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Example Questions
Example Question #1 : Oxidation Reduction Reactions
The following ReDox reaction takes place in acidic solution:
Fe2+ + Cr2O72– → Fe3+ + Cr3+
What is the sum of coefficients in this redox reaction?
33
35
37
34
36
36
When you balance the redox reaction in acidic conditons, there are 6Fe2+, 1 Cr2O72–, 14 H+, 6 Fe3+, 2 Cr3+, and 7 H2O. Don't forget to add the 1 in front of the Cr2O72–
Example Question #2 : Oxidation Reduction Reactions
For the following unbalanced redox reaction, how many electrons are transferred and which chemical species is being oxidized?
Two electrons are transferred; P is oxidized
One electron is transferred; Hg is oxidized
Two electrons are transferred; Hg is oxidized
One electron is transferred; P is oxidized
Two electrons are transferred; Hg is oxidized
To begin, we will need to separate the given reaction into the two half-reactions by identifying changes in oxidation number. In this case, mercury (Hg) and phosphorus (P) show a change in oxidation number. Mercury begins with an oxidation number of zero, and ends with an oxidation number of . Phosphorus begins with an oxidation number of and ends with an oxidation number of . Note that the oxidation numbers for fluorine and iodine reamain constant at for each.
Now we can begin to look at the half-reactions.
Balance the atoms.
Now balance the electrons. We know that each mercury atom loses one electron and each fluorine atom gains one electron.
We can see that two electrons are tranferred. To identify the element being oxidized, we must find the element that is losing electrons. In this case, mercury is being oxidized.
Example Question #1 : Balancing Oxidation Reduction Reactions
How many electrons are involved in the following reaction?
5 e-
3 e-
2 e-
1 e-
4 e-
5 e-
The common factor between 1 e- and 5 e- is 5. Therefore the number of electrons involved is 5 e-.
Example Question #1 : Balancing Oxidation Reduction Reactions
How many electrons are involved in the following reaction?
4 e-
10 e-
5 e-
1 e-
2 e-
10 e-
The common factor between 2 e- and 5 e- is 10. Therefore the number of electrons involved is 10 e-.
Example Question #1 : Balancing Oxidation Reduction Reactions
What is the balanced coefficient on OH- for the following reaction:
(under basic conditions)
4
5
1
2
3
2
Add them together:
Simplify:
Add Hydroxides to each side to counter H+.
Simplify:
Example Question #4 : Oxidation Reduction Reactions
What is the sum of all the balanced coefficients in the following reaction:
(basic conditions)
16
12
10
8
14
14
Add the equations together
Simplify
Add 2 OH- to each side to cancel out the H+.
Simplify:
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