Chem 109 - General Chemistry - Spring 2013

Lecture Notes 12: 18 February

Balancing Redox Equations, cont.

Basic Solution:

balance as in acid, then:
1. Add enough OH^- (equal numbers to both sides) to cancel the H⁺. (This is necessary because there will not be protons present in a basic solution!). (There are a couple of other conventions for balancing in basic solutions. If you are familiar with another and prefer it, you may use it instead.)
2. Combine the H⁺ and OH^- on the appropriate side of the equation to give waters.
3. Go back and cancel waters which appear on both sides to give the final equation.

Example: Balance the equation above in basic solution (I don't believe this reaction actually occurs in basic solution, but due to our time constraints we'll pretend it does.)

MnO₄^- + Cl^- Mn²⁺ + Cl₂

First we balance as above to give:

16 H⁺ + 2 MnO₄^- + 10 Cl^-

2 Mn²⁺ + 8 H₂O + 5 Cl₂

Next add 16 OH^- to each sides to cancel H⁺

16 OH^- + 16 H⁺ + 2 MnO₄^- + 10 Cl^-

2 Mn²⁺ + 8 H₂O + 5 Cl₂ + 16 OH^-

Combine OH^- and H⁺ to give 16 H₂O

16 H₂O + 2 MnO₄^- + 10 Cl^-

2 Mn²⁺ + 8 H₂O + 5 Cl₂ + 16 OH^-

canceling waters then gives the final equation:

8 H₂O + 2 MnO₄^- + 10 Cl^- 2 Mn²⁺ + 5 Cl₂ + 16 OH^-

(Additional examples for balancing Redox equations in acidic solution can be found in the Discussion Module.)

Acid-Base Reactions

Neutralization

When we combine equal numbers of moles of hydrogen ion and hydroxide ion a neutralization occurs. That is, there is no reactive component left, all of the acid has been consumed by all of the base, and water has been synthesized.

Consider the reaction of 50.0 mL of 0.25 M hydrochloric acid with 25.0 mL of 0.50 M sodium hydroxide.

Write a net ionic equation for this reaction:

H⁺ + Cl^- + Na⁺ + OH^- H₂O + Cl^- + Na⁺

Giving: H⁺ + OH^- H₂O

How much acid will be left over? Base? How much water was made (synthesized)?

First find out the moles of each:

acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10^-2moles;
base = (25.0 mL)(1 L/1000 mL)(0.50 mole/L) = 1.25 x 10^-2moles.

Since the amounts are identical they will completely neutralize each other. That is they will react completely and both acid and base will be consumed with none left over.
Since the ratio in the equation is 1:1:1 acid:base:water, 1.25 x 10^-2moles of water are produced.

Consider the reaction of a weak acid and strong base: 50.0 mL of 0.25 M acetic acid is reacted with 18.0 mL of 0.50 M sodium hydroxide. Find the number of moles of each of the reactants and products after reaction.

First recall that for the reaction of a weak acid and a strong base the reaction will go until one of the reactants is completely consumed.

Writing the net ionic reaction we get (note that since it is a weak acid, we write it in the undissociated state):

CH₃COOH + OH^- H₂O + CH₃COO^-

First find moles of each reactant:

acid = (50.0 mL)(1 L/1000 mL)(0.25 mole/L) = 1.25 x 10^-2moles

base = (18.0 mL)(1 L/1000 mL)(0.50 mole/L) = 0.900 x 10^-2moles

After reaction all of the base is consumed, so:

base = 0

acetate = 0.900 x 10^-2moles

acid = 1.25 x 10^-2moles - 0.900 x 10^-2moles = 0.35 x 10^-2moles

Water synthesized = 0.900 x 10^-2moles.

Oxidation Numbers

For simple elemental ions it is easy to determine the charge on an atom, but in many other circumstances this is not the case. In order to name compounds and understand reactions we frequently need this information which is obtained from oxidation numbers.

Oxidation numbers are in essence an electronic accounting method in which electrons are assigned to a particular atom in a bond or interaction. As such they give an approximate picture of where electrons actually reside in compounds. We will find this information very useful later when we look at particular types of chemical reactions. Oxidation numbers are essential for nomenclature.

Oxidation numbers are most readily assigned using a simple set of rules:

In the formula for any substance the sum of the oxidation numbers of all the atoms in the formula is equal to the charge shown. Thus:
- For elements, such as Ar, O₂, S₈, etc. in the uncombined state the oxidation number for each atom must be 0, since no charge is shown and the atoms are equal to each other.
- For monoatomic ions the oxidation number equals the charge.
- For a compound the sum of the oxidation numbers of the atoms equals 0.
- For a polyatomic ion the sum of the oxidation numbers of the atoms equals the charge on the ion.
In compounds fluorine is always assigned an oxidation number of -1.
Alkali metals in compounds will always (for our class) be assigned an oxidation number of +1.
Alkaline-earth metals in compounds will always (for our class) be assigned an oxidation number of +2
In compounds oxygen is usually assigned an oxidation number of -2.
- Exception 1: in peroxides it is -1 while in superoxides it is -1/2. These will generally be obvious due to other rules (or the names).
- Exception 2: in combination with fluorine oxygen can be positive due to Rule 2 above, thus for OF₂ oxygen is assigned an oxidation number of +2.
In compounds hydrogen is usually assigned an oxidation number of +1

Exception: in metallic hydrides hydrogen is assigned an oxidation number of -1. These exceptions will be fairly obvious: NaH, CaH₂, etc.

Aluminum will always (for our class) be assigned an oxidation number of +3, other elements in this Group will usually be assigned an oxidation number of +3.

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© R A Paselk

Last modified 18 February 2013

Humboldt State University ® Department of Chemistry

Richard A. Paselk