Unit 5 Electrochemistry 5.1 Introduction Electrochemistry

• Branch of chemistry concerned with the conversion of chemical energy to electrical energy and vice versa

• Explains Copper in different acids

• Demo Cu and Concentrated HCl and HNO3 Why ships apply current to their hulls Cleaning silverware with aluminum foil Why the waste from copper refining is so valuable

Electrochemistry has lots of overlap with acid base chemistry • Similar table (see bible)

If you put copper in a clear silver nitrate solution, the solution turns blue and pure silver crystals appear on the coil as electrons are transfered.

2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+

(aq)

You can also use reactions to make electricity using the electrochemical cell

The voltmeter will show that electrons are flowing through the wire from the zinc strip to the copper strip. The reaction Cu2+

(aq) + Zn(s) Cu(s) + Zn2+(aq) only involved the

loss or gain of electrons. The cell split the reaction in half so that the copper is reacting on the left half

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

and the zinc is reacting on the right half Zn(s) Zn2+

(aq) + 2e- and the electrons moved through the wire. Each half of an electrochemical cell is called a half-cell. The reaction in each half is called a half-reaction. The salt bridge is crucial to the cell functioning. Take it out and the reaction stops. Oxidation reaction

• The half reaction in which a species loses electrons Reduction reaction

• The half reaction in which a species gains electrons “LEO the lion says GER”

• Loss of Electrons is Oxidation • Gain of Electrons is Reduction

Redox reactions are reactions involving the loss and gain of electrons In 2Ag+

(aq) + Cu(s) 2Ag(s) + Cu2+(aq) ,

• Ag+ is the agent which causes Cu to be oxidized Oxidizing agent

• Cu is the agent which causes Ag+ to be reduced Reducing agent

If a species changes its charge then the reaction is a redox reaction

• If a species becomes more positive or less negative (loss of electrons) it is oxidation

• If a species becomes less positive or more negative (gain of electrons) it is reduction

Ex Zn2+ + Mg Zn + Mg2+ The half reactions are

Zn2+ + 2e- Zn Mg 2e- + Mg2+

Zn2+ is reduced and is the oxidizing agent Mg is oxidized and is the reducing agent Assign 1-2

5.2 Oxidation Numbers Oxidation number

• The charge that an atom would possess if the species containing the atom were made up of ions

• Imaginary, since most molecules are held together with covalent bonds

• The sum of the positive and negative charges must equal the overall charge.

• Column 1 = +1 • Column 2 = +2 • Oxygen = -2 • Halogens usually = -1

Ex What is the oxidation # of P in H4P2O7 ? H4 P2 O7 = Individual charge for an atom +1 x -2 Total charge +4 2x -14 0 4+2x-14=0 x=+5 The oxidation # of P is +5 Ex What is the oxidation # of P in P4? 4x = 0 (the molecule is neutral) …x=0 Ex What is the oxidation # of Cr3+? X=+3 (the oxidation # of a monatomic ion is the charge of the ion)

Ex What is the oxidation # of S in SO4

2-? S O4 2- Individual charge x -2 Total charge x -8 -2 x-8=-2 x=+6 Trick They call it oxidation because oxygen is the most common oxidizing agent

• If more oxygen has been attached oxidation has occurred

• If oxygen has been removed, reduction has occurred • Demo (NH4)2Cr2O7 --> 4H2O + N2 + Cr2O3

Cr+6 is reduced… N-3 is oxidized Assign 3-6

5.3 Predicting the spontaneity of a redox reaction On the table of Standard Reduction Potentials

• On left tendency to reduce (strength as an oxidizing agent) increases up

• On the right tendency to oxidize (strength as a reducing agent) increases down

• Metals tend to be on the bottom half on the right Except Cu, Ag, Hg, Au (clustered on periodic

table) • Halogens and oxyanions are on the top at the left • Some metals have more than one common oxidation #

Fe, Sn, Cr, Hg, Cu More than one half reaction on the table Can be found on both sides of the table

• H2O2 is at the top left and bottom right • Species on the top left have a tendency to go forward

(reduction) • Species on the bottom right have a tendency to go

backward (oxidation) • All of the reactions can go forward or backward

Zn2+ + 2e- --> Zn (written as a reduction) Zn --> Zn2+ + 2e- (written as an oxidation)

• Use single arrows if you know it is reduction or oxidation. Use double arrows if aren’t specifying direction

If you have two half-cells joined, you need to figure out which one is reduction and which one is oxidation.

• Cu in a Cu2+ solution and Zn in a Zn2+ solution • Cu2+

+ 2 e- <--> Cu • Zn2++ 2 e- <--> Zn

• Use the table to find which reaction is more likely to go forward (top left) Cu2+ is higher so the reduction is

• Cu2+ + 2 e- --> Cu

Zn is lower on the right so the oxidation is • Zn --> Zn2++ 2 e-

If two half-cells are joined the higher half reaction will undergo reduction and the lower half reaction will undergo oxidation. Cu2+

+ 2 e- --> Cu Zn2++ 2 e- <-- Zn The species in bold will react to give the species on the other side of the arrow.

• The reactants are used up and the products are created • If only one species of a half reaction is present, the

other species is not present unless you have been told that it is there.

If you are given two potential reactions you can use the table to determine if a reaction will occur:

• If both reactants are found only on the left or only on the right then no reaction is possible Cu and Zn = no reaction

• If one reactant is on the left and the other is on the right there are two possibilities The reactant to be reduced (left side) is higher

than the reactant to be oxidized (right side) • Spontaneous reaction goes to 100%

completion • Cu2+ and Zn

The reactant to be reduced (left side) is lower than the reactant to be oxidized (right side)

• No reaction occurs

• Cu and Zn2+ A reaction will be spontaneous only if there is a reactant to be reduced (left) above a reactant to be oxidized (right) Ex

• Cl2 and Br- ? Left above right…spontaneous

• Sn and Mn Two rights …no reaction

• Ni2+ and Pb Right above left … no reaction

H+ is a reactant for many half reactions and therefore the reaction will only go if H+ is present (acidic solution)

• H+ can be reduced to H2 Assign 7-18

5.4 Balancing half reactions Half reactions must be balanced for mass and charge

• Those who are sloppy about including charges for ions are in trouble!

• You will be given a skeleton equation (major atoms) and then you must fill in the rest.

1. Balance atoms other than O and H 2. Balance oxygen atoms by adding H2O (most reactions occur

in water) 3. Balance H by adding H+ (we start with the assumption of

working in acid) 4. Balance the charge by adding electrons

Never vary the order because each step is affected by the proceeding step “Major Hydroxide” mnemonic:

• MAJOR OH- Major species Oxygen Hydrogen e-

Ex Balance RuO2 ↔Ru

1. RuO2 ↔Ru 2. RuO2 ↔Ru + 2H2O 3. RuO2 + 4H+ ↔Ru + 2H2O 4. RuO2 + 4H+ + 4e-↔Ru + 2H2O

Note: By adding negative charges (electrons) the charge on a side can only be reduced

• Find the charge on both sides • Find the difference in charge on both sides • To the side with the greater charge add a number of

electrons equal to the difference in charge Ex Z6+ --> Z2+

• The difference is 4 so add 4 e- to the left side • Z6+ + 4 e- --> Z2+

Ex X2- --> X+ • The difference is 3 so add 3 e- to the right • X2- --> X+ + 3 e-

Ex Balance Cr2O7

2- ↔ Cr3+ in acidic solution

1. Cr2O72-

↔ 2Cr3+ 2. Cr2O7

2- ↔ 2Cr3+ + 7H2O

3. Cr2O72-

+ 14H+ ↔ 2Cr3+ + 7H2O 4. Cr2O7

2- + 14H+ + 6e- ↔ 2Cr3+ + 7H2O

Ex Balance Pb ↔HPbO2

- in basic solution 1. Pb ↔HPbO2

- 2. Pb + 2H2O ↔HPbO2

- 3. Pb + 2H2O ↔HPbO2

- + 3H+ 4. Pb + 2H2O ↔HPbO2

- + 3H+ + 2e- Use the self ionization of water to cancel all the H+

• Pb + 2H2O ↔HPbO2- + 3H+ + 2e- plus

• 3H+ + 3OH- ↔ 3H2O equals • Pb + 3OH-↔ H2O + HPbO2

-+ 2e- Ex balance H2 ↔

1. H2 ↔ 2. H2 ↔ 3. H2 ↔ 2H+ 4. H2 ↔ 2H+ + 2e-

Assign 19 ace…20-23all

5.5 Balancing redox equations using half reactions

There are two ways of balancing redox equations

• Half reactions Longer but easier (especially on more

complicated equations) • Oxidation numbers

Faster (for simple equations) but harder • You can use whichever you want.

Separate the reaction into separate reduction and oxidation half reactions. Ex: Balance Os + IO3

- --> OsO4 + I2 (acidic solution) 1. break into 2 half reactions:

Os --> OsO4 IO3

- --> I2 2. Balance the individual half reactions:

Os + 4H2O --> OsO4 + 8H+ + 8e- 2 IO3

- + 12H+ +10e- --> I2 + 6H2O 3. Multiply the half reactions by whole numbers to get equal

numbers of electrons on both sides: 5X [Os + 4H2O --> OsO4 + 8H+ + 8e-] 4X [2 IO3

- + 12H+ +10e- --> I2 + 6H2O] 4. Add the two half reactions together and cancel off anything

that appears on both sides: 5Os + 20H2O --> 5OsO4 + 40H+ + 40e-

+ 8 IO3- + 48H+ +40e- --> 4I2 + 24H2O

5Os + 8 IO3- + 8H+ --> 5OsO4+ 4I2 + 4H2O

Ex: Balance MnO4

- + C2O42- --> MnO2 + CO2 (basic solution)

1. MnO4---> MnO2

C2O42- --> CO2

2. 3e- + 4H+ + MnO4---> MnO2 + 2H2O

C2O42- --> 2CO2 + 2e-

3. 2x [3e- + 4H+ + MnO4---> MnO2 + 2H2O ]

3x [C2O42- --> 2CO2 + 2e-]

4. 6e- + 8H+ + 2MnO4---> 2MnO2 + 4H2O

3C2O42- --> 6CO2 + 6e-

8H+ + 2MnO4- +3C2O4

2- -->2MnO2 + 4H2O + 6CO2 5. Convert the resulting redox equation to basic conditions 8H+ + 2MnO4+3C2O4

2- -->2MnO2 + 4H2O + 6CO2 8H2O --> 8H+ + 8OH-

2MnO4+3C2O42-+4H2O -->2MnO2 + 6CO2 + 8OH-

Disproportionation

• Reaction in which the same species is both oxidized and reduced

Ex: Balance ClO2- --> ClO3

- + Cl- (basic solution) 1. Break into half reactions (use the reactant twice)

ClO2- --> ClO3

-

ClO2- --> Cl-

2. Balance

H2O+ ClO2- --> ClO3

- + 2H+ + 2e-

4e- + 4H+ + ClO2- --> Cl- + 2H2O

3. Cross multiply 2x [H2O+ ClO2

- --> ClO3- + 2H+ + 2e-]

4e- + 4H+ + ClO2- --> Cl- + 2H2O

4. Add 2H2O+ 2ClO2

- --> 2ClO3- + 4H+ + 4e-

4e- + 4H+ + ClO2- --> Cl- + 2H2O

3ClO2--->2ClO3

- + Cl-

(didn’t have any H+ to turn to OH-)

Assign 24 (a, d, g, j…)

5.6 Balancing redox equations using oxidation numbers The gain or loss of electrons is directly related to a decrease or increase in oxidation number and the total number of electrons lost must equal the total number of electrons gained. Therefore:

• an increase in the oxidation number in one species must be balanced by a decrease in another

• The overall change in oxidation numbers must be zero in any redox equation

Ex: Balance U4+ + MnO4

- --> Mn2+ + UO22+

1. Assign oxidation numbers to all atoms involved in a change of oxidation # (∆ON)

U4+ + MnO4- --> Mn2+ + UO2

2+ U4+ MnO4

- Mn2+ UO22+

+4 +7 +2 +6

∆ON =-5 ∆ON =+2 2. Since the ∆ON must equal 0, cross multiply each half reaction

by the changes in oxidation numbers.

5x [U4+ --> UO22+] : ∆ON = 5 x (+2) = +10

2x [MnO4- --> Mn2+] : ∆ON = 2 x (-5) = -10

Total ∆ON = 10 -10 =0 3. Add the reactions

5U4+ + 2MnO4- --> 2Mn2+ + 5UO2

2+ 4. Balance the O’s by adding H2O and the H's by adding H+.

5U4+ + 2MnO4- + 2H2O --> 2Mn2+ + 5UO2

2+ + 4H+

Ex: Balance Zn + As2O3 --> AsH3 + Zn2+ in basic solution

1. Balance the major atoms Zn + As2O3 --> 2AsH3 + Zn2+

2. Assign oxidation numbers and look at change in oxidation number • Zn goes from 0 to 2+…∆ON = +2 • As goes from +3 to –3 but there are two of them so

∆ON = 2(-6) = -12 • In order to cancel the change in oxidation number the

first reaction must happen 6 times and the second happens once

6x [Zn --> Zn2+] ∆ON =+12 1x [As2O3 --> 2AsH3] ∆ON =-12

6Zn + As2O3 --> 2AsH3 + 6Zn2+ ∆ON =0

There is a shortcut for making a basic solution • Instead of adding H+ and then canceling with the self

ionization of water just add OH- right away to balance the charge

6Zn + As2O3 --> 2AsH3 + 6Zn2+ + 12OH- • Now when you add water you will balance both H and

O (if you didn’t make a mistake to this point) 6Zn + As2O3 + 9H2O --> 2AsH3 + 6Zn2+ + 12OH-

Ex: Balance P4 --> H2PO2

- + PH3 in acid • P4 must be both reduced and oxidized

0 0 +1 -3 P4 + P4 --> 4H2PO2

- + 4PH3 ∆ON=4(+1)=4 ∆ON=4(-3)=-12

• The oxidation needs to happen 3 times for each reduction

4P4 --> 12H2PO2- + 4PH3

• Which simplifies to P4 --> 3H2PO2

- + PH3

• Balance O and H with water and H+ 6H2O + P4 --> 3H2PO2

- + PH3 + 3H+

• Check the charge 0-->-3+3=0 Assign 25 a, d, g…

5.7 Redox titrations Redox titrations allow the accurate determination of an unknown concentration of a species that is either undergoing oxidation or reduction. Oxidizing agents

• acidic KMnO4 is one of the most useful as it is reduced to Mn2+ it changes colour

• it acts as it’s own indicator it can oxidize many substances

MnO4- + 8H+ + 5e- Mn2+ + 4H2O; Eo = 1.49V

Purple Colourless Ex To find the [Fe2+] in a solution you can use the fact that it oxidizes to Fe3+. So…

5x [ Fe2+ Fe3+ + e-] MnO4

- + 8H+ + 5e- Mn2+ + 4H2O MnO4

- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+ • As KMnO4 is added the MnO4

- is destroyed until all the last of the Fe2+ is used up

o A tiny bit of MnO4- will remain at the end point and the

solution will have a hint of purple Ex When 25.00mL of an unknown Fe2+ solution is titrated with 17.52mL of 0.1000M acidic KMnO4. What is the [Fe2+]?

MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + Fe3+

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redoxNew/redox.html Reducing agents • The iodide ion can reduce many substances

2I- I2 + 2e-

colorless yellow • The titration is more complicated

o First oxidize the I- to I2 o Second, the I2 is reduced back by a second reducing

agent (S2O32-)

Ex the reduction of laundry bleach NaOCl 2I- I2 + 2e-

2e- + 2H+ + OCl- Cl- + H2O 2H+ + OCl- + 2I- Cl- + H2O + I2

• a deliberate excess of I- is used to make sure that all of the OCl- is used up.

• The second reaction is between I2 and S2O32-

I2 + 2e-2I-

2S2O32- S4O6

2- + 2 e-

I2 +2S2O32- S4O6

2- + 2I- • When most of the I2 is used up there is only a faint

brown colour. Starch is added which goes deep blue with the iodine and the titration is continued until the blue colour disappears

Ex 25.00 mL of bleach is reacted with excess KI according to

2H+ + OCl- + 2I- Cl- + H2O + I2 the I2 produced required 46.84mL of 0.7500M Na2S2O3 to reach the end point according to

2S2O32- + I2 S4O6

2- + 2I-

using starch as an indicator. What is the [OCl-] in the bleach?

The I2 produced in the first reaction is used up in the second reaction.

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redoxNew/redox.html Assign 26-32 even

5.8 The Electrochemical Cell Electrode • a conductor at which a half reaction occurs Anode • the electrode at which oxidation occurs (vowels) • the electrode receiving electrons from the substance being

oxidized • the electrode towards which anions travel Cathode • the electrode at which reduction occurs (consonants) • the electrode supplying electrons to the substance being reduced • the electrode towards which cations travel An ox cared

• Anode is where oxidation occurs and • cathode is where reduction occurs

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/volticCell.html Look at page 215 • Starting with Ag↔Ag++ e- and Cu↔Cu2+ + 2e- • Copper oxidizes more easily

o More likely to lose it’s electrons and form the ion o Copper electrode is the anode

• As electrons flow through the wire towards Ag the equilibrium is upset and shifts

o AgAg++ e- o Reduction occurs and the silver electrode is the

cathode • Electrons flow from the anode to the cathode (alphabetical)

• Water and some ions can pass through the salt bridge but there is not a free mixing of the solutions

o Electrons can’t pass directly from the copper to the Ag+ and are forced through the wire

• As Cu2+ ions are formed they accumulate around the anode o Depleted by random motion and repulsion of like

charges • Negative ions are attracted to the anode (NO4

2-) • As [Ag+] is depleted at the cathode the net positive charge is

decreased around the cathode o Ag+ ions migrate towards the cathode due to random

motion o Negative ions (NO3

-) move away from the cathode (towards the anode)

Only ions flow in the solution (no electrons) The number of electrons in oxidation = number of electrons involved in reduction Complete the example cell labeling on page 216 together Assign 34-35

5.9 Standard reduction potentials The tendency of electrons to flow in an electrochemical cell is called the voltage or electrical potential to do work Voltage • work done per electron transferred Electrons can’t flow in an isolated half cell • an individual half cell voltage can’t be determined • the difference in electrical potentials for different half cells can

be determined o you may not know your mass in kg but if I give you a

25kg object you know the difference between your current and previous masses is 25kg.

A zero point is arbitrarily defined on the voltage scale • the hydrogen half cell is defined as

2H+(aq) + 2e- H2(g) ; Eo= 0.000V

• Eo is the standard reduction potential o The o implies this is a standard state

25oC all gases at 101.3kPa (atm) all elements are in their standard state

• normal phase at 25oC All reactants and products are at 1M

When all of the reactants in a half reaction are ions, liquids or gases an inert electrode like platinum or carbon is used. ClO4

- + 8H+ + 8e- ↔ Cl- + 4H2O has a standard reduction potential of 1.39 V. But this voltage only applies if all the reactants are 1M, regardless of the coefficients in the balanced equation, otherwise the voltage will differ and the o is removed from the E terms.

The standard reduction potentials are measured relative to the hydrogen half cell. http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/SHECu.html http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/SHEZnV7.html All of the voltages are for reductions, if the half reaction is reversed (oxidation) then reverse the sign of the voltage

Zn2+ + 2e- Zn ; Eo= -0.76V Zn Zn2+ + 2e- ; Eo= +0.76V

What happens when the following half cells are joined?

Hg2+ + 2e- Hg ; Eo= 0.85V Cu2+ + 2e- Cu ; Eo= 0.34V

• The first half reaction is higher up on the left and goes as written and the second one will be reversed

• The voltage is also added to give the voltage for the redox equation

Hg2+ + 2e- Hg ; Eo= 0.85V Cu Cu2+ + 2e- ; Eo= -0.34V

Hg2+ + Cu Hg + Cu2+ ; Eo= 0.51V • In general

o Eocell = Eo

red - Eoox

o The potential of a cell is just the difference between half cell voltages for the reduction and oxidation half reactions

Note that if the Eocell is positive then the reaction is spontaneous

(top left and bottom right) and can do work Ex Calculate the potential of Ni2+ + Fe Ni + Fe2+ The half reactions are

Ni2++ 2e- Ni ; Eo = -0.26V Fe Fe2+ + 2e- ; Eo = +0.45V (reversed from table)

Ni2+ + Fe Ni + Fe2+ ; Eo = +0.19V

Or Eo

cell = Eored - Eo

ox Eo

cell = (-0.26 V) – (-0.45V)= 0.19V The reaction is spontaneous because it has a positive Eo

cell Calculate potential of Ni + Fe2+ Ni2+ + Fe

Fe2+ +2e- Fe ; Eo=-0.45V Ni Ni2+ +2e- ; Eo= +0.26V

Ni + Fe2+ Ni2+ + Fe ; Eo= -0.19V

The reaction is not spontaneous • Negative Eo

cell • It will take work to make the cell operate

Calculate the potential for 3Ag+ + Al 3Ag + Al3+

Ag+ + e- Ag ; Eo=0.80V Al Al3+ + 3e- ; Eo= 1.66V

• Do you multiply the voltage by 3 when you multiply the first half reaction by 3? No. Voltage gives work per electron Work/electron = 0.80V (3X work)/ 3 electrons = 0.80 V Eo

cell=2.46 V Note that just because a reaction is spontaneous does not mean that it is necessarily fast

• This reaction is normally slow unless you follow the procedure on page 222

Water is almost always present and must be considered

• +0.82V for the oxidation • -0.41 V for the reduction (there are 2 ways of writting

the reduction) 2H+ (10-7) + 2e- ↔ H2(g) ; Eo=-0.41 V

2H2O +2e-↔ H2(g) + OH- (10-7) ; Eo=-0.41 V If a reaction occurs in acidic solution the reduction of H+ (0.00V) must be considered The surface area of the electrode has not effect on the cell voltage

• the concentration of a solid does not change as the solid is added or removed

• The surface area does affect the rate of the reaction More sites for electron exchange means a faster

reaction More amps

If cell are not at standard conditions you can use Le Chatelier’s Principle to predict in general the effect on the potential.

• Cu2+ + 2e- Cu Eo = 0.34V If the [Cu2+] is increased to 2M then the tendency

to reduce is stronger and the E > 0.34V A working cell is not at equilibrium. As the reactants are used up the concentrations of the oxidizing agents drop and the reducing agents rise. This brings the voltages closer together until the difference in reduction potentials is zero. The cell is used up when the cell reached equilibrium. Assign 36-46 even

gZn Zn

Ag

Cu

5.10 Selecting Preferred Reactions When a cell contains a mixture of reactants, several different reactions might seem possible at first. Draw page 226 Zn2+ Ag+

NO2- Cu2+

NO3-

There are three possible half reactions

Ag+ + e- Ag(s) ; Eo = 0.80V Cu2+ + 2e- Cu(s); Eo = 0.34V Zn2+ + 2e- Zn(s) ; Eo = -0.76V

• When several different reduction half reactions can occur the half reaction with the highest tendency to accept electrons (highest reduction potential) will occur preferentially • Ag+ to Ag(s)

• When several different oxidation half reactions can occur the half reaction with the highest tendency to lose electrons (lowest reduction potential) will occur preferentially • Zn(s) to Zn2+

List the species present

• Break ionic compounds into ions first Start at the upper left hand side of the table and work down until you find the first match with one of the species listed

• This is the reduction

Start at the lower right hand side of the table and work up until you find a match with one of the species listed

• This is the oxidation Ex: An iron strip is placed in a mixture of Br2(aq) and I2(aq). What is the preferred reaction? The species present are Br2 + 2e- 2Br- ; Eo = 1.09V I2 + 2e- 2I- ; Eo = 0.54V Fe2+ + 2e- Fe ; Eo = -0.45V The iron is the only species that can be oxidized and the Br2 has a greater tendency to be reduced so

Br2 + 2e- 2Br- ; Eo = +1.09V Fe Fe2+ + 2e-; Eo = +0.45V

Fe + Br22Br- + Fe2+ ; Eo = 1.54V

Ex: A beaker contains an iron nail wrapped in a piece of copper wire and a piece of magnesium ribbon, immersed in a solution containing CuSO4 and some dissolved Cl2. What is the overall reaction? The species present are Fe, Cu, Mg, Cu2+, SO4

2-, Cl2, H2O • Looking down the left the highest reduction reaction is

Cl2 + 2e- 2Cl- ; Eo = 1.36V • Looking up the right the lowest oxidation reaction is

Mg2+ + 2e- Mg ; Eo = -2.37V • Adding the reactions gives

Cl2 + 2e- 2Cl- ; Eo = 1.36V Mg Mg2+ + 2e- ; Eo = +2.37V

Cl2 + Mg 2Cl- + Mg2+; Eo = 3.73V

Spectator ions • An ion capable of being reduced, with another ion that is more

easily reduced in the same solution

• An ion capable of being oxidized, with another ion that is more easily oxidized in the same solution

• Common examples • Na+, K+, Ca2+, Mg2+, SO4

2- (neutral), Cl- Assign 47-48

5.11 Applied Electrochemistry The Breathalyser

• ethanol can be reduced by acidic dichromate ions C2H5OH+ K2Cr2O7 +H2SO4CH3COOH+Cr2(SO4)3+K2SO4+H2O

• determines the amount of alcohol consumed based on the colour of the reaction Cr2O7

2- is orange yellow Cr3+ is green A spectrophotometer determines exactly how

much of each colour is present • Some of the alcohol in your blood stream diffuses from

your capillaries to your alveoli and is breathed out It is an equilibrium process

• The breath alcohol content (BAC) is a very good measure of blood alcohol content

• Except…some other substances are also oxidized by the dichromate ion o Methanol (death or blindness) o Isopropyl alcohol (leathal) o Butanol (toxic) o If you consumed these you’d be dead

so we assume the reason the test came out positive is that you’ve been drinking

Assign 49-51 Batteries and fuel cells Lead acid storage battery http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/PbbatteryV9web.html The anode reaction is

Pb + SO42- PbSO4 + H++ 2e- or

Simply PbPb2+ + 2e- The cathode reaction is

PbO2 + HSO42- + 3H+ + 2e- PbSO4 + 2H2O or

Pb4+ + 2e- Pb2+ Over all PbO2 + HSO4

2- + 2H+ + Pb + HSO42- 2PbSO4 + 2H2O or

Pb + Pb4+ 2Pb2+ As the cell is discharged lead sulphate coats the electrodes. To recharge the cell simply apply a voltage in the opposite direction and the reverse reaction occurs. 2PbSO4 + 2H2O PbO2 + HSO4

2- + 2H+ + Pb + HSO42- or

2Pb2+ Pb + Pb4+ • Some of the lead sulphate can fall off the electrode

leaving less to be recharged and the cell loses power over time

• The relative concentrations of H2SO4 and H2O can be used to determine the condition of a battery Look at the density

Assign 52-53 Zinc carbon batteries (common dry cell) http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/ZnCbatteryV8web.html

• The zinc liner is used up and forms Zn(NH3)42+ at the

anode Voltage drops then battery dies

• It is cheap • Can’t be recharged • Short shelf life

Assign 54 Alkaline Dry Cell

• Similar to the zinc carbon battery but the paste (KOH) is alkaline or basic

• Can deliver greater current and gives a more constant voltage

Assign 55 Fuel cells

• A device into which fuel is constantly fed and electricity is continuously obtained

• Pollution free • Incredibly efficient (70-80% vs 30% for a diesel

generator) • The most common kind is the hydrogen fuel cell • Expensive!

• 2H2 + 4OH- 4H2O + 4e- O2 + 2H2O + 4e- 4OH-

2H2 + O2 2 H2O Assign 56

5.12 Corrosion of metals: causes and prevention Rusting is the most common example

• oxidation of iron • needs oxygen and water

within a drop of water there is an oxygen rich area (cathode) and an oxygen poor area (anode)

• See page 233 In the oxygen poor region

• FeFe2+ + 2e- After being formed the Fe2+ ions migrate away

from the anode due to random motion • Exposes fresh Fe

In the oxygen rich area • ½ O2 + H2O + 2e- 2OH-

When the iron ion reaches the outside solid Fe(OH)2 precipitates out

• Leaves a ring of rust around the location of the drop

• Fe(OH)2 is eventually oxidized to Fe2O3 and H2O by the air

• Rust is Fe(OH)2.XH2O o Different amounts of water changes

the colour of the rust (red, yellow, brown and black)

A metal can corrode if it touches another metal in the presence of an electrolyte solution

• One metal will be oxidized Ex Iron touches copper and they get wet

• Fe Fe2+ + 2e- (iron has a greater tendency to oxidize than copper)

• The copper conducts the electrons away and makes them available to the oxygen and water touching the wire

• ½ O2 + H2O + 2e- 2OH-

Assign 57-59 Preventing corrosion

• it can be slowed down or stopped By preventing the corrosive environment from contacting the metal

• apply a protective layer like paint, plastic or oil to the surface oxygen can’t get through

• Apply a metal which is corrosion resistant to the surface of the metal Tin plated steel

• Tin oxide bonds to the steel • protects the metal beneath the surface

By using electrochemical principles to prevent corrosion in an otherwise corrosive environment

• cathodic protection protecting a substance from oxidation by

connecting it to a substance with a higher tendency to oxidize

if magnesium oxidizes more easily than iron then if they are in contact only the magnesium will oxidize (the iron will be the cathode and remain intact in its reduced form)

Zinc strips on iron hulls • Or hook up the hull to a generator…it can’t

lose electrons to oxidize Galvanized iron is coated in zinc which oxidizes

• zinc oxide is very hard and protects the steel Buried metal tanks are attached to a post made of

magnesium or zinc • Easier to replace a stake at the surface than

dig up the tank

• Change the conditions in the surroundings to lower the tendency of the surroundings to reduce

When iron is placed in oxygenated water the following half reaction oxidizes the iron

½ O2 + 2H+ (10-7) + 2e- 2H2O ; E = 0.82V If oxygen is removed from the water then this reaction with a much lower reduction potential occurs

2H+ (10-7) + 2e- H2 ; E = -0.41V Which is barely above the iron half cell

Fe2+ + 2e- Fe ; Eo = -0.45V Instead of removing oxygen you could remove the H+ needed in the first reaction. By adding OH- you can stop the rusting of iron.

2H2O + 2e- H2 + 2OH- ; Eo = -0.83V Assign 60-63

5.13 Electrolysis Electrolysis

• the process of supplying electrical energy to a molten ionic compound or a solution containing ions so as to produce a chemical change

Electrolytic cell or electrolysis cell • apparatus in which electrolysis can occur

Electrolysis supplies energy to drive non spontaneous reactions (Eo<0) When a molten salt like NaCl is electrolyzed its ions are free to move

• Opposites attract

At the anode

2Cl- Cl2 + 2e- ; Eo = -1.36V At the cathode

Na++ e- Na ; Eo = -2.71V 2Cl- + 2 Na+ Cl2 + 2Na; Eo = -4.07V

The battery must supply at least +4.07V to work • There is internal resistance and the cell is not at

standard conditions…really it’s around 10V

• The real cell is more complex because Chlorine and Sodium react so violently

Assign 64 Electrolysis of aqueous solutions Ex: NaI(aq) contains Na+, I- and H2O

• You must consider the possible oxidation or reduction of water

Two possible reductions • 2H2O + 2e- H2(g) + 2OH-(10-7M); Eo = -0.41V • Na+ + e- Na ; Eo = -2.71V

Two possible oxidations • H2O ½ O2 + 2H+(10-7M) +2e- Eo = -0.82V • 2I- I2 + 2e- ; Eo = -0.54V

When you hook up the cell and start applying the voltage, the first reaction to occur will be the one that requires the least voltage (-0.41 + -0.54)

• The half reactions having the greatest tendency to reduce (top left) and the greatest tendency to oxidize (bottom right) are preferred Same as for spontaneous reactions

2H2O + 2e- H2(g) + 2OH-(10-7M); Eo = -0.41V 2I- I2 + 2e- ; Eo = -0.54V

2H2O +2I- H2(g) + 2OH-(10-7M) + I2 ; Eocell = -0.95V

It will require at least +0.95V to operate Notes

• Don’t worry about the concentrations changing the results

• Most questions will be in neutral solutions If it is in acid just substitute the H+ for any metal

ions • Neutral solutions require you to consider the oxidation

of water at 0.82V and the reduction of water at –0.41V • Acidic solutions require you to consider the oxidation

of water at 1.23V and the reduction of hydrogen at 0.00V

• Water doesn’t behave quite as predicted The overpotential effect

• It takes more energy than expected to oxidize water

• Electrolysis of aqueous solutions of Cl- or Br- will produce Cl2 or Br2 not O2 at the anode.

Ex: What products are formed and what is the overall reaction when NiSO4(aq) is electrolyzed using inert electrodes? Determine the minimum voltage needed.

1. Establish the species present a. Ni2+, SO4

2-, H2O 2. Acid or neutral?

a. Neutral 3. Find the highest reduction

a. Ni2++2e- Ni ; Eo= -0.26V 4. Find the lowest oxidation

a. H2O ½ O2 + 2H+(10-7M) +2e- Eo = -0.82V 5. Determine the overall reaction

a. Ni2++ H2O ½ O2 + 2H+(10-7M)+ Ni ; Eo= -1.08V b. At least +1.08v must be applied

Ex: What is the overall reaction when 1MHCl(aq) is electrolyzed using carbon electrodes?

1. Establish the species present a. H+, Cl-, H2O

2. Acid or neutral? a. Acid

3. Find the highest reduction a. 2H++2e- H2 ; Eo= 0.00V

4. Find the lowest oxidation a. 2Cl- Cl2 +2e- Eo = -1.36V

5. Determine the overall reaction a. 2H++2Cl- Cl2 + H2 ; Eo= -1.36V

At least +1.36v must be applied Assign 65-72ace Electroplating

• electrolytic process in which an electrode is plated out at the cathode

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/electrolysis10.html

• the cathode is made of the material that will receive the plating supply electrons to the cathode

• the electroplating solution contains ions of the metal to be plated

• The anode usually made out of the metal that will be plated to recharge the ions used up

Assign 73-76 Electrorefining

• Purifies a metal (copper) by electrolysis • Impure metal is the anode

Only the copper is oxidized (lower on the right) and moves as ions to the cathode where pure copper forms

Impurities flake of and fall to the bottom of the cell (Gold, silver and platinum)

Production of aluminum

• Bauxite is converted to aluminum oxide trihydrate • The water is driven off • Cryolite is added to lower the melting point to 1000o • The molten aluminum oxide is electrolyzed

Anode C + O2 CO2 + 4e-

Cathode 2Al3+ + 6e- 2Al

A voltage of 4-6 volts is needed but requires a huge current (mainly to keep everything hot enough) Assign 78-80