chapter 19...salt bridge -- a device used to maintain electrical neutrality in a galvanic cell. this...

Electrochemistry

Chapter 19

Electrochemistry—the study of the interchange of chemical and

electrical energy

OIL RIG – oxidation is loss, reduction is gain (of electrons)

Oxidation – the loss of electrons, increase in charge

Reduction – the gain of electrons, reduction of charge

Oxidation number – the assigned charge on an atom

Oxidizing agent (OA) – the species that is reduced and thus

causes oxidation

Reducing agent (RA) – the species that is oxidized and thus

causes reduction

Vocabulary

ELECTROCHEMISTRY INVOLVES

TWO MAIN TYPES OF PROCESSES:

A. Galvanic (voltaic) cells – which are spontaneous

chemical reactions (battery)

B. Electrolytic cells – which are non-spontaneous

and require external e− source (DC power source)

C. BOTH of these fit into the category entitled

Electrochemical cells

2Mg (s) + O2 (g) 2MgO (s)

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2-

Oxidation half-reaction (lose e-)

Reduction half-reaction (gain e-)

19.1

Electrochemical processes are oxidation-reduction reactions

in which:

• the energy released by a spontaneous reaction is

converted to electricity or

• electrical energy is used to cause a nonspontaneous

reaction to occur

0 0 2+ 2-

Oxidation number

The charge the atom would have in a molecule (or an

ionic compound) if electrons were completely transferred.

1. Free elements (uncombined state) have an oxidation

number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

2. In monatomic ions, the oxidation number is equal to

the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2

3. The oxidation number of oxygen is usually –2. In H2O2

and O22- it is –1.

4.4

4. The oxidation number of hydrogen is +1 except when

it is bonded to metals in binary compounds. In these

cases, its oxidation number is –1.

6. The sum of the oxidation numbers of all the atoms in a

molecule or ion is equal to the charge on the

molecule or ion.

5. Group IA metals are +1, IIA metals are +2 and fluorine is

always –1.

HCO3-

O = -2 H = +1

3x(-2) + 1 + ? = -1

C = +4

Oxidation numbers of all

the atoms in HCO3- ?

4.4

Balancing Redox Equations

19.1

1. Write the unbalanced equation for the reaction ion ionic form.

The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution?

Fe2+ + Cr2O72- Fe3+ + Cr3+

2. Separate the equation into two half-reactions.

Oxidation:

Cr2O72- Cr3+

+6 +3

Reduction:

Fe2+ Fe3++2 +3

3. Balance the atoms other than O and H in each half-reaction.

Cr2O72- 2Cr3+


4. For reactions in acid, add H2O to balance O atoms and H+ to

balance H atoms.

Cr2O72- 2Cr3+ + 7H2O

14H+ + Cr2O72- 2Cr3+ + 7H2O

5. Add electrons to one side of each half-reaction to balance the

charges on the half-reaction.

Fe2+ Fe3+ + 1e-

6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

6. If necessary, equalize the number of electrons in the two half-

reactions by multiplying the half-reactions by appropriate

coefficients.

6Fe2+ 6Fe3+ + 6e-

6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

19.1


7. Add the two half-reactions together and balance the final

equation by inspection. The number of electrons on both

sides must cancel.

6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

6Fe2+ 6Fe3+ + 6e-Oxidation:

Reduction:

14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O

8. Verify that the number of atoms and the charges are balanced.

14x1 – 2 + 6x2 = 24 = 6x3 + 2x3

19.1

9. For reactions in basic solutions, add OH- to both sides of the

equation for every H+ that appears in the final equation.

Anode--the electrode where oxidation occurs. After a period of time, the

anode may appear to become smaller as it falls into solution.

Cathode-- the anode where reduction occurs. After a period of time it may

appear larger, due to ions from solution plating onto it.

Inert electrodes -- used when a gas is involved OR ion to ion involved such

as Fe3+ being reduced to Fe2+ rather than Fe0. Made of Pt or graphite.

Salt bridge -- a device used to maintain electrical neutrality in a galvanic cell.

This may be filled with agar which contains a neutral salt or it may be

replaced with a porous cup.

Electron flow -- always from anode to cathode. (through the wire)

Standard cell notation (line notation) - anode/solution// cathode solution/

cathode Ex. Zn/Zn2+ (1.0 M) // Cu2+ (1.0 M) / Cu

Voltmeter - measures the cell potential (emf) . Usually is measured in volts.

GALVANIC CELLS

Parts of the voltaic or galvanic cell:

MnO4- + Fe2+ → Mn2+ + Fe3+ [acidic]

Why separate beakers?

Why connected by wire?

What do we need?

-Salt Bridge

-Cell potential -Volts (V) J/Coulomb)

-Volt meter

Galvanic Cells

19.2

spontaneous

redox reaction

anode

oxidation

cathode

reduction

Galvanic Cells

19.2

The difference in electrical

potential between the anode

and cathode is called:

• cell voltage

• electromotive force (emf)

• cell potential

Cell Diagram

Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)

[Cu2+] = 1 M & [Zn2+] = 1 M

Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)

anode cathode

Standard Reduction Potentials

19.3

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

2e- + 2H+ (1 M) H2 (1 atm)

Zn (s) Zn2+ (1 M) + 2e-Anode (oxidation):

Cathode (reduction):

Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)


19.3

Standard reduction potential (E0) is the voltage associated

with a reduction reaction at an electrode when all solutes

are 1 M and all gases are at 1 atm.

E0 = 0 V

Standard hydrogen electrode (SHE)

2e- + 2H+ (1 M) H2 (1 atm)

Reduction Reaction

19.3

E0 = 0.76 Vcell

Standard emf (E0 )cell

0.76 V = 0 - EZn /Zn0

2+

EZn /Zn = -0.76 V02+

Zn2+ (1 M) + 2e- Zn E0 = -0.76 V

E0 = EH /H - EZn /Zncell0 0

+ 2+2


E0 = Ecathode - Eanodecell0 0

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

STANDARD REDUCTION POTENTIALS• Each half-reaction has a cell potential

• Each potential is measured against a standard which is the

standard hydrogen electrode [consists of a piece of inert

Platinum that is bathed by hydrogen gas at 1 atm]. The

hydrogen electrode is assigned a value of ZERO volts.

• Standard conditions—1 atm for gases, 1.0M for solutions

and 25°C for all (298 K)

• naught, ° -- we use the naught to symbolize standard

Conditions. That means Ecell , Emf , or εcell become Ecello , Emf

o ,

or εcello when measurements are taken at standard conditions.

The diagram to the right

illustrates what really

happens when a Galvanic

cell is constructed from zinc

sulfate and copper(II) sulfate

using the respective metals

as electrodes.

Notice that 1.0 M solutions of each salt

are used

Notice an overall voltage of 1.10 V for

the process


19.3

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)

2e- + Cu2+ (1 M) Cu (s)

H2 (1 atm) 2H+ (1 M) + 2e-Anode (oxidation):


H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)


E0 = 0.34 Vcell

Ecell = ECu /Cu – EH /H 2+ +2

0 0 0

0.34 = ECu /Cu - 002+

ECu /Cu = 0.34 V2+0

19.3

• E0 is for the reaction as

written

• The more positive E0 the

greater the tendency for the

substance to be reduced

• The half-cell reactions are

reversible

• The sign of E0 changes

when the reaction is

reversed

• Changing the stoichiometric

coefficients of a half-cell

reaction does not change

the value of E0

Elements that have the most positive reduction

potentials are easily reduced (in general, non-metals)

Elements that have the least positive reduction potentials

are easily oxidized (in general, metals)

The table can also be used to tell the strength of various

oxidizing and reducing agents.

Can also be used as an activity series. Metals having

less positive reduction potentials are more active and will

replace metals with more positive potentials.

HOW CAN WE DETERMINE WHICH SUBSTANCE IS

BEING REDUCED AND WHICH IS BEING OXIDIZED??

The MORE POSITIVE reduction potential gets to be

reduced - IF you are trying to set up a cell that can act as

a battery.

How to Calculate “Standard Cell Potential”, Symbolized

by E°cell OR Emf° OR εcell° (terms are interchangable)

1. Decide which element is oxidized or reduced using the table of

reduction potentials.

Remember: THE MORE POSITIVE REDUCTION POTENITAL

GETS TO BE REDUCED.

2. Write both equations AS IS from the chart with their voltages.

3. Reverse the equation that will be oxidized and change the sign of

the voltage - This is now: E°oxidation

4. Balance the two half reactions **do not multiply voltage values**

5. Add the two half reactions and the voltages together.

6. E°cell = E°oxidation + E°reduction (where ° means standard conditions: 1atm, 1M, 25°C)

▪AN OX – oxidation occurs at the anode (may show mass decrease)

▪RED CAT – reduction occurs at the cathode (may show mass increase)

▪FAT CAT – The electrons in a voltaic or galvanic cell ALWAYS flow

“From the Anode To the CAThode”

▪Ca+hode – the cathode is + in galvanic (voltaic) cells

▪EPA--in an electrolytic cell, there is a positive (+) anode.

▪Salt Bridge – bridge between cells whose purpose is to provide ions

to balance the charge. Usually made of a salt filled agar (KNO3) or a

porous cup.

Neumonic Devices - Handy when constructing a

“spontaneous cell” --one that can act as a battery:

What is the standard emf of an electrochemical cell made

of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr

electrode in a 1.0 M Cr(NO3)3 solution?

Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V

Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V

Cd is the stronger oxidizer

Cd will oxidize Cr

2e- + Cd2+ (1 M) Cd (s)

Cr (s) Cr3+ (1 M) + 3e-Anode (oxidation):


2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)

x 2

x 3


E0 = -0.40 – (-0.74) cell

E0 = 0.34 V cell

19.3

Sample Problem:

a) Calculate the cell voltage for the following reaction.

b) Draw a diagram of the galvanic cell for the reaction and

label completely.

Fe3+ (aq) + Cu(s) → Cu2+ (aq) + Fe2+ (aq)

Exercise 2

a) Calculate the cell voltage for the galvanic cell that would

utilize silver metal and involve iron(II) ion and iron(III) ion.

b) Draw a diagram of the galvanic cell for the reaction and

label completely. Given: Fe3+ + e-

Fe2+ Eo = 0.77 V

Ag+ + e- Ag(s) Eo = 0.80 V

E°cell= 0.03 V

Reaction Quotient• Q: equal to equilibrium expression but does not have to

be at equilibrium

• used to tell if a reaction is at equilibrium or not

• relationship between Q and K tells which way the

reaction will shift

– Q=K: at equilibrium, no shift

– Q > K: too large, forms reactants, shift to left

– Q < K: too small, forms products, shift to right

A B

13.1

rate = -D[A]

Dt

rate = D[B]

Dt

time

Law of Mass Action

• created in 1864 by Guldberg and Waage

(Norweigen)

• For a reaction: jA + kB ⇄ lC + mD

– equilibrium constant: K

Law of Mass Action

• because

– Ratef = kf[A]j[B]k

– Rater = kr[C]l[D]m

– so if Ratef = Rater

– kf[A]j[B]k = kr[C]l[D]m

19.4

Spontaneity of Redox Reactions

DG = -nFEcell

DG0 = -nFEcell0

n = number of moles of electrons in reaction

F = 96,500J

V • mol = 96,500 C/mol

DG0 = -RT ln K = -nFEcell0

Ecell0 =

RT

nFln K

(8.314 J/K•mol)(298 K)

n (96,500 J/V•mol)ln K=

=0.0257 V

nln KEcell

0

=0.0592 V

nlog KEcell

0

Spontaneity of Redox Reactions

19.4

DG0 = -RT ln K = -nFEcell0

2e- + Fe2+ Fe

2Ag 2Ag+ + 2e-Oxidation:

Reduction:

What is the equilibrium constant for the following reaction

at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)

=0.0257 V

nln KEcell

0

19.4

E0 = -0.44 – (0.80)

E0 = -1.24 V

0.0257 V

x nE0cell

expK =

n = 2

0.0257 V

x 2-1.24 V= exp

K = 1.23 x 10-42

E0 = EFe /Fe – EAg /Ag0 0

2+ +

The Effect of Concentration on Cell Emf

DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE 0

-nFE = -nFE0 + RT ln Q

E = E0 - ln QRT

nF

Nernst equation

At 298

19.5

-0.0257 V

nln QE0E = -

0.0592 Vn

log QE0E =

CELL POTENTIAL, ELECTRICAL WORK & FREE ENERGY

Combining thermodynamics and electrochemistry, and a bit of

physics:

• The work that can be accomplished when electrons are

transferred through a wire depends on the“push” or emf which

is defined in terms of a potential difference [in volts] between

two points in the circuit.

(V = J/C)

Thus “one joule of work” is produced [or required] when one

coulomb of charge is transferred between two points in the

circuit that differ by a potential of one volt- IF work flows OUT it is assigned a MINUS sign

When a cell produces a current, the cell potential is positive

and the current can be used to do work THEREFORE “ε”and “work” have opposite signs!

A “faraday” (F) - the charge on one MOLE of electrons =

96,485 coulombs (AP Exam uses 96,500) And: q = # moles of

electrons × F

For a process carried out at constant temperature and

pressure, wmax [neglecting the very small amount of energy that is lost as friction or heat] is

equal to ΔG, therefore….

ΔGo = −nFEo

G = Gibb’s free energy

n = number of moles of electrons

F = Faraday constant 9.6485309 × 104 J/V • mol

So it follows that:

−Eo implies nonspontaneous.

+Eo implies spontaneous (would be a good battery!)

.

Will the following reaction occur spontaneously at 250C if

[Fe2+] = 0.60 M and [Cd2+] = 0.010 M?

Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)

2e- + Fe2+ Fe

Cd Cd2+ + 2e-Oxidation:

Reduction:n = 2

E0 = -0.44 – (-0.40)

E0 = -0.04 V

E0 = EFe /Fe – ECd /Cd0 0

2+ 2+

-0.0257 V

nln QE0E =

-0.0257 V

2ln -0.04 VE =

0.010

0.60

E = 0.013

E > 0 Spontaneous

19.5

Bell Work - 9-27-2013 (Exercise 3)Using the table of standard reduction potentials, calculate ΔG° for the

reaction

Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq)

Exercise 4Using the table of standard reduction potentials, predict whether 1 M

HNO3 will dissolve gold metal to form a 1 M Au3+ solution.

DEPENDENCE OF CELL POTENTIAL ON CONCENTRATION

Voltaic cells at NON-standard conditions: LeChatlier’s

principle can be applied. An increase in the concentration

of a reactant will favor the forward reaction and the cell

potential will increase. (The converse is also true!)

Exercise 5 - For the cell reaction:

2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) E°cell = n/a

predict whether Ecell is larger or smaller than E°cell for the

following cases.

a. [Al3+ ] = 2.0 M, [Mn2+ ] = 1.0 M

b. [Al3+ ] = 1.0 M, [Mn2+] = 3.0 M

Bell Work Oct 19, 2013

For the cell reaction:Al(s) + Mn2+(aq) → Al3+ (aq) + Mn(s)

1) Balance the equation

2) Calc E°cell

3) Predict whether Ecell is larger or smaller than E°cell for the following

cases.

a. [Al3+ ] = 2.0 M, [Mn2+ ] = 1.0 M

b. [Al3+ ] = 1.0 M, [Mn2+] = 3.0 M

4) Calc. the Ecell for each of the above solutions.

5) Calc ΔG° for the “std conditions rxn” (#2) and ΔG for each of the

different concentrations from #3

For a more quantitative approach….. When cell is not at

standard conditions, use Nernst Equation:

R = Gas constant 8.315 J/K• mol

F = Faraday constant

Q = reaction quotient [products]coefficient/[reactants]coefficient

E = Energy produced by reaction

T = Temperature in Kelvins

n = # of electrons exchanged in BALANCED redox equation

As E declines with reactants converting to products, E

eventually reaches zero.

Zero potential means reaction is at equilibrium [dead

battery]. Also, Q = K AND ΔG = 0 as well.

CONCENTRATION CELLSWe can construct a cell where both compartments contain the

same components BUT at different concentrations.

Notice the difference in the concentrations pictured at

left. Because the right compartment contains 1.0 M

Ag+ and the left compartment contains 0.10 M Ag+,

there will be a driving force to transfer electrons from

left to right. Silver will be deposited on the right

electrode, thus lowering the concentration of Ag+ in

the right compartment. In the left compartment the

silver electrode dissolves [producing Ag+ ions] to raise

the concentration of Ag+ in solution.

Exercise 6

Determine the direction of electron flow and designate the

anode and cathode for the cell represented here.

Exercise 7

Determine Eocell and Ecell based on the following half-reactions:

VO2+ + 2H+ + e- → VO2+ + H2O E°= 1.00 V

Zn2+ + 2e- → Zn E° = -0.76V

Where:

T = 25°C

[VO2+] = 2.0 M

[H+] = 0.50 M

[VO2+] = 1.0 x 10-2 M

[Zn2+] = 1.0 x 10-1 M

Eocell = 1.76 V

Ecell = 1.89 V

SUMMARY OF GIBB’S FREE ENERGY AND CELLS−Eo implies NON-spontaneous.

+Eo implies spontaneous (would be a good battery!)

E = 0, equilibrium reached (dead battery)

The larger the voltage, the more spontaneous the reaction

ΔG will be negative in spontaneous reactions

K > 1 are favored (product favored)

Two important equations:

ΔG = −nFE° -- [“minus nunfe”]

ΔG = −RTlnK -- [“ratlink”]

G = Gibbs free energy [Reaction is spontaneous if ΔG is negative]

n = number of moles of electrons.

F = Faraday const. 9.6485309×104 J/V (1 mol of electrons carries 96,500C )

E = cell potential

R = 8.31 J/mol•K

T = Kelvin temperature

K = equilibrium constant -- [products]coeff /[reactants]coeff

**Favored conditions: Ecell > 0 ΔG < 0 K > 1**

Exercise 8

For the oxidation-reduction reaction:

S4O62- (aq) + Cr2+ (aq) → Cr3+ (aq) + S2O3

2-(aq)

The appropriate half-reactions are:

S4O62- + 2e- → 2 S2O3

2- E° = 0.17V

Cr3+ + e- → Cr2+ E° = -0.50V

Balance the redox reaction, and calculate E° and K (at 25°C).

E° = 0.67 V

K = 1022.6 = 4 × 1022

Batteries

19.6

Leclanché cell

Dry cell

Zn (s) Zn2+ (aq) + 2e-Anode:

Cathode: 2NH4 (aq) + 2MnO2 (s) + 2e- Mn2O3 (s) + 2NH3 (aq) + H2O (l)+

Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

Batteries

Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-Anode:

Cathode: HgO (s) + H2O (l) + 2e- Hg (l) + 2OH- (aq)

Zn(Hg) + HgO (s) ZnO (s) + Hg (l)

Mercury Battery

19.6

Batteries

19.6

Anode:

Cathode:

Lead storage

battery

PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e- PbSO4 (s) + 2H2O (l)4

Pb (s) + SO2- (aq) PbSO4 (s) + 2e-4

Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) 2PbSO4 (s) + 2H2O (l)4

Batteries

19.6Solid State Lithium Battery

Batteries

19.6

A fuel cell is an

electrochemical cell

that requires a

continuous supply of

reactants to keep

functioning

Anode:

Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq)

2H2 (g) + 4OH- (aq) 4H2O (l) + 4e-

2H2 (g) + O2 (g) 2H2O (l)

Chemistry In Action: Bacteria Power

CH3COO- + 2O2 + H+ 2CO2 + 2H2O

Corrosion

19.7

Cathodic Protection of an Iron Storage Tank

19.7

Electrolysis

19.8

Electrolysis is the process in which electrical energy is used

to cause a nonspontaneous chemical reaction to occur.

ELECTROLYSIS AND ELECTROLYTIC CELLS

Electrolysis- the use of electricity to bring about chemical

change. Literal translation “split with electricity”

Electrolytic cells [NON spontaneous cells]:

Used to separate ores or plate out metals.

Important differences between a voltaic/galvanic cell and an

electrolytic cell:1) Voltaic cells are spontaneous and electrolytic cells are forced to occur

by using an electron pump or battery or any DC source.

2) A voltaic cell is separated into two half cells to generate electricity; an

electrolytic cell occurs in a single container.

3) A voltaic [or galvanic] cell IS a battery, an electrolytic cell NEEDS a

battery

4) AN OX and RED CAT still apply BUT the polarity of the electrodes is

reversed. The cathode is Negative and the anode is Positive (remember

E.P.A – electrolytic positive anode). Electrons

still flow FATCAT.

5) Usually use inert electrodes

Predicting the Products of Electrolysis:

If NO water is present and you have a pure molten ionic

compound, then:

the cation will be reduced (gain electrons/go down in charge)

the anion will be oxidized (lose electrons/go up in charge)

If water IS present and you have an aqueous solution of

the ionic compound, then:

you’ll need to figure out if the ions are reacting or the water

is reacting. You can look at a reduction potential table to

figure it out but, OR as a rule of thumb:

•No group IA or IIA metal will be reduced in an aqueous

solution

• Water will be reduced instead.

•No polyatomic will be oxidized in an aqueous solution

• Water will be oxidized instead.

Electrolysis of Water

19.8

Calculating the Electrical Energy of ElectrolysisHow much metal could be plated out?

How long would it take to plate out?

Faraday’s Law: The amount of a substance being oxidized

or reduced at each electrode during electrolysis is directly

proportional to the amount of electricity that passes through

the cell.

Use dimensional analysis (unit factor, T-chart) for these

calculations, remembering: # coulombs = It where:

1 Volt = 1 Joule/Coulomb

1 Amp = 1 Coulomb/second (current is measured in amp, but

symbolized by I)

Faraday = 96,500 Coulombs/mole of electrons

Balanced redox equation gives #moles of e−/mole of

substance

Formula weight gives grams/mole

Electrolysis and Mass Changes

charge (C) = current (A) x time (s)

1 mole e- = 96,500 C

19.8

How much Ca will be produced in an electrolytic cell of

molten CaCl2 if a current of 0.452 A is passed through the

cell for 1.5 hours?

Anode:

Cathode: Ca2+ (l) + 2e- Ca (s)

2Cl- (l) Cl2 (g) + 2e-

Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)

2 mole e- = 1 mole Ca

mol Ca = 0.452C

sx 1.5 hr x 3600

s

hr 96,500 C

1 mol e-

x2 mol e-

1 mol Cax

= 0.0126 mol Ca

= 0.50 g Ca

19.8

Exercise 9

How long must a current of 5.00 A be applied to a solution of

Ag+ to produce 10.5 g silver metal?

= 31.3 min

Exercise 10

An acidic solution contains the ions Ce4+, VO2+, and Fe3+ .

Using the E° values [listed in Table 17.1, Zumdahl] give the order of

oxidizing ability of these species and predict which one will be

reduced at the cathode of an electrolytic cell at the lowest

voltage.

Ce4+ > VO2+ > Fe3+

Applications of electrolytic cells:▪Production of pure forms of elements from mined ores

Aluminum from Hall-Heroult process

▪Separation of sodium and chlorine (Down's cell)

▪Purify copper for wiring

▪Electroplating—applying a thin layer of an expensive metal

to a less expensive one Jewelry --- 14 K gold plated

▪Bumpers on cars --- Chromium plated

▪Charging a battery --- i.e. your car battery when the

alternator functions

Corrosion – The process of returning metals to their

natural state, the ores.

▪Involves oxidation of the metal which causes it to lose its

structural integrity and attractiveness.

▪The main component of steel is iron.

▪20% of the iron and steel produced annually is used to

replace rusted metal!

▪most metals develop a thin oxide coating to protect them,

patina’s, tarnish, rust, etc.

chapter 19...salt bridge -- a device used to maintain electrical neutrality in a galvanic cell. this...

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