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can you give a balanced equation which is an example of redox reaction???
Example 1

Balance the equation from the following two half-reactions:
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O
H2C2O4 --> 2 CO2 + 2 H+ + 2 e-
Solution
Multiply the second equation by 3 and then add them algebraically so that the electrons in the two half-reaction equations cancel completely.

Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O
3 H2C2O4 --> 6 CO2 + 6 H+ + 6 e-
add the two equations and cancel the electrons to give the overvall equation
Cr2O72- + 8 H+ + 3 H2C2O4 --> 2 Cr3+ + 7 H2O + 6 CO2
Discussion
This example illustrate balancing redox equations in acid solutions.

Example 2

Balance the equation from the two half-reactions:
Cd --> Cd2+ + 2 e-
4 H+ + NO3- + 3 e- --> NO + 2 H2O
Solution
The first half-reaction has 2 electrons, whereas the second one has 3. The lowest common multiple of 2 and 3 is 6. Thus, you multiply the first equation by 3 and the second one by 2. The half-reaction equations become

3 Cd --> 3 Cd2+ + 6 e-
8 H+ + 2 NO3- + 6 e- --> 2 NO + 4 H2O
add the two equations and cancel the electrons to give the over all equation
3 Cd + 8 H+ + 2 NO3- --> 3 Cd2+ + 2 NO + 4 H2O
Discussion
The last step in balancing oxidation and reduction reactions is simple.

Example 3

In a basic solution, Fe(OH)2 and Fe(OH)3 are solids. The former may be oxidized by H2O2.
Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O.
Balance this equation.
Solution
The balanced half-reactions are:

Fe(OH)2 + OH- --> Fe(OH)3 (s) + e-
H2O2 + 2 e- --> 2 OH-
Thus, the balanced equation is
2 Fe(OH)2 + H2O2 --> 2 Fe(OH)3
Discussion
Note that the balanced equation does not have an H2O in it.

Example 4

Balance the following reaction, which is carried out in an acidic solution:
I- + IO3- --> I2
Solution
The half-reactions are:

2 I- --> I2 + 2 e- (oxidized)
2 IO3- + 10 e- --> I2 (reduced)
The balanced equation is
6 H+ + 5 I- + IO3- ® 3 I2 + 3H2O
Discussion
You may study Electro chemistry to gain better insight to the oxidation reduction process.


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