﻿ Appendixes - 5 Steps to a 5: AP Chemistry 2017 (2016) ﻿

## 5 Steps to a 5: AP Chemistry 2017 (2016)

### Appendixes

SI Units

Balancing Redox Equations Using the Ion-Electron Method

Common Ions

Bibliography

Websites

Glossary

Avoiding “Stupid Mistakes” on the Free-Response Section

Exam Resources

SI UNITS

SI Prefixes

SI Base Units and SI/English Conversions

Length

The base unit for length in the SI system is the meter.

1 kilometer (km) = 0.62 mile (mi)

1 mile (mi) = 1.61 kilometers (km)

1 yard (yd) = 0.914 meters (m)

1 inch (in) = 2.54 centimeters (cm)

Mass

The base unit for mass in the SI system is the kilogram (kg).

1 pound (lb) = 454 grams (g)

1 metric ton (t) = 103 kg

Volume

The unit for volume in the SI system is the cubic meter (m3 ).

1 dm3 = 1 liter (L) = 1.057 quarts (qt)

1 milliliter (mL) = 1 cubic centimeter (cm3 )

1 quart (qt) = 0.946 liters (L)

1 fluid ounce (fl oz) = 29.6 milliliters (mL)

1 gallon (gal) = 3.78 liters (L)

Temperature

The base unit for temperature in the SI system is kelvin (K).

Celsius to Fahrenheit: °F = (9/5)°C + 32

Fahrenheit to Celsius: °C = (5/9)(°F − 32)

Celsius to kelvin: K = °C + 273.15

Pressure

The unit for pressure in the SI system is the pascal (Pa).

1 millimeter of mercury (mm Hg) = 1 torr

1 Pa = 1 N/m2 = 1 kg/m s2

1 atm = 1.01325 × 105 Pa = 760 torr

1 bar = 1 × 105 Pa

Energy

The unit for energy in the SI system is the joule (J).

1 J = 1 kg m2 /s2 = 1 coulomb volt

1 calorie (cal) = 4.184 joules (J)

1 food Calorie (Cal) = 1 kilocalorie (kcal) = 4,184 joules (J)

1 British thermal unit (BTU) = 252 calories (cal) = 1,053 joules (J)

BALANCING REDOX EQUATIONS USING THE ION-ELECTRON METHOD

The following steps may be used to balance oxidation–reduction (redox) equations by the ion-electron (half-reaction) method. While other methods may be successful, none is as consistently successful as this particular method. The half-reactions used in this process will also be necessary when considering other electrochemical phenomena; thus, the usefulness of half-reactions goes beyond balancing redox equations.

The basic idea of this method is to split a “complicated” equation into two parts called half-reactions. These simpler parts are then balanced separately, and recombined to produce a balanced overall equation. The splitting is done so that one of the half-reactions deals only with the oxidation portion of the redox process, whereas the other deals only with the reduction portion. What ties the two halves together is the fact that the total electrons lost by the oxidation process MUST equal the total gained by the reduction process (step 6).

It is very important that you follow each of the steps listed below completely, in order; do not try to take any shortcuts. There are many modifications of this method. For example, a modification allows you to balance all the reactions as if they were in acidic solution followed by a step, when necessary, to convert to a basic solution. Switching to a modification before you completely understand this method very often leads to confusion, and an incorrect result.

1. Assign Oxidation Numbers and Begin the Half-Reactions, One for Oxidation and One for Reduction

Beginning with the following example (phases are omitted for simplicity):

(For many reactions, the substance oxidized and the substance reduced will be obvious, so this step may be simplified. However, to be safe, at least do a partial check to confirm your predictions. Note: One substance may be both oxidized and reduced; do not let this situation surprise you—it is called disproportionation.)

Review the rules for assigning oxidation numbers if necessary, in the Basics chapter. These numbers are only used in this step. Do not force them into step 5 .

Start the half-reactions with the entire molecules or ions from the net ionic form of the reaction. Do not go back to the molecular form of the reaction or just pull out atoms from their respective molecules or ions. Thus from the example above, the initial half-reactions should be:

The carbon is oxidized (C2− to C2+ ) and the chromium is reduced (Cr6+ to Cr3+ ). Check to make sure you get the same oxidation numbers for the carbon and the chromium (hydrogen and oxygen are +1 and −2 respectively).

1. Balance All Atoms Except Oxygen and Hydrogen

(In many reactions this will have been done in step 1; because of this many people forget to check this step. This is a very common reason why people get the wrong result.)

In the above example, carbon (C) and chromium (Cr) are the elements to be considered. The carbon is balanced, so no change is required in the first half-reaction. The chromium needs to be balanced, and so the second half-reaction becomes:

Note: To carry out the next two steps correctly, it is necessary to know if the solution is acidic or basic. A basic solution is one that you are specifically told is basic, or one that contains a base of OH anywhere within the reaction. Assume that all other solutions are acidic (even if no acid is present).

1. Balance Oxygen Atoms
2. In Acidic Solutions Add 1 H2O/O to the Side Needing Oxygen
3. In Basic Solutions Add 2 OHfor Every Oxygen Needed on the Oxygen-Deficient Side, Plus 1 H2O/O on the Opposite Side

Do not forget that two things (OH and H2 O) must be added in a basic solution. Also these must be added to opposite sides.

1. Balance Hydrogen Atoms
2. In Acidic Solutions Add H+(aq)
3. In Basic Solutions Add 1 H2O/H Needed, Plus 1 OH/H on the Opposite Side

Again, do not forget that two things must be added in basic solutions (OH and H2 O). In this case, they are still added to opposite sides, but with a different ratio.

If the basic step is done correctly, the oxygens should remain balanced. This may be used as a check at this point.

1. Balance Charges by Adding Electrons

The electrons must appear on opposite sides of the two half-reactions. They will appear on the left for the reduction, and on the right for the oxidation. Once added, make sure you check to verify that the total charge on each side is the same. Not being careful on this step is a major cause of incorrect answers. Do not forget to use both the coefficients and the overall charges on the ions (not the oxidation numbers from step 1).

1. Adjust the Half-Reactions So That They Both Have the Same Number of Electrons

(Find the lowest common multiple, and multiply each of the half-reactions by the appropriate factor to achieve this value. This is the key step, as the number of electrons lost MUST equal the number gained.)

1. Add the Half-Reactions and Cancel

(The electrons must cancel.)

1. Check to See If All Atoms Balance and That the Total Charge on Each Side Is the Same

This step will let you know whether you have done everything correctly.

If all the atoms and charges do not balance, you have made a mistake. Look over your work. If you have made an obvious mistake, then you should correct it. If the mistake is not obvious, it may take less time to start over from the beginning. The most common mistakes are made in steps 2 and 5, or step 3 in a basic solution.

Make sure you learn to apply each of the preceding steps. Look over the individual examples and make sure you understand them separately. Then make sure you learn the order of these steps. Finally, balance redox reactions; this will take a lot of practice. Make sure that you reach the point of being able to consistently balance equations without looking at the rules.

COMMON IONS

Ions Usually with One Oxidation State

Cations with More than One Oxidation State

Polyatomic Ions and Acids

Oxyhalogen Acids

Br or I can be substituted for chlorine Cl. F may form hypofluorous acid and the hypofluorite ion.

Other Ions

Ligands

Colors of Common Ions in Aqueous Solution

Most common ions are colorless in solution; however, some have distinctive colors. These colors have appeared in questions on the AP exam.

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