Question

1. Balance the following reactions:

a) PbO₂ +Mn²⁺ =Pb²⁺ + MnO^-₄   (acidic medium)

b) Ce⁴⁺ + AsO^-₂ =Ce³⁺ + AsO^3-₄ (basic medium)

c) C₂O^2-₄ +MnO^-₄ = Mn²⁺ +CO₂ (acidic medium)

d) CN^- +MnO^-₄ = CNO^- + MnO₂ (acidic medium)

e) CrO^-₂ + ClO^- = CrO^2-₄ + Cl^- ( basic medium)

Answer 1

(a) 5PbO2 + 2Mn2+ + 4H+ → 2MnO4- + 5Pb2+ + 2H2O

PbO2 + Mn2+ → MnO4- + Pb2+

Separate the process into half reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

Pb+4O-22 + Mn+22+ → Mn+7O-24- + Pb+22+

b) Identify and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Write down the transfer of electrons. Carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples.

O:

Mn+22+ → Mn+7O-24- + 5e-

(Mn)

R:

Pb+4O-22 + 2e- → Pb+22+

(Pb)

c) Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions).

O:

Mn+22+ → Mn+7O-24- + 5e-

R:

Pb+4O-22 + 2e- → Pb+22+

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas to balance the number of atoms. Never change any formulas.

a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.

O:

Mn+22+ → Mn+7O-24- + 5e-

R:

Pb+4O-22 + 2e- → Pb+22+

b) Balance the charge. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a H+ ion to the side deficient in positive charge.

O:

Mn+22+ → Mn+7O-24- + 5e- + 8H+

R:

Pb+4O-22 + 2e- + 4H+ → Pb+22+

c) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O:

Mn+22+ + 4H2O → Mn+7O-24- + 5e- + 8H+

R:

Pb+4O-22 + 2e- + 4H+ → Pb+22+ + 2H2O

Balanced half-reactions are well tabulated in handbooks and on the web in a 'Tables of standard electrode potentials'. These tables, by convention, contain the half-cell potentials for reduction. To make the oxidation reaction, simply reverse the reduction reaction and change the sign on the E1/2 value.

Step 4. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:

Mn+22+ + 4H2O → Mn+7O-24- + 5e- + 8H+

| *2

R:

Pb+4O-22 + 2e- + 4H+ → Pb+22+ + 2H2O

| *5

O:

2Mn+22+ + 8H2O → 2Mn+7O-24- + 10e- + 16H+

R:

5Pb+4O-22 + 10e- + 20H+ → 5Pb+22+ + 10H2O

Step 5. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

2Mn+22+ + 5Pb+4O-22 + 8H2O + 10e- + 20H+ → 2Mn+7O-24- + 5Pb+22+ + 10e- + 10H2O + 16H+

Step 6. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

2Mn+22+ + 5Pb+4O-22 + 4H+ → 2Mn+7O-24- + 5Pb+22+ + 2H2O

Finally, always check to see that the equation is balanced. First, verify that the equation contains the same type and number of atoms on both sides of the equation.

ELEMENT	LEFT	RIGHT	DIFFERENCE
Mn	2*1	2*1	0
Pb	5*1	5*1	0
O	5*2	2*4 + 2*1	0
H	4*1	2*2	0

Second, verify that the sum of the charges on one side of the equation is equal to the sum of the charges on the other side. It doesn't matter what the charge is as long as it is the same on both sides.

2*2 + 5*0 + 4*1 = 2*-1 + 5*2 + 2*0

8 = 8

Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms on the right side, and since the charges on both sides are equal we can write a balanced equation.

5PbO2 + 2Mn2+ + 4H+ → 2MnO4- + 5Pb2+ + 2H2O

(b) 2Ce4+ + AsO2- + 4OH- → 2Ce3+ + AsO34- + 2H2O

Ce4+ + AsO2- → Ce3+ + AsO34-

Step 2. Identify the redox couples in the reaction.

a) Assign the oxidation numbers for each atom in the equation (see: Rules for assigning oxidation numbers). The use of the oxidation numbers greatly simplifies identifying which element in a reaction is oxidized and which element is reduced.

Ce+44+ + As-1O-12- → Ce+33+ + As+2O-234-

b) Identify and write out all redox couples of atoms that have been oxidized (their oxidation number has increased) or reduced (their oxidation number has decreased). Determine the transfer of electrons for each redox couple, but make sure that the number of atoms that have been oxidized (or reduced) is equal on both sides of the equation. If necessary, write down the stoichiometric coefficients in front of the species.

When one member of the redox couple is a species containing oxygen with an oxidation state of -2 or hydrogen with an oxidation state of +1, it's best to replace it with a water molecule.

Oxidation reactions

As-1O-12- → As+2O-234- + 3e-

(As)

Reduction reactions

Ce+44+ + e- → Ce+33+

(Ce)

As-1O-12- + e- → H+12O-2

(O)

c) Remove unnecessary redox couples. Some atoms, even though they have different oxidation numbers on the left and right side of the equation, do not have to be redox atoms. It's necessary to know the chemical reaction described by the equation in order to know which redox couples should be retained and which rejected. The program has automatically marked redox couples that need to be removed (grayed out equations). Rules for deciding which redox pairs should be kept and which removed can be found on the Divide the redox reaction into two half-reactions page. You can change the selection by clicking on the check box before the desired redox couples.

Step 3. Aggregate the redox species into one equation (ARS equation). Species with redox atoms are combined in such a way that all redox atoms are balanced in the ARS equation.

a) Balance redox atoms that aren't part of a redox couple. When more than one atom of an element that changes oxidation number is present in a formula, we must calculate the number of electrons transferred per formula unit. Because of this it is necessary to combine all redox pairs with species containing several different redox atoms into one equation.

O:

As-1O-12- → As+2O-234- + H+12O-2 + 2e-

R:

Ce+44+ + e- → Ce+33+

b) Balance non-redox atoms that are present only in redox species. When the non-redox atom is found only in redox species it must be balanced before the electron transfer is equalized. Since the redox atoms are already balanced, non-redox atoms can only be balanced by multiplying the entire equation by some factor and then summing it with another oxidation or reduction equation.

This step can be skipped. There is no non-redox atoms present only in redox species.

c) Combine the remaining reactions into two partial reactions. The remaining equations are combined into two partial equations: one for oxidation and one for reduction. Take special care to ensure that the balance of non-redox atoms is not disrupted.

This step can be skipped. We already have two partial equations.

d) Equalize the electron transfer in the two partial reactions. Now that all redox atoms are balanced we can make the number of electrons lost in the oxidation equal to the number of electrons gained in the reduction.

O:

As-1O-12- → As+2O-234- + H+12O-2 + 2e-

| *1

R:

Ce+44+ + e- → Ce+33+

| *2

O:

As-1O-12- → As+2O-234- + H+12O-2 + 2e-

R:

2Ce+44+ + 2e- → 2Ce+33+

e) Sum up the partial equations and determine the 'free' species. Both partial equations are summed into the ARS equation. The electrons are cancelled. In the ARS equation, all redox and non-redox atoms that appear only in redox species are balanced. All stoichiometric coefficients of species in the ARS equation are 'frozen', except those containing redox atoms with the same oxidation number on both sides of the equation. Its coefficients can be changed independently of other species in the ARS equation.

As-1O-12- + 2Ce+44+ → As+2O-234- + H+12O-2 + 2Ce+33+

Step 4. Balance the charges and other atoms. The ARS equation is viewed as one whole and all mathematical operations that change one species must be applied to all other species in the ARS equation. If you wish, the ARS equation can be divided with the largest common divisor in order to minimize the coefficients.

a) Add non-redox species and balance all atoms besides hydrogen and oxygen. In the 'frozen' ARS equation add species that don't contain redox atoms (marked blue). Balance all other atoms besides hydrogen and oxygen. The resulting equation will often be simple enough that it can be easily balanced by inspection. In this step the program uses the Gauss method of elimination.

AsO2- + 2Ce4+ → AsO34- + H2O + 2Ce3+

b) Balance the charges. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding an H+ ion to the side deficient in positive charge.

AsO2- + 2Ce4+ → AsO34- + H2O + 2Ce3+ + 4H+

c) Balance the oxygen atoms. Check if the number of oxygen atoms on the left side of the equation is equal to their number on the right side. If they aren't equal, equilibrate the atoms by adding water molecules to the side lacking oxygen atoms.

AsO2- + 2Ce4+ + 3H2O → AsO34- + H2O + 2Ce3+ + 4H+

Step 5. Simplify the equation. The same species on opposite sides of the arrow can be canceled. If necessary, the whole equation can be divided with the largest common divisor in order to make the coefficients as small as possible.

AsO2- + 2Ce4+ + 2H2O → AsO34- + 2Ce3+ + 4H+

Step 6. Check the balance of charges and elements. Like any chemical reaction, a redox reaction must be balanced by mass and charge. Check if the sum of each type of atom on one side of the equation is equal to the sum of the same atoms on the other side. Check if the sum of electrical charges on the left side of the equation is equal to those on the right side. It doesn't matter what the sum of the charges is as long as it's equal on both sides.

ELEMENT	LEFT	RIGHT	DIFFERENCE
As	1*1	1*1	0
O	1*1 + 2*1	1*3	0
Ce	2*1	2*1	0
H	2*2	4*1	0
Charge	1*-2 + 2*4	1*-4 + 2*3 + 4*1	0

Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms on the right side, and since the charges on both sides are equal, we can write a balanced equation.

2Ce4+ + AsO2- + 2H2O → 2Ce3+ + AsO34- + 4H+

(c) 5C2O42- + 2MnO4- + 16H+ → 10CO2 + 2Mn2+ + 8H2O

(d) 2MnO4- + 3CN- + H2O → 2MnO2 + 3CNO- + 2OH-

(e) 2CrO2- + 3ClO- + 2OH- → 2CrO42- + 3Cl- + H2O

Separate the redox reaction into half-reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

Cr+3O-22- + Cl+1O-2- → Cr+6O-242- + Cl-1-

b) Identify and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down).

O:

Cr+3O-22- → Cr+6O-242-

(Cr)

R:

Cl+1O-2- → Cl-1-

(Cl)

c) Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions).

O:

Cr+3O-22- → Cr+6O-242-

R:

Cl+1O-2- → Cl-1-

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas. Never change a formula when balancing an equation. Balance each half reaction separately.

a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.

O:

CrO2- → CrO42-

R:

ClO- → Cl-

b) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O:

CrO2- + 2H2O → CrO42-

R:

ClO- → Cl- + H2O

c) Balance the hydrogen atoms. Check if there are the same numbers of hydrogen atoms on the left and right side, if they aren't equilibrate these atoms by adding protons (H+).

O:

CrO2- + 2H2O → CrO42- + 4H+

R:

ClO- + 2H+ → Cl- + H2O

d) For reactions in a basic medium, add one OH- ion to each side for every H+ ion present in the equation. The OH- ions must be added to both sides of the equation to keep the charge and atoms balanced. Combine OH- ions and H+ ions that are present on the same side to form water.

O:

CrO2- + 2H2O + 4OH- → CrO42- + 4H2O

R:

ClO- + 2H2O → Cl- + H2O + 2OH-

Step 4. Balance the charge. To balance the charge, add electrons (e-) to the more positive side to equal the less positive side of the half-reaction. It doesn't matter what the charge is as long as it is the same on both sides.

O:

CrO2- + 2H2O + 4OH- → CrO42- + 4H2O + 3e-

R:

ClO- + 2H2O + 2e- → Cl- + H2O + 2OH-

Step 5. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:

CrO2- + 2H2O + 4OH- → CrO42- + 4H2O + 3e-

| *2

R:

ClO- + 2H2O + 2e- → Cl- + H2O + 2OH-

| *3

O:

2CrO2- + 4H2O + 8OH- → 2CrO42- + 8H2O + 6e-

R:

3ClO- + 6H2O + 6e- → 3Cl- + 3H2O + 6OH-

Step 6. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

2CrO2- + 3ClO- + 10H2O + 8OH- + 6e- → 2CrO42- + 3Cl- + 11H2O + 6e- + 6OH-

Step 7. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

2CrO2- + 3ClO- + 2OH- → 2CrO42- + 3Cl- + H2O

Finally, always check to see that the equation is balanced. First, verify that the equation contains the same type and number of atoms on both sides of the equation.

ELEMENT	LEFT	RIGHT	DIFFERENCE
Cr	2*1	2*1	0
O	2*2 + 3*1 + 2*1	2*4 + 1*1	0
Cl	3*1	3*1	0
H	2*1	1*2	0

Second, verify that the sum of the charges on one side of the equation is equal to the sum of the charges on the other side. It doesn't matter what the charge is as long as it is the same on both sides.

2*-1 + 3*-1 + 2*-1 = 2*-2 + 3*-1 + 1*0

-7 = -7

Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms on the right side, and since the charges on both sides are equal we can write a balanced equation.

2CrO2- + 3ClO- + 2OH- → 2CrO42- + 3Cl- + H2O

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