+, which is
abundant in an acidic solution.
If your answer is...H
Excellent...
If your answer is...I'm lost!
Consider...
If your answer is...20
Excellent...
The balanced half-reaction equation is: