Oxidation - Reduction Reactions
Many atoms, molecules, and ions in solution participate in reactions in which there is a transfer of electrons. Such rxns are classified as oxidation-reduction reactions.
1. Zn(s) + 2HCl (aq) ® ZnCl2(aq) + H2 (g)
or
2. Zn(s) + 2H+ (aq) ® Zn2+ (aq) + H2(g)
The above equation can be written as half-reactions.
3. Zn(s) ® Zn2+ (incomplete)
4. 2H+(aq) ® H2(g) (incomplete)
For electrical neutrality, we have
5. Zn(s) ® Zn2+ (aq) + 2e-
6. 2H+ (aq + 2e-®H2(g)
Notice that the addition of 5 and 6 yields the overall reaction 2.
Oxidation is defined as the loss of electrons. Reduction is defined as a gain of electrons. The substance that gains electrons is reduced and oxidizes another substance; it is called an oxidizing agent. The substance that loses electrons is oxidized and reduces another substance; it is called a reducing agent.
Zn loses electrons
Zn is oxidized to Zn2+
Oxidation
¯
Zn(s) + 2H+(aq) ® Zn2+(aq) + H2(g)
Reduction
Zn is the reducing agent; H+ is the oxidizing agent; reduction
it reduces the H+ as it is it oxidizes the Zn as it is H+ gains electrons
oxidized reduced H+ is reduced to H2
Oxidation : Increase in oxidation number
Reduction: Decrease in oxidation number
Balancing Redox Equations by the Ion-electron method
Ion-electon method involves breaking the overall reaction into two half-reactions. One for the oxidation step and the other for reduction. Each half-reaction is first balanced materially (that is, in terms of atoms), and then electrically by adding electrons to the side of the half-reaction deficient in negative charge. Finally, the balanced half-reactions are added together in such a way that the electrons cancel from both sides of the final equation.
Example 1
Sn2+ + Hg2+ + Cl-® Hg2 Cl2 + Sn4+
1st step: Divide the equation into 2 half-rxns. Note that the atoms on each side of each half-rxn should be of the same kind.
Sn2+ ® Sn4+
Hg2+ + Cl-® Hg2Cl2
2nd Step: Balance each half-reaction in terms of atoms
Sn2+ ® Sn4+
2Hg2+ + 2Cl- ® Hg2Cl2
3rd Step: Balance the charge by adding electrons to the more positive (or less negative) side. In the first half-reaction, the net charge on the left is 2+ and on the right it is 4+, so we add 2e- to the right so that the net charge on both sides is the same. In the second half-reaction, the net charge is 2+ on the left [2 x (2+) + 2 (1-) = 2+] and zero on the right. Therefore, we add 2e- to the left. This gives
Sn2+ ® Sn4+ + 2e-
2e- + 2Hg2+ + 2Cl-® Hg2 Cl2
4th Step: Make the number of electrons gained equal to the number lost. Notice that in this example, this condition is already fulfilled. Now add the half-reactions together.
Sn2+ ® Sn4+ 2e-
Add 2e-+ 2Hg2+ + 2C-® Hg2Cl2
Sn2+ + 2Hg2+ + 2Cl- ® Sn4+ + Hg2Cl2
Example 2
Fe3+(aq) + Cl- (aq) ® Fe(s) + Cl2(g)
Fe3+(aq) ® Fe(s) ---(1)
Cl-(aq) ® Cl2 (g) ---(2)
Fe3+(aq) ® Fe(s)
2Cl- (aq) ® Cl2 (g)
x2 Fe3+(aq) + 3e-® Fe(s)
x3 2Cl-(aq) ® Cl2(g) + 2e-
2Fe3+ (aq) + 6e-® 2Fe(s)
Add 6Cl- (aq) ® 3Cl2 (g) + 6e-
2Fe3+(aq) + 6Cl-(aq) ® 2Fe(s) + 3Cl2(g))
In many oxidation-reduction rxns that take place in aqueous solution, water plays an active role. Any aqueous solution contains the species H2O, H+, and OH- In acidic solutions, the predominant species are H2O and H+; in basic solutions they are H2O and OH-. When balancing half-reactions that occur in solution, we can use these species to achieve material balance.
Example 3
Consider the rxn between dichromate ion, Cr2O72-, and hydrogen sulfide in acidic solution to produce chromium (III) ion and elemental sulfur.
Cr2O7 2- + H2S ® Cr3+ + S
H2S ® S
The half-rxn is balanced materially by adding two H+ ions to the right
H2S ® S + 2H+
Electrical balance is achieved by adding two electrons to the right so that the net charge on both sides is zero.
1. H2S ® S + 2H+ + 2e-
For chromium,
Cr2O2-® Cr3+
Cr2O72- ® 2Cr3+
For the oxygen on the left, add 7H2O to the right and hydrogen imbalance that results is removed by placing 14H+ on the left.
14H+ + Cr2O72- ® 2Cr3+ + 7H2O
6e- + 14H+ + Cr2O72-® 2Cr3+ + 7H2O
1x3 3H2S ® 3S + 6H+ + 6e-
Add 6e- +14H++ Cr2O72-®2Cr3++ 7H2O
3H2S + 8H+ + Cr2O72-® 3S + 2Cr3+ + 7H2O
Note: When balancing half-reactions in acid soln;
(a) To balance a hydrogen atom we add a hydrogen ion, H+, to the other side of the equation.
(b) To balance an oxygen atom we add a water molecule to the side deficient in oxygen and then two H+ ions to the opposite side to remove the hydrogen imbalance.
Redox rxns in basic solutions
In basic solution the dominant species are H2O and OH-, so these are the species that should be used to achieve material balance.
Suppose we wished to balance the following half-reaction taking place in a basic soln. Pb ® PbO
First we balance it as if it occurred in an acidic soln. H2O + Pb ® PbO + 2H+ + 2e-
The conversion to basic solution follows these three steps:
Step 1: For each H+ that must be eliminated from the equation, add an OH-to both sides of the equation.
H2O + Pb + 2OH-® PbO + 2H+ + 2OH-+ 2e-
Step 2: Combine H+ and OH- to form H2O 2H+ and 2OH- on the right give 2H2O
H2O + Pb + 2OH-® PbO + 2H2O + 2e-
Step 3: Cancel any H2O that are the same on both sides.
Pb+ 2OH-® PbO + H2O + 2e-
Example 4
Balance the following reaction for the oxidation of plumbite ion, Pb(OH)3- to lead dioxide by hypochlorite ion in basic solution.
Pb(OH)3-+ OCl- ® PbO2 + Cl-
Solution: First , the equation is balanced as though it occurred in acidic solution.
Half-rxns: Pb(OH)3- ® PbO2
OCl-® Cl-
Now we balance them according to atoms
Pb(OH)3-® PbO2 + H2O + H+
2H+ + OCl-® Cl-+ H2O
Then we balance the charge by adding electrons.
Pb(OH)3- ® PbO2 + H2O + H+ + 2e-
Add 2e-+ 2H+ + OCl- ® Cl-+ H2O
H+ + OCl-+ Pb(OH)3-® Cl-+ 2H2O + PbO2
Next we perform the three-step conversion to basic solution.
First we add to each side the same number of OH- as there are H+ in the equation
OH- + H+ + OCl-+ Pb(OH)3- ® Cl- + 2H2O + PbO2 + OH
H2O
H2O + OCl- + Pb(OH)3-® Cl-+ 2H2O + PbO2 + OH-
OCl-+ Pb(OH)3- ® Cl-+ H2O + PbO2 + OH-
Note: To balance half-reactions in basic solution, the following rules can be used:
(a) To balance a hydrogen atom we add one H2O molecule to the side of the half-reaction deficient in hydrogen, and to the other side we add one hydroxide ion.
(b) To balance one oxygen atom we add two hydroxide ions to the side deficient in oxygen and one H2O molecule to the other side.
Oxidation number change method
HCl + K2Cr2O7 ® KCl + CrCl3 + Cl2 + H2O
Step 1: Assign oxidation numbers to all the atoms in the equation.
H Cl + K2 Cr2 O7 ® K Cl + Cr Cl3 + Cl2 H2 O
+1 -1 +1 +6 -2 +1 -1 +3 -1 0 +1 -2
Step 2: Identify which atoms change oxidation number and insert temporary coefficients so that we have the same number of each on both sides.
Oxidation No.Reactants / Oxidation No
Products
H / 1+ / 1+
Cl / 1- / 1-
K / 1+ / 1+
Cr / 6+ / 3+ / ®Reduced
O / 2- / 2-
Cl / 1- / 0 / ®Oxidized
Notice that Cr goes from 6+ ® 3+. Some Cl 1- to 0 and some -1 to -10 put 2 in front of HCl and 2 in front of CrCl3
2H Cl + K2 Cr27 ® KCl + 2Cr Cl3 + Cl2 + H2O
2e- lost (total) 2 x 3e-
6e- gained (total) 2 x 3e-
2H Cl + K2Cr2 O7 ® K Cl + 2Cr Cl3 + Cl2 + H2O
-1 +6 +3 0
Step 3: Compute the total change in oxidation number for both oxidation and reduction.
Step 4: Make the total loss and gain of electrons the same by multiplying coefficients by appropriate factors.
3 X 2e- = 6e-lost
6e- gained
6H Cl + K2 Cr2 O7 ® K Cl + 2Cr Cl3 + 3Cl2 + H2O
Step 5: Finally, balance the rest of the equation by inspection.
14HCl + K2Cr2O7 ® 2KCl + 2CrCl3 + 3Cl2 + 7H2O
Example 2
MnO4-(aq) + Cl- (aq) + H+ (aq) ® Mn2+ (aq) + Cl2(g) + H2O
Oxid No (Reactants) Oxid. No (Products)
Mn +7 +2 ® reduced
O -2 -2
Cl -1 0 ® oxidized
H +1 +1
Mn changes by 5 units (decrease)
Cl changes by 1 unit (increase)
To make the increase in oxidation number equal to the decrease, there must be 5 Cl atoms oxidized for every Mn reduced.
MnO4- + 5Cl-® Mn2+ + 5/2 Cl2
2MnO4- + 10Cl-® 2Mn2+ + 5Cl2
2MnO4- (aq) + 10 Cl- (aq) + 16H+ (aq) ® 2Mn2+ (aq) + 5Cl2 (g) + 8H2O