Weak Acids

Weak acids are only partially ionized in aqueous solution.

There is a mixture of ions and un-ionized acid in solution.

Therefore, weak acids are in equilibrium:

HA(aq) + H2O(l) H3O+(aq) + A–(aq) Or:

HA(aq) H+(aq) + A–(aq)

We can write an equilibrium constant expression for this dissociation:

Ka =

or

Ka =

Ka is called the acid-dissociation constant.

Note that the subscript a indicates that this is the equilibrium constant for the dissociation of an acid.

Note that [H2O] is omitted from the Ka expression. (H2O is a pure liquid.)

The larger the Ka, the stronger the acid.

Ka is larger since there are more ions present at equilibrium relative to un-ionized molecules.

If Ka >> 1, then the acid is completely ionized, and the acid is a strong acid.

Weak Bases

Weak bases remove protons from substances.

There is an equilibrium between the base and the resulting ions:

Weak base + H2O(l) conjugate acid + OH–(aq) Example:

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq).

The base-dissociation constant, Kb, is defined as

[NH 4+][OH −] Kb

=

[NH 3]

The larger the Kb, the stronger the base.

Conjugate Acid-Base Pairs

Whatever is left of the acid after the proton is donated is called its conjugate base.

Similarly a conjugate acid is formed by adding a proton to the base.

Consider HA(aq) + H2O(l) H3O+(aq) + A–(aq):

HA and A– differ only in the presence or absence of a proton.

They are said to be a conjugate acid-base pair.

A– is called the conjugate base.

When HA (acid) loses its proton it is converted into A– (base).

Therefore HA and A– are a conjugate acid-base pair.

When H2O (base) gains a proton it is converted into H3O+ (acid). H3O+

is the conjugate acid.

Therefore, H2O and H3O+ are a conjugate acid-base pair.

Relative Strengths of Acids and Bases

The stronger an acid is, the weaker its conjugate base will be.

We can categorize acids and bases according to their behavior in water.

1. Strong acids completely transfer their protons to water.

No undissociated molecules remain in solution.

Their conjugate bases have negligible tendencies to become protonated.

Example: HCl.

2. Weak acids only partially dissociate in aqueous solution.

They exist in solution as a mixture of molecules and component ions.

Their conjugate bases show a slight tendency to abstract protons from water.

These conjugate bases are weak bases.

Example: Acetic acid is a weak acid; acetate ion (conjugate base) is a weak base.

3. Substances with negligible acidity do not transfer a proton to water.

Their conjugate bases are strong bases.

Example: CH4.

In every acid-base reaction, the position of the equilibrium favors the transfer of a proton from the stronger acid to the stronger base.

H+ is the strongest acid that can exist in equilibrium in aqueous solution.

OH– is the strongest base that can exist in equilibrium in aqueous solution.

Calculating Ka from pH

In order to find the value of the Ka, we need to know all of the equilibrium concentrations.

The pH gives the equilibrium concentration of H+.

Thus, to find Ka, we use the pH to find the equilibrium concentration of H+ and then the stoichiometric coefficients of the balanced equation to help us determine the equilibrium concentrations of the other species.

We then substitute these equilibrium concentrations into the equilibrium constant expression and solve for Ka.

Using Ka to Calculate pH

Using Ka, we can calculate the concentration of H+ ( and hence the pH ).

Write the balanced chemical equation clearly showing the equilibrium.

Write the equilibrium expression. Look up the value for Ka (in a table).

Write down the initial and equilibrium concentrations for everything except pure water.

It is customary to use x for the change in concentration of H+.

Substitute into the equilibrium constant expression and solve.

Remember to convert x to pH if necessary.

What do we do if we are faced with having to solve a quadratic equation in order to determine the value of x?

Often this cannot be avoided.

However, if the Ka value is quite small, we find that we can make a simplifying assumption.

Assume that x is negligible compared with the initial concentration of that species.

This will simplify the calculation.

It is always necessary to check the validity of any assumption.

Once you have the value of x, check to see how large it is compared with the initial concentration.

If x < 5% of the initial concentration, the assumption is probably a good one.

If x > 5%of the initial concentrtion, then it may be best to solve the quadratic equation or use successive approximations.

[H+]equilibrium

% ionization = ×100

[HA]initial

Weak acids are only partially ionized.

Percent ionization is another way to assess acid strength.

For the reaction HA(aq) H+(aq) + A–(aq)

Percent ionization relates the equilibrium H+ concentration, [H+]equilibrium, to the initial HA concentration, [HA]initial.

The higher the percent ionization, the stronger the acid.

However, we need to keep in mind that percent ionization of a weak acid decreases as the molarity of the solution increases.

Acid-Base Titrations

The plot of pH versus volume during a titration is called a titration curve.

Strong Acid–Strong Base Titrations

Consider adding a strong base (e.g., NaOH) to a solution of a strong acid (e.g.,

HCl).

We can divide the titration curve into four regions.

1. Initial pH (before any base is added).

The pH is determined by the concentration of the strong acid solution.

Therefore, pH < 7.

2. Between the initial pH and the equivalence point (see next slide).

When base is added, before the equivalence point, the pH is given by the

amount of strong acid in excess.

Therefore, pH < 7.

3. At the equivalence point.

The amount of base added is stoichiometrically equivalent to the amount of acid originally present.

Therefore, the pH is determined by the hydrolysis of the salt in solution.

Therefore, pH = 7.

4. After the equivalence point.

The pH is determined by the excess base in the solution.

Therefore, pH > 7.

How can we analyze the titration (i.e., how will we know when we are at the

equivalence point)?

Consider adding a strong base (e.g., NaOH) to a solution of a strong acid (

e.g., HCl ).

We know that the pH at the equivalence point is 7.00.

To detect the equivalence point, we use an indicator that changes color

somewhere near 7.00.

Usually, we use phenolphthalein, which changes color between pH 8.3 and

10.0.

In acid, phenolphthalein is colorless.

As NaOH is added, there is a slight pink color at the addition point.

When the flask is swirled and the reagents mix, the pink color disappears.

At the end point, the solution is light pink.

If more base is added, the solution turns darker pink.

The equivalence point in a titration is the point at which the acid and base

are present in stoichiometric quantities.

The end point in a titration is the point where the indicator changes color.

The difference between equivalence point and end point is called the titration error.

The shape of a strong base–strong acid titration curve is very similar to a strong

acid–strong base titration curve.

Initially, the strong base is in excess, so the pH > 7.

As acid is added, the pH decreases but is still greater than 7.

At the equivalence point, the pH is given by hydrolysis of the salt solution (

i.e., pH = 7).

After the equivalence point, the pH is given by the strong acid in excess, so

the pH < 7.

Weak Acid–Strong Base Titration

Consider the titration of acetic acid, HC2H3O2, with NaOH.

Again, we divide the titration into four general regions:

1. Before any base is added :

The solution contains only weak acid.

Therefore, pH is given by the equilibrium calculation.

2. Between the initial pH and the equivalence point.

As strong base is added it consumes a stoichiometric quantity of weak

acid:

HC2H3O2(aq) + OH–(aq) C2H3O2–(aq) + H2O(l)

However, there is an excess of acetic acid.

Therefore, we have a mixture of weak acid and its conjugate base.

Thus the composition of the mixture is that of a buffer.

The pH is given by the buffer calculation.

First, the amount of C2H3O2– generated is calculated, as well

as the amount of HC2H3O2 consumed. (stoichiometry.) Then

the pH is calculated using equilibrium conditions.

(Henderson-Hasselbach equation)

3. At the equivalence point, all the acetic acid has been consumed and all the

NaOH has been consumed.

However, C2H3O2– has been generated.

Therefore, the pH is given by the C2H3O2– solution.

This means pH > 7.

More importantly, the pH of the equivalence point > 7 for a weak acid–

strong base titration.

4. After the equivalence point:

The pH is given by the concentration of the excess strong base.

The pH curve for a weak acid–strong base differs significantly from that of a strong acid–strong base titration.

For a strong acid–strong base titration:

The pH begins at less than 7 and gradually increases as base is added.

Near the equivalence point, the pH increases dramatically.

For a weak acid–strong base titration:

The initial pH rise is steeper than that for a strong acid–strong base

titration.

However, then there is a leveling off due to buffer effects.

The inflection point is not as steep for a weak acid–strong base titration.

The shape of the two curves after the equivalence point is the same because

pH is determined by the strong base in excess.

The pH values at the equivalence points differ also:

The pH = 7.00 for the strong acid–strong base equivalence point.

The pH >7.00 for the weak acid–strong base equivalence point.

How to choose indicators: we select the appropriate indicator based upon the pH

of the salt solution formed at the equivalence point.

The pH curve for the titration of a weak base with a strong acid also differs significantly from that of a strong base-strong acid titration.

Consider the titration of NH3 with HCl.

The equivalence point occurs at pH 5.28 so phenolphthalein should not be used for this titration.

The color change for methyl red occurs in the pH range from 4.2 to 6.0 so it is a good indicator to use for this titration.

Acid-Base Properties of Salt

Solutions

Nearly all salts are strong electrolytes.

Therefore, salts in solution exist entirely as ions.

Acid-base properties of salts are a consequence of the reaction of their ions in solution.

Many salt ions can react with water to form OH– or H+.

This process is called hydrolysis.

Anions from weak acids are basic.

Anions from strong acids are neutral.

Anions with ionizable protons (e.g., HSO4–) are amphoteric.

They are capable of acting as an acid or a base.

All cations, except those of the alkali metals or heavier alkaline earth metals, are weak acids.

The pH of a solution may be qualitatively predicted using the following

guidelines:

Salts derived from a strong acid and a strong base are neutral.

Examples: NaCl, Ca(NO3)2.

Salts derived from a strong base and a weak acid are basic.

Examples: NaClO, Ba(C2H3O2)2.

Salts derived from a weak base and a strong acid are acidic.

Example: NH4Cl.

Salts derived from a weak acid and a weak base can be either acidic or basic.

Equilibrium rules apply!

We need to compare Ka and Kb for the hydrolysis of the anion and the cation.

For example, consider NH4CN.

Both ions undergo significant hydrolysis.

Is the salt solution acidic or basic?

The Ka of NH4+ is smaller than the Kb of CN–, so the solution should be

basic.

Buffered Solutions

A buffered solution or buffer is a solution that resists a change in pH upon addition of

small amounts of acid or base.

Composition and Action of Buffered Solutions

A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X–).

Thus a buffer contains both:

An acidic species (to neutralize OH–) and A

basic species (to neutralize H+).

When a small amount of OH– is added to the buffer, the OH– reacts with HX to produce X–

and water.

But the [HX]/[X–] ratio remains more or less constant, so the pH is not significantly

changed.

When a small amount of H+ is added to the buffer, X– is consumed to produce HX.

Once again, the [HX]/[X–] ratio remains more or less constant, so the pH does not

change significantly.

Buffer capacity is the amount of acid or base that can be neutralized by the

buffer before there is a significant change in pH.

Buffer capacity depends on the concentrations of the components of the buffer.

The greater the concentrations of the conjugate acid-base pair, the greater the

buffer capacity.

The pH of the buffer is related to the Ka and on the relative concentrations of the

acid and base.

We can derive an equation that shows the relationship between conjugate

acidbase concentrations, pH, and the Ka.

By definition:

Ka =

Rearranging, we get:

[H+]=Ka[HA− ]

[A ]

If we take the negative log of each side of the equation we get:

− log[H+ ] =−log Ka − log

By definition:

pH = pKa −log

An alternative form of this equation is:

The preceding equation is the Henderson-Hasselbach equation.

Note that this equation uses equilibrium concentrations of acid and conjugate base.

However, if Ka is sufficiently small (i.e., if the equilibrium concentration of undissociated acid is close to the initial concentration), then we can use the initial values of the acid and base concentrations in order to get a good estimate of the pH.

Addition of Strong Acids or Bases to Buffers

We break the calculation into two parts:

A stoichiometric calculation.

An equilibrium calculation.

The addition of strong acid or base results in a neutralization reaction:

X– + H3O+ ↔ HX + H2O

[ ]

[ ]

[ ]

[ ] acid

base log pK

HA

A log pK pH a a + = + =

−

HX + OH– ↔ X– + H2O

By knowing how much H3O+ or OH– was added, we know how much HX or X– is formed.

This is the stoichiometric calculation.

From the concentrations of HX and X– (note the change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation:

pH = pKa + log

This is the equilibrium calculation.

Experiment 7: Hydrolysis and Buffers

A. Determination of acidity

Salt Solution Hydrolysis reaction pH Explanation

0.1M NH4OAc NH4

+ + H2O → NH4OH + H+

OAc- + H2O → HOAc + OH- ~7

Both ions hydrolyze water to the

same extent (equivalent Ka and Kb)

0.1M NaCl NR 7 Neither ion hydrolyzes H2O

0.1M NaOAc OAc- + H2O → HOAc + OH- >7 OAc- hydrolyze water

0.1M NH4Cl NH4+ + H2O → NH4OH + H+ <7 NH4

+ hydrolyze water

0.1M NaHCO3 HCO3- + H2O → H2CO3 + OH- >7 HCO3

- hydrolyze water

0.1M Na2CO3

CO32 + H2O → HCO3

- + OH-

HCO3- + H2O → H2CO3 + OH-

>7 CO32- hydrolyze water

Solution Color with Methyl

orange

Color with

Phenolphthalein

0.1M HOAc Red Colorless

0.1M HOAc + 0.1M NaOAc Red Colorless

0.1M NH4OH Yellow Red

0.1M NH4OH + 0.1M NH4Cl Yellow Red

Buffer solutions

Color with methyl orange Color with phenolphthalein

+ HCl + NaOH + HCl + NaOH

Weak acid

HOAc Red X Colorless X

HOAc X Yellow X Pink

Buff er of weak acid and it s salt

HOAc + NaOAc Red X Colorless X

HOAc + NaOAc X Red / yellowish

red X

Colorless / light

pink

Weak base

NH4OH Yellow X Colorless X

NH4OH X Red X Pink

Buff er of weak base and it s salt

NH4OH + NH4Cl Yellow / Reddish

yellow X Light pink X

NH4OH + NH4Cl X Yellow X Pink

Answers to questions

1. A little bit because the presence of the salt lowered the concentration of the

H+ or the OH- due to common ion effect.

2. There is a great change in color upon addition of a neutralizing reagent to

the unbuffered solutions. Due to the presence of the salt and the common

ion effect, the most of the weak electrolyte will be in unionized form.

Upon addition of a neutralizing reagent, the unionized form of the weak

electrolyte will dissociate up until the H+ or OH- in solution are

replenished.

3. Buffer solutions are mixtures of solutions of weak electrolytes and the salts

of their conjugate ions. They resist pH changes as explained in #2.

4. pH before addition of HCl:

pH = pKa(HOAc) + log ([OAc-] ÷[ HOAc ]) pH = -log(1.75 × 10-

5) + log ((0.05L × 0.1M) ÷(0.05L × 0.1M)) pH after addition

of 0.01mole HCl:

pH = -log(1.75 × 10-5) + log (0.05L × 0.1M) – 0.01mole) ÷(0.05L × 0.1 M )+ 0.01mole ))

Oxidation-Reduction Reactions

Chemical reactions in which the oxidation state of one or more substances changes are called oxidation-reduction reactions (redox reactions).

Recall: OILRIG

Oxidation involves loss of electrons (OIL).

Reduction involves gain of electrons (RIG).

Electrochemistry is the branch of chemistry that deals with relationships between electricity and chemical reactions.

Consider the spontaneous reaction that occurs when Zn is added to HCl.

Zn(s) + 2H+(aq) →

Zn2+(aq) + H2(g) The

oxidation numbers of Zn and H+ have changed.

The oxidation number of Zn has increased from 0 to +2.

The oxidation number of H has decreased from +1 to 0.

Therefore, Zn is oxidized to Zn2+ while H+ is reduced to H2.

H+ causes Zn to be oxidized. Thus, H+ is the oxidizing agent or oxidant.

Zn causes H+ to be reduced. Thus, Zn is the reducing agent or reductant.

Note that the reducing agent is oxidized, and the oxidizing agent is reduced.

Balancing Oxidation-Reduction

Equations

Recall the law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end.

Conservation of charge: electrons are not lost in a chemical reaction.

Some redox equations may be easily balanced by inspection.

However, for many redox reactions, we need to look carefully at the transfer of electrons.

Half-reactions are a convenient way of separating oxidation and

reduction reactions. Consider the reaction:

Sn2+(aq) + 2Fe3+(aq) →

Sn4+(aq) + 2Fe2+(aq)

The oxidation half-reaction is:

Sn2+(aq) →

Sn4+(aq) +2e–

Note that electrons are a product here.

The reduction half-reaction is:

2Fe3+(aq) + 2e– → 2Fe2+(aq)

Note that electrons are a reactant here.

Balancing Equations by the Method of Half-Reactions (in acidic medium)

Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colorless) with KMnO4 ( deep purple ).

MnO4– is reduced to Mn2+ (pale pink) while the C2O4

2– is oxidized to CO2.

The equivalence point is given by the presence of a pale pink color.

If more KMnO4 is added, the solution turns purple due to the excess

KMnO4.

What is the balanced chemical equation for this reaction?

We can determine this using the method of half-reactions:

Write down the two incomplete half-reactions:

MnO4–(aq)

→ Mn2+(aq)

C2O42–(aq)

→ CO2(g)

Balance each half-reaction:

First, balance elements other than H and O.

MnO4–(aq)

→ Mn2+(aq)

C2O42–(aq)

→ 2CO2(g)

Then balance O by adding water.

MnO4–(aq)

→ Mn2+(aq) + 4H2O(l) C2O4

2–

(aq) →

2CO2(g)

Then balance H by adding H+.

8H+(aq) + MnO4–(aq)

→ Mn2+(aq) + 4H2O(l)

C2O42–(aq)

→ 2CO2(g)

Finish by balancing charge by adding electrons.

This is an easy place to make an error!

For the permanganate half-reaction, note that there is a charge of 7+ on the left and 2+ on the right.

Therefore, 5 electrons need to be added to the left:

5e– + 8H+(aq) + MnO4–(aq)

→ Mn2+(aq) + 4H2O(l)

In the oxalate reaction, there is a 2- charge on the left and a 0 charge on the right, so we need to add two electrons:

C2O42–(aq)

→ 2CO2(g) + 2e–

Multiply each half reaction to make the number of electrons gained equal to the number of electrons lost.

To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplication gives:

10e– + 16H+(aq) + 2MnO4–(aq)

→ 2Mn2+(aq) + 8H2O(l)

5C2O42–(aq)

→ 10CO2(g) + 10e–

Now add the reactions and simplify.

16H+(aq) + 2MnO4–(aq) + 5C2O4

2–(aq) →

2Mn2+(aq) + 8H2O(l) + 10CO2(g)

The equation is now balanced!

Confirm by checking that there are equal numbers of each

atom on both sides of the reaction (law of conservation of mass). And confirm that both sides have equal charge (law

of conservation of charge).

Note that all the electrons have canceled out!

Balancing Equations for Reactions Occurring in Basic Solution

We use OH– and H2O rather than H+ and H2O.

The same method is used as in an acid solution, but OH– is added to “neutralize” the H+ used.

The equation must again be simplified by canceling like terms on both sides of the equation.

Voltaic Cells

The energy released in a spontaneous redox reaction may be used to perform electrical work.

Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit.

Voltaic cells utilize spontaneous reactions.

If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+.

Zn(s) + Cu2+(aq) →

Zn2+(aq) + Cu(s)

Zn is spontaneously oxidized to Zn2+ by Cu2+.

The Cu2+ is spontaneously reduced to Cu0 by Zn.

The entire process is spontaneous.

This voltaic cell consists of:

An oxidation half-reaction:

Zn(s) →

Zn2+(aq) + 2e–

Oxidation takes place at the anode.

A reduction half-reaction:

Cu2+(aq) + 2e– → Cu(s)

Reduction takes place at the cathode.

A salt bridge (used to complete the electrical circuit).

Cations move from anode to cathode.

Anions move from cathode to anode.

The two solid metals are the electrodes ( cathode and anode ).

As oxidation occurs, Zn is converted to Zn2+ and 2e–.

The electrons flow toward the cathode, where they are used in the reduction reaction.

We expect the Zn electrode to lose mass and the Cu electrode to gain mass.

Electrons flow from the anode to the cathode.

Therefore, the anode is negative and the cathode is positive.

Electrons cannot flow through the solution; they have to be transported through an external wire.

Anions and cations move through a porous barrier or salt bridge.

Cations move into the cathodic compartment to neutralize the excess negatively charged ions

(Cathode: Cu2+ + 2e– → Cu, so the counter ion of Cu is in excess).

Anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by oxidation.

Cell EMF

The flow of electrons from anode to cathode is spontaneous.

What is the “driving force”?

Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode.

Potential difference: difference in electrical potential.

The potential difference is measured in volts.

One volt (V) is the potential difference required to impart one joule ( J) of energy to a charge of one coulomb (C ):

1V = 1 J / C

Electromotive force (emf) is the force required to push electrons through the external circuit.

Cell potential: Ecell is the emf of a cell.

This is known as the cell voltage.

Ecell > 0 for a spontaneous reaction.

For 1 M solutions or 1 atm pressure for gases at 25°C (standard conditions), the standard emf (standard cell potential) is called E°cell.

For example, for the reaction:

Zn(s) + Cu2+(aq) →

Zn2+(aq) + Cu(s)

E°cell = +1.10 V.

Standard Reduction Potentials

We can conveniently tabulate electrochemical data.

Standard reduction potentials, E°red, are measured relative to a standard.

The emf of a cell can be calculated from standard reduction potentials:

E°cell = E°red(cathode) - E°red( anode )

We use the following half-reaction as our standard:

2H+(aq, 1M) + 2e– → H2(g, 1 atm) E°cell = 0 V.

This electrode is called a standard hydrogen electrode (SHE).

The SHE is assigned a standard reduction potential of zero.

Consider the half-reaction:

Zn(s) →

Zn2+(aq) + 2e–

We can measure E°cell relative to the SHE.

In this cell the SHE is the cathode.

It consists of a Pt electrode in a tube placed in 1 M H+ solution. H2

is bubbled through the tube.

E°cell = E°red(cathode) - E°red( anode )

0.76 V = 0 V - E°red( anode ).

Therefore E°red( anode) = -0.76 V.

Standard reduction potentials must be written as reduction reactions:

Zn2+(aq, 1 M) + 2e– → Zn(s) E°red = -0.76 V.

Since E°red = -0.76 V, we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous.

However, the oxidation of Zn with the SHE is spontaneous.

The standard reduction potential is an intensive property.

Therefore, changing the stoichiometric coefficient does not affect

E°red.

2Zn2+(aq) + 4e– → 2Zn(s) E°red = -0.76 V

Reactions with E°red > 0 are spontaneous reductions relative to the SHE.

Reactions with E°red < 0 are spontaneous oxidations relative to the SHE.

The larger the difference between E°red values, the larger the E°cell.

The more positive the value of E°red, the greater the driving force for reduction.

Oxidizing and Reducing Agents

Consider a table of standard reduction potentials.

We can use this table to determine the relative strength of reducing (and oxidizing ) agents.

The more positive the E°red, the stronger the oxidizing agent (written in the table as a reactant).

The more negative the E°red, the stronger the reducing agent (written as a product in the table).

We can use this table to predict if one reactant can spontaneously oxidize another.

Example:

F2 can oxidize H2 or Li. Ni2+

can oxidize Al(s).

We can use this table to predict if one reactant can spontaneously reduce another.

Example:

Li can reduce F2.

Spontaneity of Redox Reactions

For any electrochemical process

E° = E°red(reduction process) - E°red(oxidation process).

A positive E° indicates a spontaneous process (voltaic cell).

A negative E° indicates a non-spontaneous process.

This equation is used to understand the activity series of metals.

Consider the reaction of nickel with silver ion:

Ni(s) + 2Ag+(aq) →

Ni2+(aq) + 2Ag(s)

The standard cell potential is:

E°cell = E°red(Ag+/Ag) - E°red(Ni2+/Ni) =

(0.80 V) - (-0.28 V)

= 1.08 V

This value indicates that the reaction is spontaneous.

EMF and Free-Energy Change

We can show that:

∆G = -nFE

Where ∆G is the change in free energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell.

We define: C J

F = 96,500 mol e- = 96,500 (V) (mol e- )

Since n and F are positive, if ∆G < 0, then E > 0, and the reaction will be spontaneous.

Effect of Concentration on Cell

EMF A voltaic cell is functional until E = 0, at which point equilibrium has been reached.

The cell is then “dead”.

The point at which E = 0 is determined by the concentrations of the species

involved in the redox reaction.

We can calculate the cell potential under nonstandard conditions.

Recall that: ∆G = ∆G° + RT ln Q

We can substitute in our expression for the free energy change:

-nFE = -nFE° + RT ln Q

Rearranging, we get the Nernst equation:

Or

Note the change from natural logarithm to log base-10.

The Nernst equation can be simplified by collecting all the constants together using a temperature of 298 K:

0.0592

E = E°− log Q

n

Example: If we have the reaction:

Q nF

RT E E ln − ° =

Q nF

RT E E log

303 2. − ° =

Zn(s) + Cu2+(aq) →

Zn2+(aq) + Cu(s)

If [Cu2+] = 5.0 M, and [Zn2+] = 0.050 M:

0 .

Ecell = 1.10V −= 1.16V

Cell EMF and Chemical Equilibrium

A system is at equilibrium when ∆G = 0.

From the Nernst equation, at equilibrium;

nE°

logK =

.0592

Thus, if we know the cell emf, we can calculate the equilibrium constant.

Experiment 8: Electrochemistry

Spontaneous redox reactions occur in Galvanic cells producing electricity in the process

∆G = –nFEcell

Ecell = Ecathode – Eanode

= Ereduction – Eoxidation

= Epositive – Enegative

For a redox reaction to be spontaneous: Ecell > 0

When ∆G = 0, Ecell = 0, the cell is in equilibrium (that is, dead )

Non-spontaneous redox reactions (Ecell < 0) are made to occur in electrolytic cells by applying electrical work (e.g. recharging batteries)