balancing oxidation- reduction reactions any reaction involving the transfer of electrons is an...

Balancing Balancing Oxidation-Oxidation-Reduction Reduction ReactionsReactionsAny reaction involving the Any reaction involving the

transfer of electrons is an transfer of electrons is an oxidation-reduction (or redox) oxidation-reduction (or redox)

reactionreaction

Definitions:Definitions:

Oxidation is Oxidation is the loss of electronsthe loss of electrons.. Reduction is Reduction is the gain of electrons.the gain of electrons.

Oxidation cannot take place without Oxidation cannot take place without reduction.reduction.

During a redox reaction, the During a redox reaction, the oxidation numbers of reactants will oxidation numbers of reactants will change.change.

For any equation to be For any equation to be balanced:balanced:

1. The number of atoms of each type 1. The number of atoms of each type on the left side of the arrow must on the left side of the arrow must equal the number of atoms of each equal the number of atoms of each type to the right of the arrow.type to the right of the arrow.

2. The total charges of all the ions on 2. The total charges of all the ions on the left side of the arrow must the left side of the arrow must equal the total charges of all the equal the total charges of all the ions to the right of the arrow.ions to the right of the arrow.

In addition, for redox In addition, for redox reactions:reactions:

3. The electrons lost (during 3. The electrons lost (during oxidation) must equal the electrons oxidation) must equal the electrons gained (during reduction).gained (during reduction).

Balancing Oxidation-Balancing Oxidation-Reduction Reactions:Reduction Reactions:

1. Assign oxidation numbers to every 1. Assign oxidation numbers to every atom in the reaction. atom in the reaction.

CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22



CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

Oxygen, in a compound or ion, is -2Oxygen, in a compound or ion, is -2



CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

Use the combined ‘charges’ of the oxygens Use the combined ‘charges’ of the oxygens in each ion or compound to determine the in each ion or compound to determine the oxidation number of Cr or C.oxidation number of Cr or C.



CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

-14-14 -8-8 -4-4

Use the combined ‘charges’ of the oxygens Use the combined ‘charges’ of the oxygens in each ion or compound to determine the in each ion or compound to determine the oxidation number of Cr or C.oxidation number of Cr or C.



CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

-14-14 -8-8 -4-4

The sum of the oxidation numbers of the The sum of the oxidation numbers of the other element must add up to the charge other element must add up to the charge on the ion or molecule.on the ion or molecule.



CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

+12+12 -14-14 +6 +6 -8-8 +3 +3 +4 +4 -4-4

The sum of the oxidation numbers of the The sum of the oxidation numbers of the other element must add up to the charge other element must add up to the charge on the ion or molecule.on the ion or molecule.



CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

+12+12 +6 +6 +3 +3 +4 +4

Divide the sum of the charges by the Divide the sum of the charges by the number of atoms to get the oxidation number of atoms to get the oxidation number for chromium and carbon in the number for chromium and carbon in the reactants.reactants.



CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

+12/2=+6+12/2=+6 +6/2=+3 +6/2=+3 +3 +3 +4 +4

Divide the sum of the charges by the Divide the sum of the charges by the number of atoms to get the oxidation number of atoms to get the oxidation number for chromium and carbon in the number for chromium and carbon in the reactants.reactants.

Balancing Oxidation-Balancing Oxidation-Reduction Reactions: Reduction Reactions:

2. Write ‘bare bones’ half reactions.2. Write ‘bare bones’ half reactions.

Include only the atom, ion or element Include only the atom, ion or element that changes oxidation number.that changes oxidation number.

CrCr+6+6 + 3e + 3e- - Cr Cr+3+3

CC+3+3 C C+4+4 + 1e + 1e--

Remember that each half reaction must Remember that each half reaction must also be balanced for charge. The total also be balanced for charge. The total charges on the left must equal the total charges on the left must equal the total charges on the right.charges on the right.


3. Take into account any subscripts in 3. Take into account any subscripts in the formulas of reactants and the formulas of reactants and products, and multiply the half products, and multiply the half reactions accordingly.reactions accordingly.

CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

2[Cr2[Cr+6+6 + 3e + 3e- - Cr Cr+3+3]]

2[C2[C+3+3 C C+4+4 + 1e + 1e--]]


3. Take into account any subscripts in 3. Take into account any subscripts in the formulas of reactants and the formulas of reactants and products, and multiply the half products, and multiply the half reactions accordingly.reactions accordingly.

CrCr22OO772-2- + C + C22OO44

2-2- Cr Cr3+3+ + CO + CO22

2Cr2Cr+6+6 + 6e + 6e- - 2 Cr 2 Cr+3+3

2C2C+3+3 2 C 2 C+4+4 + 2e + 2e--


4. Multiply each half reaction by the 4. Multiply each half reaction by the appropriate factor so that the appropriate factor so that the number of electrons lost = number number of electrons lost = number of electrons gained.of electrons gained.

1[2Cr1[2Cr+6+6 + 6e + 6e- - 2 Cr 2 Cr+3+3]]

3[2C3[2C+3+3 2 C 2 C+4+4 + 2e + 2e--]]


4. Multiply each half reaction by the 4. Multiply each half reaction by the appropriate factor so that the appropriate factor so that the number of electrons lost = number number of electrons lost = number of electrons gained.of electrons gained.

2Cr2Cr+6+6 + 6e + 6e- - 2 Cr 2 Cr+3+3

6C6C+3+3 6 C 6 C+4+4 + 6e + 6e--


4. Add the two half reactions 4. Add the two half reactions together.together.

2Cr2Cr+6+6 + 6e + 6e- - 2 Cr 2 Cr+3+3

6C6C+3+3 6 C 6 C+4+4 + 6e + 6e--

2Cr2Cr+6+6 + 6C + 6C+3+3 2 Cr 2 Cr+3+3 + 6 C + 6 C+4+4


4. Add the two half reactions together.4. Add the two half reactions together.

2Cr2Cr+6+6 + 6e + 6e- - 2 Cr 2 Cr+3+3

6C6C+3+3 6 C 6 C+4+4 + 6e + 6e--

2Cr2Cr+6+6 + 6C + 6C+3+3 2 Cr 2 Cr+3+3 + 6 C + 6 C+4+4

At this point, the electrons lost = the At this point, the electrons lost = the electrons gained during the reaction.electrons gained during the reaction.


5. You now have the number of each atom that 5. You now have the number of each atom that undergoes oxidation or reduction in the undergoes oxidation or reduction in the balanced equation. Take any subscripts balanced equation. Take any subscripts into account when inserting coefficients.into account when inserting coefficients.

2Cr2Cr+6+6 + 6C + 6C+3+3 2 Cr 2 Cr+3+3 + 6 C + 6 C+4+4

CrCr22OO772-2- + 3C + 3C22OO44

2-2- 2Cr2Cr3+3+ + 6CO + 6CO22


6. Balance the reaction for charge, 6. Balance the reaction for charge, using OHusing OH- - (if in base) or H(if in base) or H++ (if in (if in acid). acid).

The equation below takes place The equation below takes place in in acidacid::

CrCr22OO772-2- + 3C + 3C22OO44

2-2- 2Cr2Cr3+3+ + 6CO + 6CO22

Charges: -2 + -6 = -8 (left) Charges: -2 + -6 = -8 (left) +6 (right) +6 (right)


CrCr22OO772-2- + 3C + 3C22OO44

2-2- 2Cr2Cr3+3+ + 6CO + 6CO22

Charges: -2 + -6 = -8 (left) Charges: -2 + -6 = -8 (left) +6 +6 (right)(right)

Since the reaction takes place in Since the reaction takes place in acid, you need to add acid, you need to add 14 H14 H++ to the to the left side so that the charges left side so that the charges become equal.become equal.


14 H14 H+ + + Cr+ Cr22OO772-2- + 3C + 3C22OO44

2-2- 2Cr2Cr3+3+ + + 6CO6CO22

Charges on left = +6 = Charges on rightCharges on left = +6 = Charges on right


7. Balance for H and O by adding 7. Balance for H and O by adding water to the appropriate side of the water to the appropriate side of the reaction.reaction.

14 H14 H+ + + Cr+ Cr22OO772-2- + 3C + 3C22OO44

2-2- 2Cr2Cr3+3+ + 6CO + 6CO22 + + 7 H7 H22OO


8. Check the balance for all atoms in the 8. Check the balance for all atoms in the reaction.reaction.

14 H14 H+ + + Cr+ Cr22OO772-2- + 3C + 3C22OO44

2-2- 2Cr2Cr3+3+ + 6CO + 6CO22 + 7 H + 7 H22OO

Left: Left: 14 H14 H Right: 14 HRight: 14 H 2 Cr2 Cr 2 Cr 2 Cr

19 O19 O 19 O19 O 6 C6 C 6 C 6 C

Redox StoichiometryRedox Stoichiometry

Calculations involving Calculations involving concentrations and redox reactions concentrations and redox reactions are quite common. Many ores are quite common. Many ores containing metals are analyzed using containing metals are analyzed using redox titrations. Since many redox titrations. Since many compounds change color as they are compounds change color as they are oxidized or reduced, one of the oxidized or reduced, one of the reactants may serve as the indicator reactants may serve as the indicator in the titration.in the titration.


The concentration of iron(II) can be The concentration of iron(II) can be determined by titration with determined by titration with bromate ion, in acid. The products bromate ion, in acid. The products are iron(III) ion and the bromide ion. are iron(III) ion and the bromide ion.

What is the concentration of What is the concentration of iron(II) ion if 31.50 mL of 0.105M iron(II) ion if 31.50 mL of 0.105M potassium bromate is required to potassium bromate is required to completely react with 10.00 mL of completely react with 10.00 mL of the iron solution.the iron solution.


The concentration of iron(II) can be The concentration of iron(II) can be determined by titration with bromate determined by titration with bromate ion, in acid. The products are iron(III) ion, in acid. The products are iron(III) ion and the bromide ion. ion and the bromide ion.

1. Write the balanced chemical 1. Write the balanced chemical reaction.reaction.

FeFe2+2+((aqaq) + BrO) + BrO331-1-((aqaq) ) Fe Fe3+3+((aqaq) + ) +

BrBr1-1-((aqaq))

balancing oxidation- reduction reactions any reaction involving the transfer of electrons is an...

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