Chemistry 101: Reaction Types

A Brønsted-Lowry acid is any species capable of donating a proton to the solution and results in an increase in hydronium ion concentration, hence a decrease in pH.

Note: This is the definition that the AP Chemistry test uses for any acid in general.
Example acid reaction: HA + H2O ⇒ A- + H3O+

A Brønsted-Lowry base is any species capable of accepting a proton from the solution and results in an increase in hydroxide ion concentration, hence an increase in pH.

Note: This is the definition that the AP Chemistry exam uses for any general base.
Example base reaction: NH3 + H2O ⇒ NH4+ + OH-

CH3COOH is an acid. It loses a proton to solution.

Acid Equation: CH3COOH + H2O ⇒ CH3COO- + H3O+

NH3 is a base. It accepts a proton (or creates an OH-) in solution.

Base Reaction: NH3 + H2O ⇒ NH4+ + OH-

A strong acid is one which dissociates completely in solution.

For a monoprotic acid (one proton per molecule), each mole of acid in solution results in one mole of protons in solution as well.
Ex: HCl is a classic strong acid (monoprotic).

A strong base is one which dissociates completely in solution.

For a monobasic compound (one hydroxide per molecule), each mole of base in solution results in one mole of hydroxide ions in solution as well.
Ex: NaOH is a classic strong base (monobasic).

CH4 < NH3 < H2O < HF

Acidity, in general, is a measure of how easily that substance will donate a proton into solution.

CH4 < NH3 < H2O < HF

These molecules are an increasing series going from left to right across the second row of the Periodic Table. Traveling left to right across the Periodic Table, acidity always increases.

The reason for this is polarity. Chemically, these acids are: a hydrogen atom bound to a central atom. As the central atom becomes more electronegative, the bonds with hydrogen become more polar. More polar bonds are easier to dissociate in aqueous solution. Hence, the more electronegative the central atom, the more easily it donates protons, and the more acidic it is.

HF < HCl < HBr < HI

Recall: acid strength is determined by how easily the substance will donate a proton into solution.

HF < HCl < HBr < HI

These molecules are a series going down a column of the Periodic Table. As you travel down a column in the Periodic Table, acidity increases.

The reason for this is atomic size. Larger atoms can carry negative charges more easily, so the I- ion is more stable than the F- ion. The more stable the conjugate base, the stronger the acid.

1. HI (Hydrogen Iodide)

2. HBr (Hydrogen bromide)

3. HCl (Hydrogen chloride)

4. HNO3 (Nitric Acid)

5. HClO4 (Perchloric Acid)

6. H2SO4 (Sulfuric Acid)

While there are several other acids to choose from, these are the ones most commonly used in most chemistry courses, including on the AP Chem exam.

1. NaOH (sodium hydroxide)

2. KOH (potassium hydroxide)

3. NH2- (amide ion)

4. H- (hydride ion)

5. Ca(OH)2 (calcium hydroxide)

6. Na2O (sodium oxide)

7. CaO (calcium oxide)

While there are several other bases to choose from, these are the ones most commonly used in most chemistry courses, including on the AP Chem exam.

A polyprotic acid can donate more than one proton to a solution.

Ex: H2SO4 (can donate 2 protons)

An amphoteric substance can act as an acid or a base depending on the solution.

The classic example is water.
Acting as a base: H20 + HA ⇔ A- + H30+
Acting as an acid: H20 + B- ⇔ BH + OH-

Conjugate acid-base pairs are molecules which differ via the presence or absence of a proton. The protonated form of the molecule is the conjugate acid, the deprotonated form is the conjugate base (Brønsted-Lowry acid definition)

Generic equation: HA + H2O → A- + H3O+
HA is an acid and A- is its conjugate base. Similarly, H2O is acting as a base, with H3O+ acting as its conjugate acid.

The acetate ion, CH3COO-

An acid’s conjugate base is the deprotonated remainder of the molecule’s acid reaction.
The general acid neutralization reaction is HA + OH- ⇒ A- + H2O
where HA is the acid, and A- is the conjugate base.

Salt and water, according to the general equation

HA + BOH ⇒ AB + H20

It is possible to add acid and base and NOT create water (Lewis acid/base pairs) but there will always be a salt formed. Additionally, when you use Bronsted-Lowry as the acid/base definition (like the AP Chem exam does), it is safe to assume that water is always formed as well.

The purpose of a titration experiment is to discover the concentration of an unknown acid or base, by neutralizing it with a measured quantity of a base or acid solution whose concentration is known.

In titrations, the known solution is the titrant and the unknown solution is the analyte.

The equivalence point is the point at which every molecule of acid has been neutralized by a molecule of base.

[H+ ions]original = [OH- ions]added
In a titration, this is where all of the unknown (analyte) has been completely neutralized by the known (titrant), giving a neutral pH.

More than 7, slightly basic.

The equivalence point is a point at which the amount of equivalents of acid and equivalents of base from the analyte and titrant are equal.
In the example: a weak base (acetate ion, CH3COO-) is being formed, hence the pH at the equivalence point will be above 7.

In a titration, the indicator changes the color of the solution to indicate that the equivalence point pH has been reached.

Ex: Phenolphthalein goes from colorless to fuchsia between pH 8.3-10
Ex: Methyl red goes from red to green to yellow between pH 4.4-5.2-6.2

1. An indicator is a molecule that must change color visibly in a set pH range, usually a range of about 2 pH units. To select an indicator for a specific titration, find one whose pKa is roughly equal to the pH of the titration’s equivalence point.

2. Some common indicators include:
   

   • Methyl red (pH range 4.4-6.2)

   • Thymol blue (8.0-9.6)

   • Azolitmin (4.5-8.3)

The equivalence point is a point at which the amount of equivalents of acid and equivalents of base from the analyte and titrant are equal. For a strong acid/strong base combination, this happens at a pH of 7. At the equivalence point:

VA NA = VB NB

Where
VA = volume of acid in L
NA = normality of acid in #equivalents*mol/L
VB = volume of base in L
NB = normality of base in #equivalents*mol/L

An analyte is an acid or base whose concentration is determined by a titration.

A titrant (known concentration) is used to bring the unknown (analyte) solution up to a standard pH (usually neutral 7). In this way, the concentration of the analyte can be calculated.

A titrant is a strong acid or base with a known concentration.

In a titration experiment the titrant is used to neutralize, or bring to a known pH, an unknown (analyte) solution. In this way, the concentration of the unknown can be calculated.

Point A

Point A is the first half-equivalence point, where one-half as many equivalents of base have been added as acid molecules that were in the solution to begin with, so one half of the acid molecules have been neutralized, and [H2A] = [HA-].

Point D

A2- production is not complete until 2 full equivalents of base have been added. This must be at the second equivalence point, which is Point D.

The Ag atom is undergoing oxidation.

The mnemonic to use in redox (reduction-oxidation) reactions is OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

In this case, the Ag atom is losing an electron, so it is being oxidized.

The Fe3+ ion is undergoing reduction.

The mnemonic to use in redox (reduction-oxidation) reactions is OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

In this case, the Fe3+ ion is gaining an electron, so it is being reduced.

An electrochemical cell is a set of chemical systems which either:

1. undergo a reaction and generate electric current, or

2. undergo a reaction when electric current is run through the system.

An electrochemical cell has two half-cells, each of which has an electrolyte and an electrode. The electrodes are connected via electrically conductive material (often a wire).

The half-cells can be separate, as below, or share an electrolyte, but there will always be two separate electrodes.

The anode is the electrode of the half-cell in which oxidation is taking place.

The cathode is the electrode of the half-cell in which reduction is taking place.

Electrolysis is the act of using a voltage source to drive a non-spontaneous chemical reaction.

Electrochemical cells in which electrolysis is occurring are referred to as electrolytic cells.

In any electrochemical cell, oxidation occurs at the anode, while reduction occurs at the cathode.

A simple mnemonic to remember this by: AN OX, RED CAT
ANode = OXidation, REDuction = CAThode.

Electrons always flow from Anode to Cathode.

Remember: Reduction occurs at the Cathode (RED CAT), and Reduction Is Gain of electrons (OIL RIG), so electrons must be flowing to the cathode to facilitate reduction.

The cell is an electrolytic cell.

Electrode A is the anode.

The positive end of the battery attracts electrons towards itself, and those electrons must be leaving Electrode A. Since electrons flow from anode to cathode, Electrode A must be the anode.

Note: Since the anode is attached to the positive end of the battery, in an electrolytic cell the anode is the positive (+) electrode.

Electrode B is the cathode.

The negative end of the battery rejects electrons away from itself, and those electrons must reach Electrode B. Since electrons flow from anode to cathode, Electrode B must be the cathode.

Note: Since the cathode is attached to the negative end of the battery, in an electrolytic cell the cathode is the negative (-) electrode.

O2(g) will appear at the anode.

The anode is where oxidation occurs, so look for a species which can lose electrons.

Water can be thought of as consisting of O2- ions and H+ ions. The O2- ions can lose electrons to form molecular oxygen, O2, and this will occur at the anode.

Cl2(g) will appear at the anode.

NaCl is made up of Na+ and Cl- ions. Cl- ions will lose electrons at the anode to form molecular Chlorine, Cl2.

H2(g) will appear at the cathode.

The cathode is where reduction occurs, so look for a species which can gain electrons.

Water can be thought of as consisting of O2- ions and H+ ions. The H+ ions can gain electrons to form molecular hydrogen, H2, and this will occur at the cathode.

Na(s) will appear at the anode.

NaCl is made up of Na+ and Cl- ions. Na+ ions will gain electrons at the anode to form Na metal.

An electrolyte is any species added to the cell (in which an electrochemical reaction is occurring) which allows current to flow more easily.

Ex: in the electrolysis of water, either acid or salt must be added to the water to improve current flow, since pure water is not a good conductor of electricity.

The units of current are amperes, A.

Current is defined as charge over time Q/t, so 1 A = 1 C/s.

360 C

I = 2.0 A = 2.0 C/s
Q = I x t = 2.0 3 min 60s/min
= 360 C

Faraday's Equation,
I x t = n x F

where:
I = current in A
t = seconds the current flows
n = # of moles of electrons
F = Faraday's constant, 96,485 C/mol e-

Note: for the AP Chem exam, approximate F = 105 C/mol e-

2.3x10-3 g of Na will be deposited.

Use Faraday's equation to calculate the number of moles of electrons which flow through the circuit:
n = I x T / F
= (1)10/105 = 1x10-4 mol e-

Each mole of electrons reduces one mole of Na+. Calculate the mass of 10-4 mol of Na(s)
m = 1x10-4 * 23g/mol
= 2.3x10-3 g Na

The cell is a galvanic cell.

The two main types of electrochemical cells (electrolytic and galvanic) can be easily discerned by the presence or absence of a power source. In this case, the lack of a battery indicates that this is a galvanic cell.

The Zn electrode is the anode.
In any electrochemical cell, electrons flow from anode to cathode. Electrons are leaving the Zn electrode, so it must be the anode.

The Cu electrode is the cathode.

In any electrochemical cell, electrons flow from anode to cathode. Electrons are arriving at the Cu electrode, so it must be the cathode.

The cathode has a positive (+) charge in a galvanic cell.

The cathode spontaneously attracts electrons, which is what would be expected of a positively-charged electrode. Note that this is the opposite convention from electrolytic cells, where the cathode has a negative charge.

The anode has a negative (-) charge in a galvanic cell.

The anode spontaneously discharges electrons, which is what would be expected of a negatively-charged electrode. Note that this is the opposite convention from electrolytic cells, where the anode has a positive charge.

The Zn anode will get lighter as current flows.

Oxidation occurs at the anode in any electrochemical cell. In this case, the oxidation reaction will result in Zn(s) going to Zn2+(aq). As the Zn ions move into solution, the Zn electrode will get lighter.

The cathode will get heavier as current flows.

Reduction occurs at the cathode in any electrochemical cell. In this case, the reduction reaction will result in Cu2+(aq) going to Cu(s). As the Cu2+ ions move out of solution, they will plate on the Cu electrode, which will get heavier.

The Cu cathode will get heavier as Cu ions plate onto its surface.

The oxidation half-reaction is:

Na(s) ⇒ Na+(aq) + e-

An oxidation half-reaction follows only the species that gets oxidized in a redox rection, and includes the electron or electrons that flow, as well.

The reduction half-reaction is:

Co3+(aq) + e- ⇒ Co2+(aq)

A reduction half-reaction follows only the species that gets reduced in a redox reaction, and includes the electron or electrons that flow, as well.

Eºred is the amount of voltage either required or liberated by the reduction half-reaction.

The more positive a species' reduction potential, the more strongly it will tend to be reduced in electrochemical cells.

Eºred = +1.82 V

In general, the reduction potential gives the amount of voltage released or required by a cell for the reduction reaction to occur. Since the Eºred > 0 for this cell, it is favorable to reduce Co3+, and energy is released when it happens.

Eºox = -1.82 V

An oxidation reaction is the reverse of a reduction reaction; in this case, the reverse of the reduction of Co3+. The oxidation potential Eºox is simply the reverse of Eºred. Since Eºox \< 0, the oxidation is unfavorable to occur, unless 1.82 V are supplied to oxidize this system.

The cell's full potential is:

Eºcell = Eºox + Eºred

Remember to keep track of the appropriate signs of the oxidation and reduction half-reactions; a classic way that chemistry tests will try to trick you is to give you a reduction potential, then ask about the oxidation potential for the same reaction.

Eºcell > 0 for any galvanic cell.

Since a galvanic cell is spontaneous, it must have a favorable voltage, which means it is positive in sign.

Eºcell \< 0 for any electrolytic cell.

Since an electrolytic cell is non-spontaneous, it must have an unfavorable voltage, which means it is negative in sign.

The cell's total potential is +4.53 V.

Since the cell is galvanic, its potential must be positive. So the two reactions must be combined (one oxidation, one reduction) to have an overall positive cell potential. In this case, that means that Co must be reduced (Eºred = +1.82V) while Na is oxidized (Eºox = -(-2.71V)).

So the total potential is
Eºcell = Eºred(Co) + Eºox(Na)
= 1.82 + 2.71 = 4.53 V

The cell's total potential is -4.53 V.

Note that the answer is exactly the inverse of these two cells arranged as a galvanic cell. This must be, since the electrolytic cell is just the galvanic cell run backwards. So Co is being oxidized, while Na is being reduced, and the total potential is:

Eºcell = Eºred(Na) + Eºox(Co)
= -2.71 + (-1.82) = -4.53 V

ΔGº = -nFEº

where
n = # of moles of electrons in the equation
F = 105 C/mol electrons

For an electrolytic cell, ΔGº > 0

The standard voltage Eº of an electrolytic cell is negative, which means ΔGº must be positive. This can also be derived from the fact that an electrolytic cell is non-spontaneous, and any non-spontaneous reaction has a positive ΔGº.

For a galvanic cell, ΔGº \< 0

The standard voltage Eº of a galvanic cell is positive, which means ΔGº must be negative. This can also be derived from the fact that a galvanic cell is spontaneous, and any spontaneous reaction has a negative ΔGº.

nFEº = RT ln(keq)

Where:
R = ideal gas constant in J/K*mol
T = temperature in K
n = # of moles of electrons
F = 105 C/mol electrons

For a galvanic cell, keq > 1.

For a galvanic cell, Eº is positive, so ln(keq) must be positive, hence keq > 1. This can also be derived from the fact that a galvanic cell is spontaneous, and for any spontaneous reaction, keq > 1.

For an electrolytic cell, keq \< 1.

For an electrolytic cell, Eº is negative, so ln(keq) must be negative, hence 0\eq\< 1. This can also be derived from the fact that an electrolytic cell is non-spontaneous, and for any non-spontaneous reaction, 0\eq\< 1.

Use the equation: nFE = -RT ln(Q/keq)

Note that this is similar to the equation for calculating the standard voltage. In fact, at standard conditions (where Q = 1), this equation becomes the calculation for standard voltage.

If Q \< keq, Ecell > 0.

If Q \< keq, the reaction has not yet reached equilibrium, and will move forwards to reach it. So, the reaction will proceed spontaneously, the cell is galvanic, and Ecell will be positive.

If Q > keq, Ecell \< 0.

If Q > keq, the reaction has exceeded equilibrium, and will proceed backwards to reach it. Since Ecell always refers to the forward reaction, Ecell will be negative.

If Q = keq, Ecell = 0.

If Q = keq, the reaction is at equilibrium, and will hold at its current state. Since neither the forward nor backward reaction will proceed, Ecell = 0.