SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 12

Oxidation-Reduction and Electrochemistry

BALANCING REDOX EQUATIONS

In general, many chemical reactions are simple enough that once their reactants and products are known, the equations can be readily written and balanced by inspection. Some redox reactions, however, are of such complexity that writing them requires several steps. First limit the change involved to the actual electron shifts that occur. Balance the number of electrons gained and lost. The rest of the balancing can then usually be done without too much difficulty. Several methods can be used to accomplish this. We will show two methods with several examples of each. The first is the electron shift method. The second is the ion-electron method. Since both methods assign oxidation states to the elements involved, so the assigning of values is briefly reviewed here.

The Rules for Assigning an Oxidation State

To apply these rules, remember that the sum of the oxidation states must equal zero for an electrically neutral compound. For an ion, the sum of the oxidation states must equal the charge on the ion, including polyatomic ions.

1. The oxidation state of an atom in an element is zero.
Examples: 0 for Na(s), O2(g), H(l)

2. The oxidation state of a monoatomic ion is the same as its charge.
Examples: Na+1, Cl−1

3. The oxidation state of fluorine is −1 in its compounds.
Examples: HF where 1H(+1) + 1F(−1) = 0, PF3 where 1P(+3) + 3F(−1) = 0

4. The oxidation state of oxygen is usually −2 in its compounds.
Example: H2O where 2H(+1) + 1O(−2) = 0
One exception to this rule occurs when oxygen is bonded to fluorine and the oxidation state of fluorine takes precedence. Another exception occurs in peroxide compounds, where the oxidation state is assigned the value of −1.

5. The oxidation state of hydrogen in most compounds is +1.
Examples: H2O, HCl, NH3
In hydrides, there is an exception. Oxygen is assigned the value of −1.

TIP 

Notice that electrons gained are on the left side and electrons lost are on the right side.

The Electron Shift Method

The oxidation state method breaks the reaction into two half-reactions, the reduction reaction and the oxidation reaction. This method is based on balancing the electrons gained and the electrons lost in the two half-reactions.

FIRST EXAMPLE: Balance the equation HCl(aq) + MnO2(s) → H2O(l) + MnCl2(aq) + Cl2(g)

1. The oxidation states are written above all the elements in the equation.

         

2. The oxidation state of the manganese has changed from +4 to +2, while the oxidation state of the chlorine that emerges as a gas has changed from −1 to 0. A change from +4 to +2 is a reduction in the oxidation state and a gain in electrons. A change from −1 to 0 is a gain in the oxidation state and a loss of electrons. These are written as the half-reactions of the ions involved with the electron changes. Notice that oxidation states have the + or − on the left of the number and that ionic charges are shown on the right of the number.
Mn4+(aq) + 2e → Mn2+(aq)               Reduction reaction
Cl1−(aq) → Cl(g) + 1e            Oxidation reaction

3. The next step is to assure that the number of atoms required by the formulas in the reactants equals the number of atoms in the products. Notice that the product Cl2 in the original equation is a diatomic molecule and requires 2 atoms in the molecular formula.

This means there must be 2Cl ions provided on the left side of the equation. A total of 2e are lost, one for each Cl.
Mn4+(aq) + 2e → Mn2+(aq)               Reduction reaction
2Cl(aq)                → Cl2(q) + 2e        Oxidation reaction

4. If the number of electrons gained in reduction did not equal the number of electrons lost in oxidation, we would have had to multiply either one reaction line or both by an appropriate number to assure that the number of electrons lost equaled the number of electrons gained. This will be shown in the next example.

5. Next add the two half-reactions.

7. From the numbers thus established, the remaining coefficients can be deduced by inspection. Notice that 2 more molecules of HCl are required to furnish the chlorine for the MnCl2, and the 2 atoms of oxygen in MnO2 form 2H2O.

8. The final balanced equation is

4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g)

Now consider a more complicated reaction.

SECOND EXAMPLE: Balance the equation

     HCl(aq) + KMnO4(aq) → H2O(l) + KCl(aq) + MnCl2(aq) + Cl2(g)

1. The oxidation states are written above all the elements in the equation

   

2. Inspection of oxidation states shows that the oxidation state of the manganese has changed from +7 to +2, while the oxidation state of the chlorine that emerges as a gas has changed from −1 to 0. A change from +7 to +2 is a reduction in the oxidation state and a gain in electrons. A change from −1 to 0 is a gain in the oxidation state and a loss of electrons. These are written as the half-reactions of the ions involved with the electron changes. Notice that oxidation states have the + or − on the left of the number and that ionic charges are shown on the right of the number.
Mn7+(aq) + 5e → Mn2+(aq)               Reduction reaction
Cl(aq)                    → Cl + 1e              Oxidation reaction

3. The next step is to assure that the number of atoms required by the formulas in the reactants equals the number of atoms in the products. Notice, here again, that the product Cl2 in the original equation is a diatomic molecule and requires 2 atoms in the molecular formula, so the oxidation reaction must have the diatomic molecule Cl2. This is done by multiplying the other members of the oxidation reaction by 2 to provide the 2Cl1− for the Cl2.
Mn7+(aq) + 5e → Mn2+(aq)               Reduction reaction
2Cl(aq)                → Cl2(q) + 2e        Oxidation reaction

4. Remember that the number of electrons gained in reduction must equal the number electrons lost in the oxidation reaction. In this reaction, we have to multiply the reduction reaction by 2 and the oxidation reaction by 5 so that the number of electrons lost equal the number of electrons gained.
2Mn7+(aq) + 10e → 2Mn2+(aq)                 Reduction reaction
10Cl−1(aq)                → 5Cl2(q) + 10e        Oxidation reaction

5. Next add the two half-reactions.

At this point, inspection shows that the products 2KCl(aq) + 2MnCl2(aq) + 5Cl2(g) need a total of 16Cl to be provided in the reactants. The only source of Cl in the reactants is HCl(aq). Therefore, the coefficient of HCl must be 16.

The Ion-Electron Method

The second method is more complex but seems to represent the true mechanism of such reactions more closely.

In this method, only units that actually have individual existence (atoms, molecules, or ions) in the particular reaction being studied are taken into consideration. The principal oxidizing agent and the principal reducing agent are chosen from these (by a method to be indicated later). Then the electron loss or gain of each of these two principal actors is determined by taking into consideration the fact that, since electrons can neither be created nor destroyed, electrons lost by one of these actors must be gained by the other. This is accomplished by using two separate partial equations representing the changes undergone by each of the two principal actors.

Probably the best way to show how the method operates will be to follow in detail the steps taken in balancing two actual oxidation-reduction reactions.

FIRST EXAMPLE: Assume that the equation K2CrO4(aq) + HCl(aq) → KCl(aq) + CrCl3(aq) + H2O(l) + Cl2(g) is to be balanced. Notice this is an acidic solution.

1. Determine which of the substances present are involved in the oxidation-reduction.

This is done by listing all substances present on each side of the arrow in their molecular or ionic form and then crossing out those that appear on both sides of the arrow without being changed in any way.

            
Note that Cl is not crossed out although it appears on both sides because some of the Cl from the left side appears in a changed form, namely, Cl2, on the right.

The two substances on the left side that are not crossed out are the ones involved in the oxidation-reduction.

If, as in this case, there are more than two, disregard H+, OH, or H2O; this will leave the two principal actors.

2. Indicate in two as yet unbalanced partial equations the fate of each of the two active agents thus:

          CrO42−(aq) → Cr3+(aq)
Cl(aq) → Cl2(g)

3. Balance these equations chemically, inserting any substance necessary.

CrO42−(aq) + 8H+(aq) → Cr3+(aq) + 4H2O(l)

     2Cl(aq) → Cl2(g)

In the upper partial equation, 8H+ had to be added in order to combine with the oxygen from the CrO42− ion.

H+ always is used for this purpose in acid solutions. The next example handles a basic solution problem.

4. Balance these equations electrically by adding electrons on either side so that the total electric charge is the same on the left and right sides, thus:

                              CrO42−(aq) + 8H+(aq) + 3e → Cr3+(aq) + 4H2O(l)

                              2Cl(aq)                        → Cl2(g) + 2e

5. Add these partial equations.

But before we add we must realize that electrons can neither be created nor destroyed. Electrons are gained in one of these equations and lost in the other. THOSE GAINED IN THE ONE MUST COME FROM THE OTHER. Therefore we must multiply each of these equations through by numbers so chosen that the number of electrons gained in one equation will be the same as the number lost in the other, thus:

                              2[CrO42−(aq) + 8H+(aq) + 3e → Cr3+(aq) + 4H2O(l)]

                              3[2Cl(aq)                        → Cl2(g) + 2e]

      THIS IS THE KEY TO THE METHOD!

Adding the multiplied equations, we get:

SECOND EXAMPLE: (In a basic solution)

Note that, because this reaction takes place in a basic solution, H2O was used to remove oxygen in the upper partial.

Note also that in the second partial it was necessary to ADD oxygen and that this was done by means of OH ion.

If the solution had been acid, then H2O would have been used for this purpose, thus:

I(aq) + 3H2O(l) → IO3(aq) + 6H+(aq) + 6e

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REMEMBER 
In general, you will meet three types of partial equations:

1. Where electrons only are needed to balance, as in the second partial of the first example above.

2. Where oxygen must be removed from an ion.

In acid solution, H+ is used for this purpose as in the first partial of the first example above.

In basic solution, H2O is used for this as in the first partial of the second example above.

3. Where oxygen must be added to an ion.

In basic solution, OH is used for this purpose as in the second partial of the second example above.

In acid solution, H2O is used for this as in:

SO32−(aq) + H2O(l) → SO42−(aq) + 2H+(aq) + 2e

TIP 

This one is a challenge!

TRY THESE PROBLEMS:

1. Zn(s) + HNO3(aq) → Zn(NO3)2(aq) + NH4NO3(aq) + H2O(l)
Ans. 4, 10, 4, 1, 3

2. Cu(s) + HNO3(aq) → Cu(NO3)2(aq) H2O(l) + NO(g)
Ans. 3, 8, 3, 4, 2

3. KMnO4(aq) + HCl(aq) → KCl(aq) + MnCl2(aq) + H2O(l) + Cl2(g)
Ans. 2, 16, 2, 2, 8, 5

ANSWERS

1. The half-reactions balanced chemically. (Notice this is in an acidic solution.)

Zn → Zn2+
NO3 + 10H+ → NH4+ + 3H2O

Balancing electrically and adding:

In the equation:

4Zn(s) + 10HNO3(aq) → 4Zn(NO3)2(aq) + NH4NO3(aq) + 3H2O(l)

2. The half-reactions balanced chemically. (Notice this is in an acidic solution.)

Cu0   → Cu2+
NO31− + 4H+ → NO + 2H2O

Balancing electrically and adding:

In the equation:

3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 4H2O(l) + 2NO(g)

3. The half-reactions balanced chemically. (Notice this is in an acidic solution.)

MnO4 + 8H+ → Mn2+ + 4H2O
2Cl → Cl2

Balancing electrically and adding:

In the equation: