A. Redox Reactions

electrochemistry is the branch of chemistry that studies

Electrochemistry

electron transfer in chemical reactions

is a oxidation

loss of electrons “LEO”

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

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

is a reduction

gain of electrons “GER”

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

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

oxidation and reduction reactions occur together, hence the term redox

the reduction and oxidation reactions are called the

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

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

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-)

reduced oxidizing agent OA

oxidized reducing agentRA

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.

3 Zn(s) + 2 Au(NO3)3(aq) 2 Au(s) + 3 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

B. Spontaneous Redox Reactions chemical reactions which occur on their own,

without the input of , are called

not all reactions are spontaneous

additional energy spontaneou

s

in the table of redox half reactions (see pg 7 in Data Booklet), the is at the top left and the is at the bottom right

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

the rule states that a spontaneous reaction occurs if the agent is above the agent in the table of redox half reactions

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

C. 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 compoundsacids

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 must be higher than SRA to be spontaneous)

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

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

OA/RASS

[ ]

5

MnO4-(aq) SO4

2-

(aq) OA OA/ RA

spont

D. 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 7 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

Put the oxidizing agents in order from strongest to weakest.

Put the reducing agents in order from strongest to weakest.

Ag+(aq), Cu2+(aq), Pb2+(aq), Zn2+(aq)

Zn(s), Pb(s), Cu(s), Ag(s)

Redox Table

+ e- Ag(s)

Cu2+(aq) + 2e-

Hg2+(aq) + 2e-

Zn2+(aq) + 2e-

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 reactionZn(s) + Cu2+(aq) Zn2+(aq) +

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

Ag+(aq)

Cu(s)Zn(s)

Hg(l)

Label each as OA or Ra

Example 2 (continued):

Put the oxidizing agents in order from weakest to strongest.

Put the reducing agents in order from weakest to strongest.

Zn2+(aq), Cu2+(aq), Ag+(aq), Hg2+(aq)

Hg(l), Ag(s), Cu(s), Zn(s)

Redox Table

Z2 (g) + 2e- + 2e- 2Y-

(aq)

+ 2e- 2W-

(aq)

X2 (g) + 2e-

SOA

SRA

Example 3:Generate a redox table given the following data:  2X-(aq) + Y2(g) spontaneous reaction

2Z-(aq) + Y2 (g) no reaction2Z-(aq) + W2 (g) spontaneous

reaction

Label each as OA or RA

Y2 (g)

2X-(aq)

2Z-(aq)W2 (g)

Example 3 (continued):

Put the oxidizing agents in order from strongest to weakest .

Put the reducing agents in order from strongest to weakest .

W2(g),

X-(aq),

Z2(g), Y2(g), X2(g)

Y-(aq), Z-(aq), W-(aq)

E. Oxidation Numbers (States) an 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 numberappears

molecular compound ion

charges

1. if the number then has occurred

decreases reduction

2. if the number then has occurred

increases oxidation

neutral molecule polyatomic 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 H12O6

C2 H5 O H+1

–2+1+5

+1

+12

figuring out oxidation numbers can help to identify whether a reaction is a or not

for it to be a redox reaction, there has to be an in oxidation number and a in oxidation number seen in the reaction

Ag(s) + NaNO3(aq) Na(s) + AgNO3(aq)

PbSO4(aq) + 2 KI(aq) PbI2(s) + K2SO4(aq)

eg)

redox reaction

0 +1-1 -1+10

Ag increases oxidized

Na decreases reduced…

redox!!!

nothing changes NOT a redox reaction!

bothincrease decrease

+2-2 -2+2-1-1+1 +1

electron transfer occurs in eg)

living systems photosynthesis, cellular respiration

also occurs in

eg)

non-living systems

combustion, bleaching, metallurgy

F. 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

G. Balancing Redox Reactions sometimes most reactants and products are

known but the complete reaction is not given…called a reaction skeleton

1. Half Reaction Method 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 : 

+6 2 +32

4 = 1+3+6 8 = 2

(Cr is already balanced)

+3 e– +2 H2O(l)

+4 H+(aq) CrO42-(aq) CrO2

-(aq)

net charge = –1

net charge = –1

(+6) (+3)

Example 2:Balance the following half reaction: 

+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

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 reactions H2O(l) H+(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

SO42-

(aq)

Net

2. Oxidation Number Method 1. Assign oxidation numbers.

2. Balance the substances that change in oxidation number.

3. Use a to join the reducing agent with its corresponding product (ignore the H+(aq) and H2O(l)) and a to join the oxidizing agent with its corresponding product. 4. On each line, write the in oxidation number # of atoms.

5. the RA and/or OA to balance the change in oxidation number.

6. the H2O(l) and the H+(aq).

line

line

change

Multiply

Balance

Example 1:Balance the following reaction using the oxidation number method.

__ H+(aq) + __MnO4-(aq) + __ SO3

2-(aq) __ MnO2(aq) + __ SO42-(aq) + __

H2O(l)

2 2 3 32 1+7 +4 +4 +6

= 3 1 atom =

= +2 1 atom =

3 2 =6

+2 3 =+6

Example 2:Balance the following reaction using the oxidation number method.

__ H2O(l) + __N2O4(g) + _ Br(aq) _ NO2 (aq) + _ BrO3

(aq) + __ H+(aq)

3 3 1 12 6+4 -1 +3 +5

= 1 2 atoms =

= +6 1 atom =

2 3 =6

+6

6

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/RA

SRA SOA

m = 100 gM = 52.00 g/mol

n = 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) + 5Fe3+

(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 (Cr2O7

2-(aq)) –

orange

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

permanganate ions (dichromate ions) colourless (green)

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

concentration

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

KMnO4(aq).

17.40

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 mL

3 = 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 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

chemicalelectrica

l

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 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 electrons (leaving a net charge in the electrode)

negative

anions anodeloses positive

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 anode is labelled negative

the move to the since this electrode electrons (leaving a net charge in the electrode)

positive

cations cathode

negativeaccepts

electrons flow from the to the through a connecting wire 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

anode (LEOA)cathode (GERC)

moveporous cup

salt bridge

electrical neutrality

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 cupsalt bridge half

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

you also have to show and label the reduction half reaction, oxidation half reaction and net reaction including E values, E net and 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 VAn (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 reactants are

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)

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 electrolytic cell, energy

is used to force a chemical reaction to occur (opposite of a voltaic cell)

commonly used to (eg. gold, silver, bronze, chromium etc),

these reactions have a Enet

nonspontaneous

negative

electroplate metals

(eg. H2, O2, Cl2 etc)

recharge batteries, useful gases

and split compounds into

electrical

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 or 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

Voltaic Cells Electrolytic Cells chemical to electrical

energy electrical to chemical

energy

usually contains porous cup or salt bridge

does not (usually) 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 Al

chlor-alkali plantchlorine 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/RA

SRA 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 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 our rules… chlorine (ions)

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

chloride ions

Example 4An electric current is passed through a solution of sodium chloride. Predict the anode and cathode reactions, overall reaction, and minimum voltage required.

Cathode SOA(Red):

Anode SRA(Ox):

Net:2 H2O(l) + 2 Cl-(aq) H2(g) + Cl2(g) + 2 OH- (aq)

Na+(aq) Cl-

(aq) H2O(l) OA RA OA/RA

SRA SOA

Enet = -2.19 V

2 Cl-(aq) Cl2(g) + 2e- E = -1.36

V

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

(aq)

E = -0.83 V

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 3)

(q) Coulomb (C)

1.60 x 10-19 C

mole 9.65 x 104

C

9.65 x 104

C/mol Faraday constant (F)

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 1An electrochemical cell caused a 0.0720 mol of e- to flow through a wire. Calculate the charge.

ne- = 0.0720 molF = 9.65 x 104

C/mol

ne- = q F0.0720 mol = q

9.65 x 104

C/mol q = 6948 C

= 6.95 103 C

Example 2Determine 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 3If 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. 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)

anodecathode

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)

M. Prevention of Corrosion

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

applying a coating of 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

paint

plated

stronger reducing agent

sacrificial anode

Fe

Zn coating

this method is also calledcathodic protection

has been in use before the science of electrochemistry was developed

Sir Humphrey Davy first used cathodic protection on British naval ships in 1824!

because the attached metal is a than the iron in the steel, the more active metal supplies the and therefore the steel (iron) becomes the cathode and is protected

can be used to protect any metal but steel (iron) is most commonly protected

we use steel (iron) for many structures such as buried fuel tanks, septic tanks, pipelines, hulls of ships, bridge supports etc

to protect these structures by cathodic protection, an is connected by a to the structure

active metal (eg. Mg, Zn, Al)wire

stronger reducing agent

e- for reduction

Fe (tank)

more active metal eg) Mg, Zn, Al

stainless steel contains chromium and nickel, changing steels reduction potential to one characteristic of (basically unreactive)

another protection method is alloying pure metals, which changes their reduction

potential

noble metals like gold

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

electroplating

Electrolytic Cell