A. Introduction to Electrochemistry

electrochemistry is the branch of chemistry that studies

electron transfer occurs in

Electrochemistry

eg)

electron transfer in chemical reactions

living systems

photosynthesis, cellular respiration

also occurs in

eg)

non-living systems

combustion, bleaching, metallurgy

is a

is a oxidation

reduction gain of electrons “GER”

loss of electrons “LEO”

eg) Mg(s) Mg2+(aq) + 2e

eg) Fe3+(aq) + 3e Fe(s)

2Cl(aq) Cl2(g) + 2e

Br2(l) + 2e 2Br(aq)

B. Redox Reactions

oxidation and reduction reactions occur together, hence the term

the reduction and oxidation reactions are called the

“adding” the half reactions together will give you the that takes place during the

redox reaction

redox

half reactions

net ionic equation

the e lost in the oxidation half reaction the e gained in the reduction half reaction

must equal

you may have to of the half reactions to balance the e

(ions not changing) are included!

multiply one or both

spectator ions NOT

Example 1Given the following reaction, write the half reactions and the net ionic equation.

Na(s) + LiCl(aq) Li(s) + NaCl(aq) 0 1+ 1– 0 1+ 1–

ox red Cl- is spectator

Ox:

Red:

Net:

Li+(aq) + 1e- Li(s)

Na(s) Na+(aq) + 1e-

Li+(aq) + Na(s) Li(s) + Na+

(aq)

Example 2Given the following reaction, write the half reactions and the net ionic equation.

Zn(s) + Au(NO3)3(aq) Au(s) + Zn(NO3)2(aq) 0 3+ 1– 0 2+ 1–

ox red NO3- is spectator

Ox:

Red:

Net:

Au3+(aq) + 3e- Au(s)

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

2 Au3+(aq) + 3 Zn(s) 2 Au(s) + 3 Zn2+

(aq)

[ ]

[ ]

3

2

C. Spontaneous Redox Reactions

chemical reactions which occur on their own, without the input of , are called

not all reactions are spontaneous

the substance that is is called the ( ) (it causes the oxidation by taking e-)

the substance that is is called the ( ) (it causes the reduction by

giving up e-)

additional energy spontaneous

reduced oxidizing agent OA

oxidized reducing agent RA

in the table of redox half reactions (see pg 9 in Data Booklet), the is at the top left and the is at the bottom right

the rule states that a spontaneous reaction occurs if the agent is above the

agent in the table of redox half reactions

OA + spontaneous RA reaction

RA + non- spontaneousOA reaction

strongest oxidizing agent (SOA)strongest reducing agent (SRA)

redox spontaneity oxidizing

reducing

Try These:For each of the following combinations of substances, state whether the reaction would be spontaneous or non-spontaneous:

Cr3+(aq) with Ag(s)

I2(s) with K(s)

H2O2(l) with Au3+(aq)

Sn2+(aq) with Cu(s)

Fe2+(aq) with H2O (l)

non-spontaneous

non-spontaneous

non-spontaneous (both ways)

spontaneous

spontaneous

D. Predicting Redox Reactions

we will be predicting the strongest or most dominating reaction that occurs when substances are mixed (other reactions do take place because of atomic collisions!)

Steps:

1. List all species present as reactants

dissociate and

do not dissociate include ions if it is

always include

soluble ionic compounds acids

molecular compounds

H+(aq) acidic

H2O(l)

3. Identify the and using the table.

4. Write out the for the SOA and SRA.

5. Determine the

6. Determine

SOA SRA

half reactions

net ionic reaction

spontaneity

2. Identify each as or (***some can be both so memorize them… , , , )

OA RAFe2+ Cr2+ Sn2+ H2O(l)

Example 1Predict the most likely redox reaction when chromium is placed into aqueous zinc sulphate.

SOA (Red):

SRA (Ox):

Net:

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

Cr(s) Cr2+(aq) + 2e-

Zn2+(aq) + Cr(s) Zn(s) + Cr2+

(aq)

Cr(s) Zn2+(aq) SO4

2-(aq) H2O(l)

RA OA OA with H2O(l) OA/RAS S

spont

Example 2Predict the most likely redox reaction when silver is placed into aqueous cadmium nitrate.

SOA (Red):

SRA (Ox):

Net:

Cd2+(aq) + 2e- Cd(s)

Ag(s) Ag+(aq) + e-

Cd2+(aq) + 2 Ag(s) Cd(s) + 2 Ag+

(aq)

Ag(s) Cd2+(aq) NO3

-(aq) H2O(l)

RA OA OA with H+ (aq) OA/RAS S

[ ]2

nonspont

Example 3Predict the most likely redox reaction when potassium permanganate is slowly poured into an acidic iron (II) sulphate solution.

SOA (Red):

SRA (Ox):

Net:

MnO4-(aq) + 8H+

(aq) + 5e- Mn2+(aq) + 4H2O(l)

Fe2+(aq) Fe3+

(aq) + e-

MnO4-(aq) +8H+

(aq) + 5Fe2+(aq) Mn2+

(aq) + 4H2O(l) + 5 Fe3+(aq)

K+(aq) H+

(aq) Fe2+(aq) H2O(l)

OA OA with H+ (aqOA with H+ (aq), H2O(l)

OA/RASS

[ ]5

MnO4-(aq) SO4

2-(aq)

OA OA/ RA

spont

E. Generating Redox Tables you can be given data for certain ions and elements

then be asked to generate a redox table like the one on pg 9 of you Data Booklet (a smaller version!)

you may have to generate a table from real or fictional elements and ions

the tables that we use are all written as half reactions

reduction

Example 1Generate a redox table given the following data:

  Cu2+(aq) Zn2+

(aq) Pb2+(aq) Ag+

(aq)

Cu(s)

Zn(s)

Pb(s)

Ag(s)

indicates no reaction indicates a reaction

Redox Table

Ag+(aq) + e- Ag(s)

Cu2+(aq) + 2e- Cu(s)

Pb2+(aq) + 2e- Pb(s)

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

SOA

SRA

Redox Table

Ag+(aq) + e- Ag(s)

Cu2+(aq) + 2e- Cu(s)

Hg2+(aq) + 2e- Hg(l)

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

SOA

SRA

Example 2:Generate a redox table given the following data:  Cu(s) + Ag+

(aq) Cu2+(aq) + Ag(s)

Zn2+(aq) + Ag(s) no reaction

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

(aq) + Cu(s)

Hg(l) + Ag+(aq) no reaction

Redox Table

Z2 + 2e- 2Z-(aq)

Y2 + 2e- 2Y-(aq)

W2 + 2e- 2W-(aq)

X2 + 2e- 2X-(aq)

SOA

SRA

Example 3:Generate a redox table given the following data:  2X-

(aq) + Y2 spontaneous reaction

2Z-(aq) + Y2 no reaction

2Z-(aq) + W2 spontaneous reaction

1. pg 572 Lab 13.A, 2. pg 573 # 11-143. pg 5 in workbook

F. Oxidation Numbers (States) an or is the charge an atom

to have when found in a or charged

can be used when you have a where there are no to determine if oxidation or reduction is occurring

how do you use a change in the number?

oxidation number stateappears

molecular compound ion charges

1. if the number then has occurred decreases reduction

2. if the number then has occurred increases oxidation

neutral moleculepolyatomic ion

Rules for Assigning Oxidation Numbers:

1. In a pure element, the oxidation number is .

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

zero

ion charge

3. In most compounds containing hydrogen, the oxidation number for hydrogen is . (Exception is the metal hydrides eg) LiH where the oxidation number of hydrogen is ).

+1

–1

5. The sum of oxidation numbers of all atoms in a substance must equal the of the substance. ( for compounds and of the polyatomic ion)

eg) sum of MgO = sum of PO43- =

4. In most compounds containing oxygen, the oxidation number for oxygen is . (Exception is the peroxides eg) H2O2, Na2O2 where the oxidation number of oxygen

is ) –1

–2

net chargeZero the charge

0 –3

Example What is the oxidation number (state) for the element identified in each of the following substances:

a) N in N2O N2 O

individual oxidation numbers

sum of oxidation numbers

–2

–2 = 0+2

+1

b) N in NO3-

N O3–

–2

–6 = –1+5

+5

c) C in C2H5OH –2+1

= 0–4

–2

d) C in C6H12O6 –2

–12 = 00

0C6 H12 O6

C2 H5 O H+1

–2 +1+5

+1

+12

G. Balancing Redox Reactions sometimes most reactants and products are known but

the complete reaction is not given…called a reaction

we can use the oxidation numbers to balance the equations either in or conditions:

skeleton

acidic basic

1. In an Acidic Solution

1. Assign

2. Balance the that changes in oxidation number.

3. Add to balance the change in oxidation number.

4. Balance O using

5. Balance H using

6. Check that the half-reaction is balanced with respect to

oxidation numbers (ON).

element

e– total

H2O(l).

H+(aq).

net charge.

(ON subscript coefficient)

Example 1:Balance the following half reaction assuming acidic conditions: 

+6 2 +3 2

4 = 1+3+6 8 = 2

(Cr is already balanced)

+3 e– + 2 H2O(l) +4 H+(aq) CrO4

2-(aq) CrO2

-(aq)

net charge = –1 net charge = –1

(+6) (+3)

Example 2:Balance the following half reaction assuming acidic conditions: 

+1 2 0

+1 4 = 0

+6 e– + 4 H2O(l) +6 H+(aq) HClO2(aq) Cl2(g)

net charge = 0 net charge = 0

(+6) (0)+3

+3

2

1. In a Basic Solution

1. Assign

2. Balance the that changes in oxidation number.

4. Balance equation charge

5. Balance H using

6. Check that the half-reaction is balanced with respect to

oxidation numbers.

element

OH–(aq).

H2O(l).

oxygen.

3. Add to balance the change in oxidation number.

e– total(ON subscript coefficient)

Example 1:Balance the following half reaction assuming basic conditions: 

+4 2 +6 2

8 = 2+6+4 6 = 2

(S is already balanced)

+ 2 e– + H2O(l) +2 OH–(aq) SO3

2-(aq) SO4

2-(aq)

# oxygen = 2 + 3 = 5 # oxygen = 4 + 1 = 5

(+4) (+6)

3. Putting it All Together

now we are going to combine coming up with our own half reactions with figuring out the net redox reaction

Steps

1. Assign

2. Separate the partial net equation into two (omit any , , or ).

3. Balance each half-reaction.

4. of the equations so e lost = e gained.

5. Add the equations to produce a balanced

6. Check to see if all elements and charges are balanced.

oxidation numbers.

half reactionsH2O(l) H+

(aq) OH-(aq)

net ionic equation

Simplify.

Multiply one or both

Example 1:Balance the following using oxidation numbers, assuming acidic conditions:

+42

+48 = 2

+3 e– + 2 H2O(l) +4 H+(aq) CrO4

2-(aq) CrO2

-(aq)

(+6) (+3)

+6

+6CrO4

2-(aq) + SO3

2-(aq) CrO2

-(aq) + SO4

2-(aq)

2

6 = 2

22

84 = 1 = 2+3 +6

+6+3

+ 2 e– + H2O(l) + 2 H+(aq) SO3

2-(aq) SO4

2-(aq)

(+4) (+6)

Red

Ox [ ]

[ ]2

3

8 H+(aq) + 2 CrO4

2-(aq) + 3 H2O(l) + 3 SO3

2-(aq) 2 CrO2

-(aq) + 4 H2O(l) + 3 SO4

2-(aq) + 6 H+

(aq)

2 H+(aq) + 2 CrO4

2-(aq) + 3 SO3

2-(aq) 2 CrO2

-(aq) + H2O(l) + 3 SO4

2-(aq)

Net

Example 2:Balance the following using oxidation numbers, assuming basic conditions:

+32

6 = 2

+4 e– + 6 OH–(aq) +3 H2O(l) TeO3

2-(aq) Te(s)

(+4) (0)

+4

+4TeO3

2-(aq) + Cr3+

(aq) Te(s) + Cr2O72-

(aq) 2

14 = 2+12

+60

+ 6 e– + 14 OH– (aq) + 7 H2O(l) Cr3+ (aq) Cr2O7

2-(aq)

(+6) (+12)

Red

Ox [ ]

[ ]3

2

9 H2O(l) + 3 TeO32-

(aq) + 28 OH-(aq) + 4 Cr3+

(aq) 3 Te(s) + 18 OH-(aq) + 2 Cr2O7

2-(aq) + 14 H2O(l)

3 TeO32-

(aq) + 10 OH-(aq) + 4 Cr3+

(aq) 3 Te(s) + 2 Cr2O72-

(aq) + 5 H2O(l) Net

2

4. Disproportionation

disproportionation occurs when one element is both oxidized and reduced in a reaction

eg)2 H2O2(aq) 2 H2O(l) + O2(g)

-1 -2 0

Cl2(g) + 2 OH-(aq) ClO-

(aq) + Cl-(aq) + H2O(l)

0 +1 -1

H. Redox Stoichiometry

stoichiometry can be used to predict or analyze a quantity of a chemical involved in a chemical reaction

in the past we have used balanced chemical equations to do stoich calculations

1. Calculations

we can now apply these same calculations to balanced redox equations

Example 1What is the mass of zinc is produced when 100 g of chromium is placed into aqueous zinc sulphate.

SOA (Red):

SRA (Ox):

Net:

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

Cr(s) Cr2+(aq) + 2e-

Cr(s) + Zn2+(aq) Zn(s + Cr2+

(aq) )

Cr(s) Zn2+(aq) SO4

2-(aq) H2O(l)

RA OA OA with H2O(l) OA/RASRA SOA

m = 100 gM = 52.00 g/moln = m M = 100 g 52.00 g/mol = 1.923… mol

Cr(s) + Zn2+(aq) Zn(s) + Cr2+

(aq)

m = ? M = 65.39 g/mol n = 1.923… mol x 1/1 = 1.923… mol

m = nM = (1.923…mol)(65.39 g/mol) = 125.75 g = 126 g

Example 2What volume of 1.50 mol/L potassium permanganate is needed to react with 500 mL of 2.25 mol/L acidic iron (II) sulphate solution?

SOA (Red):

SRA (Ox):

Net:

MnO4-(aq) + 8H+

(aq) + 5e- Mn2+(aq) + 4H2O(l)

Fe2+(aq) Fe3+

(aq) + e-

MnO4-(aq) +8H+

(aq) + 5Fe2+(aq) Mn2+

(aq)+ 4H2O(l) + 5Fe3+(aq)

K+(aq) MnO4

-(aq) H+

(aq) H2O(l) Fe2+(aq) SO4

2-(aq)

OA with H+(aq) OA/RA

SRASOAOA OA OA/RA OA with H+

(aq) OA with H2O(l)

[ ]5

v = ? c = 1.50 mol/L n = 1.125 mol x 1/5 = 0.225 mol v = n Cv = 0.225 mol 1.50 mol/L = 0.150 L

MnO4-(aq) +8H+

(aq) + 5Fe2+(aq) Mn2+

(aq)+ 4H2O(l) + 5 Fe3+(aq)

v = 0.500 L c = 2.25 mol/L n = cv = (2.25 mol/L)(.500L) = 1.125 mol

a titration is a lab process used to determine the of a substance needed to react completely with another substance

this volume can then be used to calculate an unknown using stoichiometry

2. Titrations

one reagent ( - ) is slowly added to another ( - ) until an abrupt change ( ) occurs, usually in colour

volume

concentration

titrant OAsample RAendpoint

in redox titrations, two common oxidizing agents are used because of their and :

1.

2.

colour strengthpermanganate ions (MnO4

-(aq)) – purple

dichromate ions (Cr2O72-

(aq)) – orange

eg)MnO4-(aq) + 8 H+

(aq) + 5 e- Mn2+(aq) + 4 H2O(l)

purple colourlesscolourless

as long as the sample (RA) in the flask is reacting with the the sample will be

permanganate ions (dichromate ions)colourless (orange)

when the reaction is complete, any unreacted permanganate ions will turn the sample (pink) (with dichromate, sample goes from orange to green)

purple

the volume of titrant (OA) needed to reach the endpoint is called the equivalence point

the of the titrant must be accurately known

permanganate solution as it oxidizes distilled water

concentration

decomposes

the concentration of the permanganate solution must be calculated against a (a solution of known concentration) before it can be used in a titration itself

this is done just prior to the titration

primary standard

ExampleFind the concentration of (standardize) the KMnO4(aq) solution

by titrating 10.00 mL of 0.500 mol/L acidified tin (II) chloride with the KMnO4(aq).

Trial 1 2 3 4

Final Volume (mL)

18.40 35.30 17.30 34.10

Initial Volume (mL)

1.00 18.40 0.60 17.30

Volume of (mL)

       

Endpoint Colour

pink light pink light pink light pink

titrant17.40 16.90 16.70 16.80

endpoint average is calculated by using 3 volumes within 0.20 mL

Endpoint average = 16.90 mL + 16.70 mL + 16.80 mL3

= 16.80 mL

SOA (Red):

SRA (Ox):

Net:

MnO4-(aq) + 8H+

(aq) + 5e- Mn2+(aq) + 4H2O(l)

Sn2+(aq) Sn4+

(aq) + 2 e-

2MnO4-(aq) + 16H+

(aq) + 5Sn2+(aq) 2Mn2+

(aq)+ 8H2O(l) + 5Sn4+(aq)

K+(aq) MnO4

-(aq) H+

(aq) H2O(l) Sn2+(aq) Cl-

(aq)

OA with H+(aq) OA/RA

SRASOAOA OA OA/RA RA

[ ]5

determine net ionic redox equation

Analysis:

[ ] 2

use net redox equation to calculate KMnO4(aq) concentration

2MnO4-(aq) + 16H+

(aq) + 5Sn2+(aq) 2Mn2+

(aq)+ 8H2O(l) +5Sn4+(aq)

v = 0.01680 L C = ? n = 0.00500 mol x 2/5 = 0.00200 mol C = n vC = 0.00200 mol 0.01680 L = 0.119 mol/L

v = 0. 01000 L c = 0.500 mol/L n = cv = (0.500 mol/L)(0.01000 L) = 0.00500 mol

I. Electrochemical Cells1. Voltaic/Galvanic Cells

are devices that convert energy into energy

in redox reactions, e- are transferred from the to the

the transfer of e- can occur through a separating the two substances in containers called

electric cells chemicalelectrical

oxidized substance reduced substance

conducting wire

half cells

a is an arrangement where are joined so that the and can move between them

are made of good conducting materials so e- can flow…can be the of the solution or inert such as

the is a solution that contains ions and will transmit ions (charged particles)

voltaic or galvanic cell two half cells ionse-

electrodes metal

carbon

electrolyte

the electrode where occurs is called the

if the anode is a metal, it mass as the cell operates

the anode is labelled as since it is the electrode where the electrons originate

oxidationanode

loses

the move to the since this electrode

negative

anions anode loses electrons (leaving a net positive charge in the electrode)

the electrode where occurs is called the

if the cathode is a metal, it mass as the cell operates

reductioncathode

gains

the cathode is labelled as since

the move to the since this electrode

positive the anode is labelled negative

cations cathodegains electrons (leaving a net negative charge in the electrode)

electrons flow from the to the

ions must be able to to their attracting electrode (either through the or a ) otherwise a buildup of charge will occur opposing the movement of e-

the flow of ions through the solution and e- through the wire maintains overall electrical neutrality

anode (LEOA)cathode (GERC)

moveporous cup salt bridge

through a connecting wire

2. Standard Reduction Potentials

are the ability of a half cell to

these potentials are measured using a

each half reaction listed in the Data Booklet has an E value measured in assigned to it

reduction potentials attract e-

voltmeter

volts

all values in the table are arbitrarily assigned based on a standard

the half reaction has been set as the standard and has an E value of

hydrogen cell 0.00 V

3. Predicting Voltage of a Voltaic Cell the standard cell potential is determined by

the for the two half reactions

the on the E value for the half reaction must be

if you multiply an equation to balance e-, you multiply the E value (voltage is independent of number of e- transferred)

(Enet) adding

sign oxidationreversed

DO NOT

a E net is a reaction positive spontaneous

a E net is a reaction negative nonspontaneous

E values

Example Calculate the E net for the reaction of Zn(s) with CuSO4(aq).

SOA (Red):

SRA (Ox):

Net:

Cu2+(aq) + 2e- Cu(s)

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

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

(aq)

Zn(s) Cu2+(aq) SO4

2-(aq) H2O(l)

RA OA OA with H2O(l) OA/RAS S

E = +0.34 V

E = +0.76 V

Enet = +1.10 V

4. Shorthand Notation

line separates

double line represents the or and separates the two

ORZn(s) / Zn2+(aq) // Cu2+

(aq) / Cu(s)

Zn(s) / Zn2+(aq) // Cr2O7

2-(aq) , H

+(aq) , Cr3+

(aq)/ C(s)

(/) phases

(//) porous cup salt bridgehalf reactions

comma separates (,) chemical species in the same phase

***anode // cathode

5. Drawing Cells when drawing a cell from the shorthand notation, you have

to be able to label the

you also have to show and label the and including and

anode, cathode, positive terminal, negative terminal, electrolytes, direction of e- flow, directions of cation and anion flow

reduction half reaction, oxidation half reaction net reaction

E values, Enet spontaneity

ExampleDraw and fully label the following electrochemical cell: Al(s)/ Al3+

(aq) // Ni2+(aq) / Ni(s)

Ni(s)Al(s)

Ni2+ (electrolyte)

Al3+ (electrolyte)

e- V

anions

cations

cathode positive terminal

anodenegative terminal

SOA (Red):

SRA (Ox):

Net:

Ni2+(aq) + 2e- Ni(s)

Al(s) Al3+(aq) + 3e-

2 Al(s) + 3 Ni2+(aq) 3 Ni(s + 2 Al3+

(aq) )

Al(s) Al3+(aq) Ni2+

(aq) H2O(l)

RA OA OA OA/RAS S

Ni(s)

RA

[ ]

[ ]

3

2

E = –0.26 V

E = +1.66 V

Enet = +1.40 V

spontaneous: yes

J. Commercial Cells are made by connecting two or more voltaic

cells in

the of the battery is the of the

batteries series (one after the other)

voltageindividual cells

sum

there are many types of batteries:

a) Dry Cell common batteries of clocks, remote

controls, noisy kids toys etc.

Cathode (Red): 2 MnO2(s) + H2O(l) + 2e- Mn2O3(aq) + 2 OH-(aq) E= +0.79 V

Anode (Oxid): Zn(s) Zn2+(aq) + 2e- E = +0.76 V

Net: 2 MnO2(s)+ H2O(l) + Zn(s) Mn2O3(aq) + 2 OH-(aq) + Zn2+

(aq) Enet = +1.55 V

the produced causes irreversible side reactions to occur making recharging impossible

OH-

1.5 V and 9 V

b) Nickel-Cadmium

one type of battery

Cat (Red): 2 NiO(OH)(s) + 2 H2O(l) + 2e- 2 Ni(OH)2(s) + 2 OH-(aq) E= +0.49 V

An (Oxid): Cd(s) + 2 OH-(aq) 2 Cd(OH)2(s) + 2e- E = +0.76 V

Net: 2 NiO(OH)(s)+ 2 H2O(l) + Cd(s) 2 Ni(OH)2(s)+ 2 Cd(OH)2(s) Enet = +1.25 V

rechargeable

c) Lead Storage Battery

where serves as the anode, and serves as the cathode

both electrodes dip into an electrolyte solution of

are connected in series

typical car battery lead lead coated with lead dioxide

sulfuric acid

six cells

Cat (Red): PbO2(s)+ HSO4-(aq) + 3 H+

(aq) + 2e- PbSO4(s)+ 2 H2O(l) E= +1.68 V

An (Oxid): Pb(s) + HSO4-(aq) PbSO4(s) + H+

(aq) + 2e- E = +0.36 V

Net: Pb(s)+ PbO2(s) + 2 H+(aq) + 2 HSO4

-(aq) 2 PbSO4(s)+ 2 H2O(l) Enet = +2.04 V

d) Fuel Cells

cells where

the energy from this reaction can be used to

one type is the

is pumped in at the while is pumped in at the (which both have a lot of surface area)

reactants are continuously supplied

run machines

hydrogen-oxygen fuel cell

hydrogen gas anodeoxygen gas cathode

pressure is used to push the H2 through a platinum

catalyst which splits the H2 into 2H+ and 2e-

Cathode (Red): O2(g) + 4 H+(aq) + 4e- 2 H2O(l) E= V

Anode (Oxid): 2 H2(g) 4 H+(aq) + 4e- E = V

Net: O2(g) + 2 H2(g) 2 H2O(l) Enet = +1.23 V

the 2e- move through an external circuit towards the cathode generating electrical energy

the O2 is also pushed through the platinum catalyst forming

two oxygen atoms

the H+ ions and oxygen atoms combine to form water

+1.23

0.00

Hydrogen-oxygen Fuel Cell

need a source of hydrogen…reformers are used to convert CH4 or CH3OH into and

unfortunately, is a

about 24-32% efficient where gas-powered car is about 20% efficient

H2 CO2

CO2 greenhouse gas

K. Electrolytic Cells1. The Basics

in an electrical energy is used to force a chemical reaction to occur (opposite

of a voltaic cell)

commonly used to

these reactions have a Enet

electrolytic cell,nonspontaneous

negative

electroplate metals (eg. gold, silver, bronze, chromium etc), recharge batteries,

useful gases (eg. H2, O2, Cl2 etc) and split compounds into

the electrolytic cell is hooked up to a instead of load or external circuit) so the flow of e-

is

the of the electrolytic cell is connected to the of the battery and therefore is

battery (power supply

“pushed” by an outside force

cathodeanode negative

the of the electrolytic cell is connected to the of the battery and therefore is

anodecathode positive

all other conventions are the as voltaic cells same

Voltaic Cells Electrolytic Cells chemical to electrical energy electrical to chemical energy

usually contains porous cup or salt bridge

does not contain a porous cup or salt bridge

e– flow from anode to cathode oxidation at anode reduction at cathode cations migrate to cathode anions migrate to anode

Enet is positive (spont) Enet is negative (nonspont) has a voltmeter or external load has a power supply

cathode + anode – cathode – anode +

some processes are used in industry to produce gases, for example:

1. the for producing …aluminum oxide is electrolyzed using carbon electrodes …liquid aluminum is collected

2. a for producing …a saturated sodium chloride solution is electrolyzed …chlorine gas is formed and collected at the anodes

http://images.google.ca/imgres?imgurl=http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter18/Text_Images/FG18_18.JPG&imgrefurl=http://wps.prenhall.com/wps/media/objects/602/616516/Chapter_18.html&h=756&w=1600&sz=183&tbnid=4cFJrFlK4noQXM:&tbnh=70&tbnw=150&hl=en&start=11&prev=/images%3Fq%3Dhall-heroult%2Bcell%26svnum%3D10%26hl%3Den%26lr%3D

http://www.cheresources.com/chloralk.shtml

Hall-Heroult cell aluminum

chlor-alkali plant chlorine gas

Example 1An electric current is passed through a solution of nickel (II) nitrate using inert electrodes. Predict the anode and cathode reactions, overall reaction, and minimum voltage required.

Cathode SOA(Red):

Anode SRA(Ox):

Net:

Ni2+(aq) + 2e- Ni(s)

2 H2O(l) O2(g) + 4 H+(aq) + 4e-

2 Ni2+(aq) + 2 H2O(l) 2 Ni(s)+ O2(g) + 4 H+

(aq)

Ni2+(aq) NO3

-(aq) H2O(l)

OA OA with H+(aq) OA/RA

SRASOA

E = -0.26 V

E = -1.23 V

Enet = -1.49 V

[ ]2

Example 2An electric current is passed through a solution of potassium iodide using inert electrodes. Predict the anode and cathode reactions, overall reaction, and minimum voltage required.

Cathode SOA(Red):

Anode SRA(Ox):

Net:

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

2 I-(aq) I2(s) + 2e-

2 H2O(l) + 2 I-(aq) H2(g) + 2 OH-

(aq) + I2(s)

K+(aq) I-

(aq) H2O(l)

OA RA OA/RASRA SOA

E = -0.83 V

E = -0.54 V

Enet = -1.37 V

Example 3An electric current is passed through a solution of copper(II) sulphate using a carbon electrode and a metal electrode. Predict the anode and cathode reactions, overall reaction, and minimum voltage required.

Cathode SOA(Red):

Anode SRA(Ox):

Net:

Cu2+(aq) + 2 e- Cu(s)

2 H2O(l) + 2 Cu2+(aq) 2 Cu(s) + O2(g) + 4 H+

(aq)

Cu2+(aq) SO4

2-(aq) H2O(l)

OA RA OA/RASRASOA

E = +0.34 V

Enet = -0.89 V

2 H2O(l) O2(g) + 4 H+(aq) + 4e- E = -1.23 V

[ ]2

*** Note: is an exception to the rule… chlorine

when water and chlorine are competing as reducing agents, water is the stronger RA but is chosen because the transfer of e- from H2O to O2 is

more difficult …called overvoltage

chlorine

2. Quantitative Study of Electrolysis

quantitative analysis (stoich) provides information on necessary quantities, current and/or time for electrolytic reactions

one e- carries of charge

the unit for charge is the

this means that one of e- carry of charge

is called the (see Data Booklet pg 4)

(q) Coulomb (C)

1.60 x 10-19 C

mole 9.65 x 104 C

9.65 x 104 C/mol Faraday constant

ne- = q

F

where: ne- = number of moles of electrons (mol)q = charge in Coulombs (C)F = Faraday constant

= 9.65 x 104 C/mol

q = It

I = current in C/s or Amperes (A)t = time in seconds (s)

ne- = It

F

the above equations can be combined into one equation:

we can use these equations in stoichiometric calculations for current, time, mass, moles of a substance and moles e-

Example 1

Determine the number of moles of electrons supplied by a dry cell supplying a current of 0.100 A to a radio for 50.0 minutes.

I = 0.100 A (C/s)t = 50.0 min 60 s/min = 3000 sF = 9.65 x 104

C/mol

ne- = It F = (0.100 C/s)(3000 s) 9.65 x 104

C/mol

= 0.00311 mol

Example 2

An electrochemical cell caused a 0.0720 mol of e- to flow through a wire during a 3.00 hour period. Calculate the average current.

ne- = 0.0720 molt = 3.00 h 3600 s/h = 10 800 sF = 9.65 x 104

C/mol

ne- = It F0.0720 mol = I(10 800 s)

9.65 x 104 C/mol

I = 0.643 A

Example 3

If a 20.0 A current flows through an electrolytic cell containing molten aluminum oxide for 1.00 hours, what mass of Al(l) will be deposited at the cathode?

ne- = It F

= (20.0 A)(1.00 h 3600 s/h) 9.65 x 104

C/mol = 0.746…mol

n = 0.746…mol 1/3 = 0.248…molM = 26.98 g/molm = nM = (0.248…mol)(26.98g/mol) = 6.71 g

Al3+(l) + 3 e- Al(l)

L. Society and Technological Connections1. Rust and Corrosion

the metal is oxidized causing the loss of

corrosion can be viewed as the process of returning metals to their natural state (ore)

structural integrity

commonly, the oxide coating will scale off leaving new metal exposed an susceptible to corrosion

salt will by acting as a speed up the oxidation salt bridge

most metals develop a thin oxide coating which then protects their internal atoms against further oxidation

Fe(s)

H2O droplet

O2(g)

anode

cathode

Fe(OH)2(s)

rust

Cathode SOA(Red):

Anode SRA(Ox):

Net:

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

O2(g) + 2H2O(l) + 2Fe(s) 4 OH-(aq) + 2 Fe2+

(aq)

Fe(s) Fe2+(aq) + 2e- [ ]2

O2(g) + 2H2O(l) + 2Fe(s) 2 Fe(OH)2(s)

2. Prevention of Corrosion

other metals (eg. Zn, Cr, Sn) can be onto metals that you don’t want to corrode (eg. steel (Fe))

to protect metal from oxidation

this coating is of a metal that is a than the metal that is to be protected…the

coating metal will react instead and is called the

applying a coat of paint

plated

stronger reducing agent

sacrificial anode

3. Cathodic Protection

an is connected by a to the pipeline

used to protect steel (iron) in buried fuel tanks and pipelines

because Mg is a than the iron in the steel, the Mg supplies the and the iron becomes the cathode and is protected

active metal (eg. Mg) wire

stronger reducing agente- for reduction

is attached to the hull of ships to perform the same function titanium

4. Alloying

stainless steel contains chromium and nickel, changing steels reduction potential to one characteristic of

(basically unreactive)

alloying pure metals changes their reduction potential

noble metals like gold

5. Electroplating plating is the process of

by metal ions in solution

an object can be plated by making it the in an containing ions of the plating metal

depositing the neutral metal on the cathode reducing

cathodeelectrolytic cell