notes electrochemistry

8/3/2019 NOTES Electrochemistry

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Electrochemistry

Electrochemistry brings together two topics you have already learned:

• Redox Reactions

• Thermodynamics

To learn about how to use chemical reactions to do work.Electrochemical cells:

• Batteries: Storing energy using electrochemical cells

• Corrosion: Unwanted electrochemical cells

• Electrolysis and electroplating calculations

Petrucci 5.4

Review of redox reactions

What you (should) know about redox reactions from Chem 2A:

• Redox reaction involves transfer of electrons

• Reduction — oxidation

• Oxidation is loss of electrons

• Reduction is gain of electrons

• Oxidizing agent: gains electrons, therefore is reduced itself (causes oxida-

tion)

• Reducing agent: losses electrons, so is oxidized (causes reduction)

• An oxidation and reduction must always occur together

• Redox reaction may all seem very different, but on a molecular level they

all involve the transfer of electrons and hence a change in oxidation states

Remember OILRIG

Oxidation Is Loss — Reduction Is Gain (of electrons)



Examples of redox reactions

• Some redox reactions have an obvious transfer of electrons

Zn (s) + 2 Ag+ (aq) Zn2+ (aq) + 2 Ag (s)

– Zn (s) looses two electrons (is oxidized )

– Each Ag+ ion gains an electron (is reduced )

• Other times it is less obvious because it happens along with breaking and

making of chemical bonds

CH4 + 2O2 CO2 + 2H2O

– Oxidation: CH4 + 2H2O CO2 + 8e− + 8H+ C looses 8 electrons

– Reduction: 8H+ + 8e− + 2O2 4H2O Each O gains 2 electrons

Petrucci 11 + 18

Thermodynamics

• Two driving forces for chemical reactions, Energy (U ) and Entropy (S )

– Minimize energy

– Maximize entropy

• Gibbs free energy

– Combine energy and entropy change

∆G = ∆H − T ∆S

– A reaction is spontaneous if

∆G < 0

and at equilibrium if ∆G = 0

∆Gand Spontaneity

aA + bB −→ cC + dD

∆Gr = ∆G◦

r + RTlnQ

∆G◦

r =

ν p∆G◦

f (products) −

ν r∆G◦

f (products)

∆G◦

r =

c∆G◦

f (C ) + d∆G◦

f (D)

−

a∆G◦

f (A) + b∆G◦

f (B)

• ∆G < 0 ⇒ Reaction is spontaneous in forward direction at current con-

centrations

• ∆G > 0 ⇒ Reaction is spontaneous in backward direction at current

concentrations

2



The reaction quotient, Q

Q ={C }c{D}d

{A}a{B}b

• {A }, {B }, {C }, {D } are the activities of A, B, C, and D, respectively

• In solution: {A} =γ [A]/c◦

• Gases: {A} = γP A/P ◦

• Solids: {A} = 1γ ≈ 1 for near ideal solutions and gases.

P ◦ is the pressure in the reference state (1 bar). Likewise c◦ = 1 M.

Thermodynamics — Work

• Chemical reactions can do work

– The energy released during a reaction is released as either heat and/or

work

∆U = q + w

– The size of q and w depends on how the reaction is carried out

– The maximum amount of work that can be done by a chemical reaction

(at constant pressure) equals −∆G

Energy transfer during a redox reaction

To harness the intrinsic chemical energy stored in these oxidizing and reducing

agents, we must separate them to force the electrons to flow though an external

circuit and do some work.

Zn(s) + Cu2+

(aq) → Zn2+

(aq) + Cu(s)

∆G◦

f =

νp∆G◦

f (products)−

νr∆G◦

f (reactants)

= (−147.1 kJ/mol + 0 kJ/mol)− (65.49 kJ/mol + 0 kJ/mol)

= −212.59kJ/mol

∆Gr = ∆G◦

r + RT lnQ

Q =

Zn 2+

Cu 2+

All energy is in this configuration released as heat

Petrucci 19.1

• Galvanic cell: Spontaneous redox reaction generate electricity

– Spontaneous ⇔ ∆G < 0.

– Produces electricity (Batteries, etc.)

3



• Electrolytic cell: Non-spontaneous redox reaction is driven by electric cur-

rent

– NON-spontaneous ⇔ ∆G > 0

– DRIVEN by electricity. (Charging of batteries, electroplating, etc)

Electrochemistry — Overview

• Chem2A knowledge: Redox reactions

– How to identify redox reactions

– Oxidation state

– How to balance redox reactions (acidic and basic environment)

• Utilizing redox reactions: Electrochemical cells

– Galvanic cells (spontaneous reaction generates electricity)– Electrolytic cell (driving non-spontaneous redox reactions with elec-

tricity)

• Chem2B knowledge: Thermodynamics and chemical equilibrium

– Gibbs free energy

– work

– Spontaneous and non-spontaneous reactions

– Equilibrium constant

– Concentration dependence of ∆G

• The connection between E cell, ∆G, and K

– Nernst equations

• USE

Petrucci 3.4

Oxidation States

• Binary ionic compounds

Metal + non-metal

– Electronegativity difference between metal and non-metal so large that

we assume that the electron is 100% located at the non-metal

4



• Molecular compounds

– The electrons are shared between atoms

Oxidation states

• Book keeping method, that is not necessarily the same as the actual electric

charge on an atom. It is only a formalism, but a very useful one!

• Imagine each atom in a molecule, formula unit, or polyatomic ion in ionic

form: The O.S. is the imaginative "‘charge"’ on each "‘ion"’

• Not necessarily the same as the actual electric charge on an atom. It is only

a formalism, but a very useful one!

• For purely ionic compounds the oxidation state does equal the formal charge

For example NaCl (Na+ and Cl−), CaBr2 (Ca2+ and Br−) not HCl and Cl2

• The oxidation number refers to an atom, not to a molecule or polyatomic

ion

Examples of Different Oxidation States of Nitrogen

N has 5 valence electrons

(Ox=0)

E.N.(N) >E.N.(H)

All shared electrons goes to N

N now has total of 8 electrons

3 more than neutral ⇒ Ox =−3

5



E.N.(O) >E.N.(N)

All shared electrons goes to ON only has 2 electrons left

3 less than neutral ⇒ Ox =+3

E.N.(O) >E.N.(N)All shared electrons goes to O

N has no electrons left

5 less than neutral ⇒ Ox =+5

Identical atoms, no "‘winner"’

Electrons are equally shared

N only has 5 electrons

Same as neutral ⇒ Ox =0

E.N.(N) >E.N.(H)

The electrons in N–H bond goes to ni-

trogen Between identical N atoms, no"‘winner"’

Electrons are equally shared in N–N

bond

Each N has 7 electrons

2 more than neutral ⇒ Ox =−2

1. In compounds, sum of oxidation numbers equals molecular charge

2. Oxidation state of free element is 0

3. For ions composed of only one atom, oxidation number equals charge

4. The alkali metals (row 1) in compounds are assigned and oxidation state of

+1

5. The alkali earth metals metals (row 2) in compounds are assigned and oxi-

dation state of +2

6. Fluorine is assigned oxidation state of −1

7. Hydrogen has an oxidation state of +1

• except when bonded to metals in binary compounds when it is −1

6



8. Oxygen is assigned oxidation state of −2

• except in peroxides when it is −1

9. In any binary (two-element) compound with metals,

• group VIIA have oxidation state of −1

• group VIA have oxidation state of −2

• group VA have oxidation state of −3

Use rules in order!

Trends in oxidation states

Oxidation States Example 1

• CO2

• NH+4

• SO2−

4

• SnBr4

• NaClO4

• P2O5

7



Petrucci 5.4

Identifying redox reactions

Assign oxidation numbers. If they change the reaction is a redox reaction

Oxidation of zinc by Copper(II) ions:

Zn (s) + Cu2+

(aq) → Zn2+

(aq) + Cu (s)

Oxidation of zinc by oxygen:

2 Zn (s) + O2

∆

→ 2ZnO (s)

Identifying Redox Reactions Example 2

Identify which of the following reactions are redox reactions

1. Ca (s) + Cl2 (g) → CaCl2 (s)

2. CaCl2 (s) → Ca2+ (aq) + 2 Cl− (aq)

3. CH3COOH (aq) + NaOH (aq) → H2O (l) + NaCl (aq)

4. 2Fe2O3 (s) + 3 C (s) → 4 Fe (s) + 3 CO2 (g)

Half reactions

A redox reaction consists of two half-reactions: An oxidation reaction that pro-

duces electrons and a reduction reaction consumes electrons

For example:

Oxidation Zn (s) → Zn2+ (aq) + 2 e−

Reduction Cu2+ (aq) + 2 e− → Cu (s)

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

Disproportionation ReactionsRedox reactions where the same compound is being oxidized and reduced.

Example: Reaction to prepare bleach

2NaOH (aq) + Cl2 (g) −→ NaCl (aq) + NaClO (aq) + H2O (l) In this case Cl2acts both as oxidizing and reducing agent

Ox. 1/2 Cl2 (g) + NaOH (aq) + OH – (aq) → NaClO (aq) + H+ (aq) + e−

Red. 1/2 Cl2 (g) + NaOH (aq) + e− → NaCl (aq) + OH – (aq)

2NaOH (aq) + Cl2 (g) → NaCl (aq) + NaClO (aq)

+ H2O (l)

8



Petrucci 5.5

How to balance Redox reactions in acidic environment

1. Write half reactions

• Identify oxidized and reduced species

2. Balance elements other than H or O, if any.

3. Balance O by adding H2O molecules.

4. Balance H by adding H+.

5. Balance charge by adding electrons.

6. Add the two half reactions

7. Cancel common terms

8. Recheck!!

Balancing redox reactions Example 3

Balance the following reation in acidic environment

Fe2+ (aq) + Cr2O72− −→ Fe3+ (aq) + Cr3+ (aq)

Ox. reaction: (Fe2+ −→ Fe3+ (aq) + e−) × 6

Red. reaction: Cr2O2−7 + 14H+ + 6e− −→ 2 Cr3+ + 7H2O

Cr2O2−7 + 14H+ + 6Fe2+ −→ 6Fe3+ + 2 Cr3+ + 7H2O

How to balance Redox reactions in basic environment

• Balance reaction as in acidic solution

• To both sides of the equation, add OH – to equal the number of H+

• Form H2O from H+ + OH – , eliminate H2O molecules that appear on both

sides of the equation.

Balance redox reaction in basic environment Example 4

Balance the following reation in basic environment

Fe2+ (aq) + Cr2O72− −→ Fe3+ (aq) + Cr3+ (aq)

Cr2O2−7 + 7 H2O + 6Fe2+ −→ 6Fe3+ + 2 Cr3+ + 14OH –

9



Balancing Disproportionation Reaction Example 5

Balance the following redox reaction is basic solution

ClO2 (aq) −→ Cl− (aq) + ClO−

3 (aq)

Ox. reaction: ClO2 + H2O−→ ClO−

3 + 2 H+ + e− add OH – : ClO2 + H2O+

2OH – −→ ClO−

3 + 2 H+ + 2 OH – + e− Neutralize: ClO2 + 2OH – −→ ClO−

3 +H2O + e−

Red. reaction: ClO2 + 4H++ 5e− −→ Cl− + 2 H2O add OH – : ClO2 + 4H++

4OH – + 5e− −→ Cl− + 2 H2O+ 4 OH – Neutralize: ClO2 + 2H2O+ 5e− −→Cl− + 4 OH –

6ClO2 + 6OH – −→ 5ClO−

3 + Cl− + 3H2O

Electrochemical Cell

Cu 2+ + 2 e – −→ Cu Zn −→ Zn 2+ + 2 e – cathode – reduction anode – oxidation

Petrucci 19.1

• An electrochemical cell consists of two electrodes, an anode and a cathode

• Oxidation occurs at the anode (by definition)

• Reduction occurs at the cathode

• Saltbridge required so "‘counter ions"’ flow into half cells to maintain neu-

tral charge as electrons flow from one cell to the other.

– Electrons move from anode (oxidation) to cathode (reduction). To

compensate for this movement of negative charge:

∗ negative counter ions move from cathode to anode

∗ positive counter ions move from anode to cathode.

– Simple salt bridge can be wet filter paper (allows diffusion)

10



Anode & Cathode

— How to remember?

Red Cat

Reduction at Cathode

An Ox

Anode — oxidation

• Galvanic cell: Spontaneous redox reaction generate electricity

– Spontaneous ⇔ ∆G < 0

∗ Electrons build up where they are generated, i.e. at the anode ⇒anode is negative for a galvanic cell

∗ Electrons are consumed in reduction reaction, i.e. at the cathode

⇒ cathode is postive for a galvanic cell

– Produces electricity (Batteries, fuelcells, etc.)

• Electrolytic cell: Non-spontaneous redox reaction is driven by electric cur-

rent

– NON-spontaneous ⇔ ∆G > 0

– DRIVEN by electricity.

∗ Electrons must be added to electrode where reduction is driven to

occur ⇒ cathode is negative for an electrolytic cell

∗ Electrons must be removed to electrode where oxidation is desired⇒ anode is positive for an electrolytic cell

– Charging of batteries, electroplating, etc

Measuring Cell Potentials

• The production and consumption of electrons at the

anode and cathode, respectively, builds up a potential

between the electrodes that can be measured with a

voltmeter

11



Measuring Height Differences

Petrucci 19.2

Cell Potential (E cell)

Cell voltage:

E cell = E ox + E red

e− pulling potential (E red)

Cu2+ + 2 e−→ Cu (s)

e

−

pushing potential (E ox)

Zn (s) → Zn2+ + 2e−

12



Standard Electrode Reduction Potentials

• It would be very convenient to be able to assign a potential to every half-

reaction. In that way we could always calculate E cell for any given reaction

as

E cell = E ox + E red

• No need to know absolute potential, we only need it relative to a standard .

• All half cell potentials are measured relative to the standard hydrogen elec-

trode (SHE)

• Half cell potentials depends on concentration. Table values are for reactions

at standard conditions.

• From table values we can calulate the the standard cell potential, E ◦cell:

E ◦cell = E ◦ox + E ◦red

Standard Hydrogen Electrode

• The half-cell potential is arbitrarily set to zero for the standard hydrogen

electrode (SHE)

• Pt electrode

• Bubble Hydrogen gas at 1 bar pressure

• H+ activity = 1 (a = 1, c ≈ 1M)

2 H+ (a = 1) + 2 e−→ H2 (g, 1 bar)

Oxidation vs. Reduction Potential

• If a reaction is inverted the (half) cell potential changes sign

Zn(s) −→ Zn2+

(aq) + 2 e−

E ◦

= 0.763VCu2+(aq) + 2 e− −→ Cu(s) E ◦ = 0.340V

Zn(s) + Cu 2+(aq) −→ Zn 2+(aq) + Cu(s) E ◦cell = 1.103V

Zn 2+(aq) + 2 e− −→ Zn(s) E ◦ = −0.763V

Cu(s) −→ Cu 2+(aq) + 2 e− E ◦ = −0.340VZn 2+(aq) + Cu(s) −→ Zn(s) + Cu 2+(aq) E ◦cell = −1.103V

13



Table values (usually) listed as reduction potentials

• It is attempted by convention to make the table values reduction potentials

only.

• To get the oxidation potential for the reaction occurring at the anode the

sign must be inverted.

• The formula

E cell = E cat − E an

uses the reduction potentials for both reactions

• The formula is hence identical to the formula

E cell = E red + E ox

which refers to the oxidation potential for the reaction occuring at the anode.

• Use whichever form you like, just make sure you understand the difference.

Measuring Half cell reduction potentials

Cu2+ (aq) + 2e−

→ Cu (s)

E ◦Cu2+/Cu

= +0.34 V

Cu2+ (aq) + H2 (g) → Cu (s) + 2 H+ (aq)

spontaneous at standard conditions

Zn2+ (aq) + 2e− → Zn (s)

E ◦Zn2+/Zn

= −0.763 V

Zn2+ (aq) + H2 (g) → Zn (s) + 2 H+ (aq)

NON-spontaneous at standard conditions

14



Redox Potentials Example 6

• What is the stronger oxidizing

agent, I2 (s) or Br2 (l)?

• Is the oxidation of Fe2+ by O2

spontaneous at standard condi-

tions?

The Behavior of Metals Toward Acids Example 7

Which metals can be oxidized by H+?

The oxidation of Ag by Nitric Acid Example 8Explain why Ag can be oxidized by HNO3, but not by HCl

Cell Diagrams

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

• Single line (|) denotes phase boundary

• Double line (||) denotes boundary between half compartments (salt bridge)

• Different species in same phase are separated by commas

16



• Anode always at the left

– Remember A(node) comes before C(athode)

Cell diagram for Calomel electrode Example 9

Write the cell diagram for the cell shown in the picture

Oxidation 2 Hg (l) + 2 Cl−

(aq) → Hg2Cl2 (s) + 2 e−

Reduction Hg2+2 (aq) + 2 e− → 2 Hg (l)

Overall

Cell diagram for standard hydrogen electrode Example 10

Write the cell diagram for the following cell:

Oxidation 1/2 H2 (g) → H+ (aq) + e−

Reduction Co3+ (aq) + e− → Co2+ (aq)

Overall 1/2 H2 (g) + Co3+ (aq) → H+ (aq) + Co2+

17



Electrochemical Cell Example 11

An electrochemical cell is based on the following reaction:

2 Ag (s) + Mg(NO3)2 (aq) → Mg (s) + 2 AgNO3 (aq)

1. What reaction occurs at the anode?

2. What reaction occurs at the cathode?

3. What is the cell potential, E ◦cell?

4. Is this an electrolytic or galvanic cell?

5. Write the cell diagram

(a) Draw the cell

(b) Show the direction of electrons

(c) Designate anode and cathode and the sign of each electrode(d) Show the direction of ions

Petrucci 19.3

Thermodynamics ⇔ Electrochemistry

• −∆Gr is the maximum amount of non-expansion work which can be ex-

tracted from a closed system

welec = nF E cell =

F is Faraday’s constant which is the charge per mol electrons (96,485 C/mol)

n is the number of electrons transferred in the reaction

∆Gr = −nF E cell

∆G◦

r = −nF E ◦cell

• ∆Gr is an extensive property (depends on the size of system, and hence

how the reaction is balanced)

• E ◦cell is an intensive property and does NOT depend on the size of the system

• To determine n look at how many electrons are transferred in the half reac-

tions that add up to the overall equation

18



How to determine n Example 12

Balance the following reactions and determine n:

1. 2 Fe2+ (aq) + Cl2 (g) → 2 Fe3+ (aq) + 2 Cl− (aq)

2. CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

Extensive vs intensive properties

Intensive properties

DOES NOT

vary with size of system

Extensive properties

DOES

vary with size of system

∆G vs. E cellGibbs free energy depends on how the reaction is balanced, the cell potential does

not.

• H2(g) + 12

O2(g) −→ H2O(g) n = 2

∆G◦

r = ∆G◦

f (H2O(g)) − (∆G◦

f (H2(g)) + 1/2∆G◦

f (O2(g)))

= −228.6 kJ/mol

E ◦

cell = −

∆G◦

r

nF = −

(−228.6 kJ/mol)

2 × 96485 C/mol = 2.36 V

• 2 H2(g) + O2(g) −→ 2 H2O(g) n = 4

∆G◦

r = 2∆G◦

f (H2O(g)) − (2∆G◦

f (H2(g)) + ∆G◦

f (O2(g)))

= −457.2 kJ/mol

E ◦cell = −∆G◦

r

nF = −

(−457.2 kJ/mol)

4 × 96485 C/mol= 2.36 V

19



Relationship between the cell potential E ◦cell, Gibbs free energy ∆G◦

r , and the

equilibrium constant K

Zinc’s reaction with Iodide Example 13

Determine the EMF (electromotoric force = cell potential) and ∆G◦

r for the fol-

lowing reaction under standard conditions

Zn (s) + I2 (s) →Zn2+ (aq) + 2I− (aq)

what is the equilibrium constant for the reaction?

Use the table of reduction potentials in your textbook.

Petrucci 19.4

What about non-standard conditions?

∆Gr = ∆G◦

r + RT ln Q

Substitute cell potential

∆G◦

r = −nF E ◦cell ∆Gr = −nF E cell

−nF E cell = −nF E ◦cell + RT ln Q

Rearrange and you get the Nernst equation:

E cell = E ◦cell −RT

nF ln Q

Change bases for the logaritm ln Q = 2.3log Q

E cell = E ◦cell −RT (2.3)

nF log Q

20



At room temperature (T = 298 K):

E cell = E ◦cell −0.0592V

nlog Q

This is the way it was originally written by Nernst in 1889.

As a cell operates its potential decreases

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

E cell = E ◦cell −0.0592V

nlog Q

Electrochemical cell at non-standard conditions Example 14

What is the cell potential for the following electrochemical cell?

Pt (s) | Fe3+ (1 M), Fe2+ (0.2 M)|| Ag+ (0.1 M) | Ag (s)

Compare with example 19–9 in book: Pt (s) | Fe3+ (0.10 M), Fe2+ (0.2 M)|| Ag+

(1.0 M) | Ag (s) E cell = 0.011V

21



Concentration Cells

• A concentration cell has the same half cells, only with different concentra-

tions

Petrucci 19.5

Batteries

Voltaic cells put in series so their individual voltage add together

Anode: Zn −→ Zn 2+ + 2 e –

Cathode: 2 H+ + 2 e – −→ H2

Egyptian Battery

Connecting Electrochemical Cells

• Connecting cells in series (+ to −) ⇒ voltages add up

22



• Connecting cells in parallel (+ to + and − to −) ⇒ current add up) Same

as making the cell bigger

Batteries

• Primary (non-rechargeable) batteries

– Alkaline, Mercury, Silver, Lithium

• Secondary (rechargeable) batteries

– Cell can be recharged by driving cell reaction backwards by applying

electrical current

– Lead-acid, Nickel-Metal Hydride, Lithium ion

• Fuel cells

– Also called flow batteries– These cells are not self contained, requires fuel to run

– Hydrogen cells

Non-rechargable Battery

— Dry cell or Leclanche CellIn an acidic dry cell, the reduction reaction occurs within the

moist paste comprised of ammonium chloride (NH4Cl) and

manganese dioxide (MnO2):

2 NH+

4

+ 2 MnO2 (s) + 2 e−→Mn2O3 (s) + 2 NH3 (aq) +

H2O (l)

The zinc cylinder serves as the anode and it undergoes oxida-

tion:

Zn (s) →Zn2+ + 2 e−

The battery cannot be recharged because of the following reaction

Zn2+ + NH3 (aq) + 2 Cl− (aq) → Zn(NH3)Cl2 (s)

In the alkaline version, the ammonium chloride is replaced by KOH or NaOH and

the half-cell reactions are:

Zn (s) + 2 OH – (aq) → ZnO + H2O (l) + 2 e−

2 MnO2 (s) + 2 e−+ H2O (l) → Mn2O3 (s) + 2 OH – (aq)

The alkaline dry cell lasts much longer as the zinc anode corrodes less rapidly

under basic conditions than under acidic conditions.

23



Lead-acid Battery

An. Pb (s) + HSO−

4 (aq) → PbSO4 (s) +

H+ (a) + 2 e− E ◦= 0.296 V

Cat. PbO2

(s) + HSO−

4(aq)

+ 3 H3O+ (aq) + 2 e− → PbSO4 (s) + 2 H2O (l) E ◦= 1.628 V

• Why does the cell not need separate anode and cathode compartments?

• Why does electrodes have large surface area?

• How come the lead-acid battery can be recharged?

Fuel Cells

Fuel + Oxygen → Oxidized product

Fuel can for example be gasoline, methane (CH4), Hydrogen, metals

Combustion of methane is a highly exothermic reaction

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) + energy

An. CH4 (g) + + 2 H2O (l) → CO2 (g) + 8 H+ (aq) + 8 e−

Cat. 2 O2 (g) + 8H+ (aq) + 8e− → 4 H2O (l)

Petrucci 19.6

Corrosion

• Unwanted voltaic cell!

• Oxidizes metal to its oxide or sulfide

• Not only iron can rust.

An. (ox.) 2 Fe (s) → 2 Fe2+ + 4 e− E ◦ = 0.440 V

Cat. (red.) O2 (g) + 2H2O (l) + 4 e− → 4 OH – (aq) E ◦ = 0.401 V

2 Fe (s) + O2 (g) + 2H2O (l) → 2 Fe2+ + 4 OH – (aq) E ◦cell = 0.841 V

24



• Rust (Fe2O3 · nH2O) is not formed by the direct reaction between iron and

air but by a series of reactions

Fe2+ (aq) + 2 OH – (aq) → Fe(OH)2 (s)

4 Fe(OH)2 (s) + O2 (g) + 2 H2O (l) → Fe(OH)3 (s)

2 Fe(OH)3 (s) → Fe2O3·H2O(s) + H2O (l)

• Iron can only rust if water is present

• Loss of iron and depositioning of rust often happens at different places on

same object

• Iron rusts faster at low pH (high [H+])

Protection from Corrosion

• Iron rusts faster in contact with a less active metal (e.g. Cu) and slower in

contact with a more active metal (e.g. Zn)

25



Petrucci 19.7

Electrolysis

Electrical energy drives non-spontaneous reverse reaction.

The applied field must be at least the cell potential

In practice, the applied voltage must be much greater than the cell potential, called

an overpotential.

The extra voltage is needed to overcome kinetic barrier.

Electrolysis of Water

• Extremely pure water is hard to

electrolyze because no ions are

present to conduct the current

• Volume of H2 at Cathode is twice

that of O2 at Anode

An. (ox.) 2 H2O (l) → O2 (g) + 4 H++ 4 e− E ◦ = −1.229 V

Cat. (red.) ( 2 H2O(l) + 2 e− → H2 (g) + 2 OH – (aq) )×2 E ◦ = −0.828 V

2 H2O (l) → H2 (g) + O2 (g) E ◦

= −2.057 V

Electroplating

Electrolysis can cause metal plating on the cathode.

26



Cat. (red.) Ag+ + e−→Ag (s) E ◦=0.800 V

Battery pushes electrons to cathode

The number of electrons "‘pushed"’ by the battery determines amount of product

plated on the electrode

ne =Current × time

Faradays constant=

I × t

F

Battery does not "‘supply"’ electrons. Oxidation reaction occurring at anode does.

Plating Copper Example 15

What is the mass of Cu(s) plated at cathode when a 10.0 amp current flows for

30.0 min in the following setup?What reaction occurs at the Pt anode?

27



Electrolysis of molten NaCl Example 16

Two inert electrodes are placed in molten NaCl (melting point 801 ◦C) a a voltage

is applied to drive the reaction

2 NaCl (l) → 2 Na (s) + Cl2 (g)

1. Write the half reactions and specify which one occurs at the anode andcathode, respectively

2. What is the minimum voltage required to drive the reaction assuming that

no over potential is needed?

3. Draw the cell and show the direction of ions and electrons

4. Label anode and cathode with signs

5. How long does it take to produce 1.00 g of Na (s) at 10.0 A?

6. What reaction would occur if an aqueous solution of NaCl was used insteadof molten NaCl?

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Cell diagram Example 17

For the cell:

Cu (s) | Cu 2+ || Br2 (g, 1 atm) | Br− (1.0 M) | Pt (s)

1. Write the anode half reaction

2. Write the cathode half reaction

3. Draw the galvanic cell and show

(a) the direction of electron flow

(b) the direction of positive and negative counter ions

(c) which electrode is positive and negative, respectively

Cell Diagram from Reaction Example 18

Write the cell diagram for a cell that uses the reaction:

Mg (s) + 2H+

(aq) →H2 (g) + Mg2+

(aq)A platinum electrode is used with the H+ /H2 couple and a salt bridge separates the

two compartments

Anode: Mg (s) → Mg2+ = 2e− Cathode: H++ 2e−→ H2 Mg(s)|Mg2+(aq)|| H+

(aq) | H2 (g) | Pt(s)

Write Reactions from Cell Diagram Example 19

Write the chemical equation for the reaction corresponding to the cell

Pt(s) | H2 (g) | H+ (aq) || Co3+ (aq), Co2+ (aq) | Pt (s)

Galvanic Cell Example 20

A working galvanic cell is based on the following half-reactions:

Fe2+ (aq) + 2e− → Fe (s) E ◦= −0.44 V

MnO−

4 (aq) + 5e− + 8H+ → Mn2+ (aq) + 4 H2O (l) E ◦= +1.51 V

1. What is the cell reaction?

2. What is the cell potential, E ◦cell?

3. Write the cell diagram

4. Draw the cell

5. Show the direction of electrons

6. Designate anode and cathode

7. Show the direction of ions

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Overview of Electrochemistry

• Review of redox reactions

– OILRIG

– Oxidation states

– half reactions

– How to balance redox reactions

in acidic and basic solution

• Electrochemical cells

– Galvanic vs electrolytic cell

– Anode & cathode

– Saltbridge

– Cell diagram

– Cell potential

– Standard half cell reduction po-

tential

• The connection between E ◦

cell, ∆G◦

r , and K

– Spontaneous change

– Electrical work

– Nernst equation

• Concentration Cells

– Non-ideal conditions – Measuring K sp

• Batteries

– Primary vs secondary cells

– Dry cells

– Lead-acid battery

– Fuel cells

• Corrosion

– Corrosion protection

• Electrolysis

– Electroplating – Quantitative calculations

notes electrochemistry

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