Transcript Balancing Oxidation-Reduction Reactions

Balancing OxidationReduction Reactions
Any reaction involving the transfer of
electrons is an oxidation-reduction (or
redox) reaction
Definitions:




Oxidation is the loss of electrons.
Reduction is the gain of electrons.
Oxidation cannot take place without reduction.
During a redox reaction, the oxidation numbers
of reactants will change.
For any equation to be balanced:
1. The number of atoms of each type on the left
side of the arrow must equal the number of
atoms of each type to the right of the arrow.
2. The total charges of all the ions on the left side
of the arrow must equal the total charges of
all the ions to the right of the arrow.
In addition, for redox reactions:
3. The electrons lost (during oxidation) must
equal the electrons gained (during reduction).
Balancing Oxidation-Reduction
Reactions:
1. Assign oxidation numbers to every atom in the
reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
Balancing Oxidation-Reduction
Reactions:
1. Assign oxidation numbers to every atom in the
reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
Oxygen, in a compound or ion, is -2
Balancing Oxidation-Reduction
Reactions:
1. Assign oxidation numbers to every atom in the
reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
Use the combined ‘charges’ of the oxygens in each
ion or compound to determine the oxidation number
of Cr or C.
Balancing Oxidation-Reduction
Reactions:
1. Assign oxidation numbers to every atom in the
reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
-14
-8
-4
Use the combined ‘charges’ of the oxygens in each
ion or compound to determine the oxidation number
of Cr or C.
Balancing Oxidation-Reduction
Reactions:
1. Assign oxidation numbers to every atom in the
reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
-14
-8
-4
The sum of the oxidation numbers of the other
element must add up to the charge on the ion or
molecule.
Balancing Oxidation-Reduction
Reactions:
1. Assign oxidation numbers to every atom in the
reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
+12
-14
+6
-8
+3
+4 -4
The sum of the oxidation numbers of the other
element must add up to the charge on the ion or
molecule.
Balancing Oxidation-Reduction
Reactions:
1. Assign oxidation numbers to every atom in the
reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
+12
+6
+3
+4
Divide the sum of the charges by the number of
atoms to get the oxidation number for chromium and
carbon in the reactants.
Balancing Oxidation-Reduction
Reactions:
1. Assign oxidation numbers to every atom in the
reaction.
Cr2O72- + C2O42-  Cr3+ + CO2
+12/2=+6
+6/2=+3
+3
+4
Divide the sum of the charges by the number of
atoms to get the oxidation number for chromium and
carbon in the reactants.
Balancing Oxidation-Reduction
Reactions:
2. Write ‘bare bones’ half reactions.
Include only the atom, ion or element that
changes oxidation number.
Cr+6 + 3e-  Cr+3
C+3  C+4 + 1eRemember that each half reaction must also be
balanced for charge. The total charges on the
left must equal the total charges on the right.
Balancing Oxidation-Reduction
Reactions:
3. Take into account any subscripts in the
formulas of reactants and products, and
multiply the half reactions accordingly.
Cr2O72- + C2O42-  Cr3+ + CO2
2[Cr+6 + 3e-  Cr+3]
2[C+3  C+4 + 1e-]
Balancing Oxidation-Reduction
Reactions:
3. Take into account any subscripts in the
formulas of reactants and products, and
multiply the half reactions accordingly.
Cr2O72- + C2O42-  Cr3+ + CO2
2Cr+6 + 6e-  2 Cr+3
2C+3  2 C+4 + 2e-
Balancing Oxidation-Reduction
Reactions:
4. Multiply each half reaction by the appropriate
factor so that the number of electrons lost =
number of electrons gained.
1[2Cr+6 + 6e-  2 Cr+3]
3[2C+3  2 C+4 + 2e-]
Balancing Oxidation-Reduction
Reactions:
4. Multiply each half reaction by the appropriate
factor so that the number of electrons lost =
number of electrons gained.
2Cr+6 + 6e-  2 Cr+3
6C+3  6 C+4 + 6e-
Balancing Oxidation-Reduction
Reactions:
4. Add the two half reactions together.
2Cr+6 + 6e-  2 Cr+3
6C+3  6 C+4 + 6e2Cr+6 + 6C+3  2 Cr+3 + 6 C+4
Balancing Oxidation-Reduction
Reactions:
4. Add the two half reactions together.
2Cr+6 + 6e-  2 Cr+3
6C+3  6 C+4 + 6e2Cr+6 + 6C+3  2 Cr+3 + 6 C+4
At this point, the electrons lost = the electrons
gained during the reaction.
Balancing Oxidation-Reduction
Reactions:
5. You now have the number of each atom that
undergoes oxidation or reduction in the balanced
equation. Take any subscripts into account when
inserting coefficients.
2Cr+6 + 6C+3  2 Cr+3 + 6 C+4
Cr2O72- + 3C2O42- 2Cr3+ + 6CO2
Balancing Oxidation-Reduction
Reactions:
6. Balance the reaction for charge, using OH- (if
in base) or H+ (if in acid).
The equation below takes place in acid:
Cr2O72- + 3C2O42- 2Cr3+ + 6CO2
Charges: -2 + -6 = -8 (left)  +6 (right)
Balancing Oxidation-Reduction
Reactions:
Cr2O72- + 3C2O42- 2Cr3+ + 6CO2
Charges: -2 + -6 = -8 (left)  +6 (right)
Since the reaction takes place in acid, you need
to add 14 H+ to the left side so that the
charges become equal.
Balancing Oxidation-Reduction
Reactions:
14 H+ + Cr2O72- + 3C2O42- 2Cr3+ + 6CO2
Charges on left = +6 = Charges on right
Balancing Oxidation-Reduction
Reactions:
7. Balance for H and O by adding water to the
appropriate side of the reaction.
14 H+ + Cr2O72- + 3C2O42- 2Cr3+ + 6CO2 + 7 H2O
Balancing Oxidation-Reduction
Reactions:
8. Check the balance for all atoms in the reaction.
14 H+ + Cr2O72- + 3C2O42- 2Cr3+ + 6CO2 + 7 H2O
Left: 14 H
2 Cr
19 O
6C
Right: 14 H
2 Cr
19 O
6C
Redox Stoichiometry

Calculations involving concentrations and redox
reactions are quite common. Many ores
containing metals are analyzed using redox
titrations. Since many compounds change color
as they are oxidized or reduced, one of the
reactants may serve as the indicator in the
titration.
Redox Stoichiometry

The concentration of iron(II) can be
determined by titration with bromate ion, in
acid. The products are iron(III) ion and the
bromide ion.
What is the concentration of iron(II) ion if
31.50 mL of 0.105M potassium bromate is
required to completely react with 10.00 mL of
the iron solution.
Redox Stoichiometry

The concentration of iron(II) can be
determined by titration with bromate ion, in
acid. The products are iron(III) ion and the
bromide ion.
1. Write the balanced chemical reaction.
Fe2+(aq) + BrO31-(aq)  Fe3+(aq) + Br1-(aq)