Chapter 18 Electrochemistry - Center for Nonlinear Sciencesflow.chem.pdx.edu/web_data/spring08/chem_223/Chapter18_LEC.pdf · Chapter 18 Electrochemistry 2007, Prentice Hall ... A

Chapter 18Electrochemistry

2007, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Tro, Chemistry: A Molecular Approach 2

Redox Reaction• one or more elements change oxidation number

all single displacement, and combustion,some synthesis and decomposition

• always have both oxidation and reductionsplit reaction into oxidation half-reaction and a reduction half-reaction

• aka electron transfer reactionshalf-reactions include electrons

• oxidizing agent is reactant molecule that causes oxidationcontains element reduced

• reducing agent is reactant molecule that causes reductioncontains the element oxidized


Oxidation & Reduction• oxidation is the process that occurs when

oxidation number of an element increaseselement loses electronscompound adds oxygencompound loses hydrogenhalf-reaction has electrons as products

• reduction is the process that occurs whenoxidation number of an element decreaseselement gains electronscompound loses oxygencompound gains hydrogenhalf-reactions have electrons as reactants


Rules for Assigning Oxidation States• rules are in order of priority1. free elements have an oxidation state = 0

Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)2. monatomic ions have an oxidation state equal

to their chargeNa = +1 and Cl = -1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0

Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0


Rules for Assigning Oxidation States3. (b) the sum of the oxidation states of all the atoms in

a polyatomic ion equals the charge on the ionN = +5 and O = -2 in NO3

–, (+5) + 3(-2) = -1

4. (a) Group I metals have an oxidation state of +1 in all their compounds

Na = +1 in NaCl

4. (b) Group II metals have an oxidation state of +2 in all their compounds

Mg = +2 in MgCl2


Rules for Assigning Oxidation States5. in their compounds, nonmetals have oxidation

states according to the table belownonmetals higher on the table take priority

NH3-3Group 5ACS2-2Group 6ACCl4-1Group 7ACO2-2OCH4+1HCF4-1F

ExampleOxidation StateNonmetal


Oxidation and Reduction

• oxidation occurs when an atom’s oxidation state increases during a reaction

• reduction occurs when an atom’s oxidation state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2

oxidationreduction


Oxidation–Reduction• oxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them

• the reactant that reduces an element in another reactant is called the reducing agent

the reducing agent contains the element that is oxidized

• the reactant that oxidizes an element in another reactant is called the oxidizing agent

the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agent


Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ → 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O


Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ → 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr → MnBr2 + Br2 + 2 H2O

+1 -2 +5 -2 +1 0 +2 -2 +1 -2

ox agred ag

+4 -2 +1 -1 +2 -1 0 +1 -2

oxidationreduction

oxidationreduction

red agox ag

11

Common Oxidizing AgentsOxidizing Agent Product when ReducedO2 O-2

H2O2 H2OF2, Cl2, Br2, I2 F-1, Cl-1, Br-1, I-1

ClO3-1 (BrO3

-1, IO3-1) Cl-1, (Br-1, I-1)

H2SO4 (conc) SO2 or S or H2SSO3

-2 S2O3-2, or S or H2S

HNO3 (conc) or NO3-1 NO2, or NO, or N2O, or N2, or NH3

MnO4-1 (base) MnO2

MnO4-1 (acid) Mn+2

CrO4-2 (base) Cr(OH)3

Cr2O7-2 (acid) Cr+3

12

Common Reducing AgentsReducing Agent Product when OxidizedH2 H+1

H2O2 O2

I-1 I2

NH3, N2H4 N2

S-2, H2S SSO3

-2 SO4-2

NO2-1 NO3

-1

C (as coke or charcoal) CO or CO2

Fe+2 (acid) Fe+3

Cr+2 Cr+3

Sn+2 Sn+4

metals metal ions


Balancing Redox Reactions1) assign oxidation numbers

a) determine element oxidized and element reduced2) write ox. & red. half-reactions, including electrons

a) ox. electrons on right, red. electrons on left of arrow3) balance half-reactions by mass

a) first balance elements other than H and Ob) add H2O where need Oc) add H+1 where need Hd) neutralize H+ with OH- in base

4) balance half-reactions by chargea) balance charge by adjusting electrons

5) balance electrons between half-reactions6) add half-reactions7) check


Ex 18.3 – Balance the equation:I−

(aq) + MnO4−

(aq) → I2(aq) + MnO2(s) in basic solution

ox:red:

Separate into half-reactions

I−(aq) + MnO4

−(aq) → I2(aq) + MnO2(s)Assign

Oxidation States

ox: I−(aq) → I2(aq)

red: MnO4−

(aq) → MnO2(s)

Separate into half-reactions

Assign Oxidation States



(aq) + MnO4−


ox: I−(aq) → I2(aq)

red: MnO4−

(aq) → MnO2(s)

Balance half-reactions by mass

ox: 2 I−(aq) → I2(aq)

red: MnO4−

(aq) → MnO2(s)

Balance half-reactions by mass

ox: 2 I−(aq) → I2(aq)

red: MnO4−

(aq) → MnO2(s) + 2 H2O(l)

Balance half-reactions by mass then O by adding H2O

ox: 2 I−(aq) → I2(aq)

red: 4 H+(aq) + MnO4

−(aq) → MnO2(s) + 2 H2O(l)

Balance half-reactions by mass then H by adding H+

ox: 2 I−(aq) → I2(aq)

red: 4 H+(aq) + MnO4

−(aq) → MnO2(s) + 2 H2O(l)

4 H+(aq) + 4 OH−

(aq) + MnO4−

(aq) → MnO2(s) + 2 H2O(l) + 4 OH−(aq)

4 H2O(aq) + MnO4−

(aq) → MnO2(s) + 2 H2O(l) + 4 OH−(aq)

MnO4−

(aq) + 2 H2O(l) → MnO2(s) + 4 OH−(aq)

Balance half-reactions by mass in base, neutralize the H+

with OH-



(aq) + MnO4−


ox: 2 I−(aq) → I2(aq) + 2 e− } x3

red: MnO4−

(aq) + 2 H2O(l) + 3 e− → MnO2(s) + 4 OH−(aq) }x2

ox: 6 I−(aq) → 3 I2(aq) + 6 e−

red: 2 MnO4−

(aq) + 4 H2O(l) + 6 e− → 2 MnO2(s) + 8 OH−(aq)

Balance electrons between half-reactions

ox: 2 I−(aq) → I2(aq) + 2 e−

red: MnO4−

(aq) + 2 H2O(l) + 3 e− → MnO2(s) + 4 OH−(aq)

Balance Half-reactions by charge



(aq) + MnO4−


Check

ox: 6 I−(aq) → 3 I2(aq) + 6 e−

red: 2 MnO4−

(aq) + 4 H2O(l) + 6 e− → 2 MnO2(s) + 8 OH−(aq)

tot: 6 I−(aq)+ 2 MnO4

−(aq) + 4 H2O(l) → 3 I2(aq)+ 2 MnO2(s) + 8 OH−

(aq)

Add the Half-reactions

2−charge2−

8H812O122Mn26I6

ProductCountElement

ReactantCount


Practice - Balance the EquationH2O2 + KI + H2SO4 → K2SO4 + I2 + H2O


Practice - Balance the EquationH2O2 + KI + H2SO4 → K2SO4 + I2 + H2O+1 -1 +1 -1 +1 +6 -2 +1 +6 -2 0 +1 -2

oxidationreduction

ox: 2 I-1 → I2 + 2e-1

red: H2O2 + 2e-1 + 2 H+ → 2 H2Otot 2 I-1 + H2O2 + 2 H+ → I2 + 2 H2O

1 H2O2 + 2 KI + H2SO4 → K2SO4 + 1 I2 + 2 H2O


Practice - Balance the EquationClO3

-1 + Cl-1 → Cl2 (in acid)


Practice - Balance the EquationClO3

-1 + Cl-1 → Cl2 (in acid)+5 -2 -1 0

oxidationreduction

ox: 2 Cl-1 → Cl2 + 2 e-1 } x5red: 2 ClO3

-1 + 10 e-1 + 12 H+ → Cl2 + 6 H2O} x1tot 10 Cl-1 + 2 ClO3

-1 + 12 H+ → 6 Cl2 + 6 H2O

1 ClO3-1 + 5 Cl-1 + 6 H+1 → 3 Cl2 + 3 H2O


Electrical Current• when we talk about the current

of a liquid in a stream, we are discussing the amount of water that passes by in a given period of time

• when we discuss electric current, we are discussing the amount of electric charge that passes a point in a given period of time

whether as electrons flowing through a wire or ions flowing through a solution


Redox Reactions & Current

• redox reactions involve the transfer of electrons from one substance to another

• therefore, redox reactions have the potential to generate an electric current

• in order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring


Electric Current Flowing Directly Between Atoms


Electric Current Flowing Indirectly Between Atoms


Electrochemical Cells• electrochemistry is the study of redox reactions

that produce or require an electric current• the conversion between chemical energy and

electrical energy is carried out in an electrochemical cell

• spontaneous redox reactions take place in a voltaic cell

aka galvanic cells• nonspontaneous redox reactions can be made to

occur in an electrolytic cell by the addition of electrical energy


Electrochemical Cells• oxidation and reduction reactions kept separate

half-cells

• electron flow through a wire along with ion flow through a solution constitutes an electric circuit

• requires a conductive solid (metal or graphite) electrode to allow the transfer of electrons

through external circuit

• ion exchange between the two halves of the systemelectrolyte


Electrodes• Anode

electrode where oxidation occursanions attracted to itconnected to positive end of battery in electrolytic cellloses weight in electrolytic cell

• Cathodeelectrode where reduction occurscations attracted to itconnected to negative end of battery in electrolytic cellgains weight in electrolytic cell

electrode where plating takes place in electroplating


Voltaic Cellthe salt bridge is required to complete the circuit and maintain charge balance


Current and Voltage• the number of electrons that flow through the system per

second is the currentunit = Ampere1 A of current = 1 Coulomb of charge flowing by each second1 A = 6.242 x 1018 electrons/secondElectrode surface area dictates the number of electrons that canflow

• the difference in potential energy between the reactants and products is the potential difference

unit = Volt1 V of force = 1 J of energy/Coulomb of chargethe voltage needed to drive electrons through the external circuitamount of force pushing the electrons through the wire is calledthe electromotive force, emf


Cell Potential• the difference in potential energy between the

anode the cathode in a voltaic cell is called the cell potential

• the cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode

• the cell potential under standard conditions is called the standard emf, E°cell

25°C, 1 atm for gases, 1 M concentration of solutionsum of the cell potentials for the half-reactions


Cell Notation• shorthand description of Voltaic cell• electrode | electrolyte || electrolyte | electrode• oxidation half-cell on left, reduction half-cell on

the right• single | = phase barrier

if multiple electrolytes in same phase, a comma is used rather than |often use an inert electrode

• double line || = salt bridge


Fe(s) | Fe2+(aq) || MnO4−(aq), Mn2+(aq), H+(aq) | Pt(s)


Standard Reduction Potential• a half-reaction with a strong tendency to

occur has a large + half-cell potential• when two half-cells are connected, the

electrons will flow so that the half-reaction with the stronger tendency will occur

• we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction

• we select as a standard half-reaction the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 v

standard hydrogen electrode, SHE



Half-Cell Potentials• SHE reduction potential is defined to be exactly 0 v• half-reactions with a stronger tendency toward

reduction than the SHE have a + value for E°red

• half-reactions with a stronger tendency toward oxidation than the SHE have a − value for E°red

• E°cell = E°oxidation + E°reductionE°oxidation = −E°reduction

when adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half-reactions to balance the equation



Ex 18.4 – Calculate E°cell for the reaction at 25°CAl(s) + NO3

−(aq) + 4 H+

(aq) → Al3+(aq) + NO(g) + 2 H2O(l)

E°ox = −E°red = +1.66 vE°red = +0.96 vE°cell = (+1.66 v) + (+0.96 v) = +2.62 v

find the E° for each half-reaction and sum to get E°cell

ox: Al(s) → Al3+(aq) + 3 e−

red: NO3−

(aq) + 4 H+(aq) + 3 e− → NO(g) + 2 H2O(l)

Separate the reaction into the oxidation and reduction half-reactions


Ex 18.4a – Predict if the following reaction is spontaneous under standard conditions

Fe(s) + Mg2+(aq) → Fe2+

(aq) + Mg(s)

red: Mg2+(aq) + 2 e− → Mg(s)

red: Fe2+(aq) + 2 e− → Fe(s)

since Mg2+ reduction is below Fe2+

reduction, the reaction is NOT spontaneousas written

look up the relative positions of the reduction half-reactions

ox: Fe(s) → Fe2+(aq) + 2 e−

red: Mg2+(aq) + 2 e− → Mg(s)

Separate the reaction into the oxidation and reduction half-reactions


the reaction is spontaneous in the reverse direction

Mg(s) + Fe2+(aq) → Mg2+

(aq) + Fe(s)ox: Mg(s) → Mg2+

(aq) + 2 e−

red: Fe2+(aq) + 2 e− → Fe(s)

sketch the cell and label the parts –oxidation occurs at the anode; electrons flow from anode to cathode


Practice - Sketch and Label the Voltaic CellFe(s) Fe2+(aq) Pb2+(aq) Pb(s) , Write the

Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.


ox: Fe(s) → Fe2+(aq) + 2 e− E° = +0.45 V

red: Pb2+(aq) + 2 e− → Pb(s) E° = −0.13 V

tot: Pb2+(aq) + Fe(s) → Fe2+(aq) + Pb(s) E° = +0.32 V


Predicting Whether a Metal Will Dissolve in an Acid

• acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H+

(aq)

• metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid


E°cell, ∆G° and K• for a spontaneous reaction

one the proceeds in the forward direction with the chemicals in their standard states∆G° < 1 (negative)E° > 1 (positive)K > 1

• ∆G° = −RTlnK = −nFE°celln is the number of electronsF = Faraday’s Constant = 96,485 C/mol e−


Example 18.6- Calculate ∆G° for the reactionI2(s) + 2 Br−(aq) →Br2(l) + 2 I−(aq)

since ∆G° is +, the reaction is not spontaneous in the forward direction under standard conditions

Answer:

Solve:

Concept Plan:

Relationships:

I2(s) + 2 Br−(aq) →Br2(l) + 2 I−(aq)

∆G°, (J)Given:

Find:

E°ox, E°red E°cell ∆G°oooredoxcell EEE += oo

cellFEG n=∆

ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 v

red: I2(l) + 2 e− → 2 I−(aq) E° = +0.54 v

tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 v

oocellFEG n=∆

( )( )( )J 101.1G

55.0485,96 mol 2G5

CJ

mol

C

×+=∆

−=∆ −−

o

o

ee


( ) ( )125.11 102.310

5.11V 0592.0

mol 2V 34.0log

−−

−

×==

−=−=

K

eK

Example 18.7- Calculate Κ at 25°C for the reactionCu(s) + 2 H+

(aq) →H2(g) + Cu2+(aq)

since Κ < 1, the position of equilibrium lies far to the left under standard conditions

Answer:

Solve:

Concept Plan:

Relationships:

Cu(s) + 2 H+(aq) →H2(g) + Cu2+

(aq)

ΚGiven:

Find:

E°ox, E°red E°cell Κoooredoxcell EEE += K

nlogV 0592.0Ecell =o

ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 v

red: 2 H+(aq) + 2 e− → H2(aq) E° = +0.00 v

tot: Cu(s) + 2H+(aq) → Cu2+

(aq) + H2(g) E° = −0.34 v

Kn

logV 0592.0Ecell =o


Nonstandard Conditions -the Nernst Equation

• ∆G = ∆G° + RT ln Q• E = E° - (0.0592/n) log Q at 25°C• when Q = K, E = 0• use to calculate E when concentrations not 1 M


E° at Nonstandard Conditions


V 41.1E.0]1[.0]2[]010.0[log

6V 0592.0V 34.1E

][H][MnO]Cu[logV 0592.0EE

cell

83

3

cell

834

32

cellcell

=

−=

−=+−

+

no

Example 18.8- Calculate Ecell at 25°C for the reaction3 Cu(s) + 2 MnO4

−(aq) + 8 H+

(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)

units are correct, Ecell > E°cell as expected because [MnO4

−] > 1 M and [Cu2+] < 1 MCheck:

Solve:

Concept Plan:

Relationships:

3 Cu(s) + 2 MnO4−

(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+

(aq) + 4 H2O(l)

[Cu2+] = 0.010 M, [MnO4−] = 2.0 M, [H+] = 1.0 M

Ecell

Given:

Find:

E°ox, E°red E°cell Ecelloooredoxcell EEE += Q

nlogV 0592.0EE cellcell −= o

ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = −0.34 v

red: MnO4−

(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = +1.68 v

tot: 3 Cu(s) + 2 MnO4−

(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+

(aq) + 4 H2O(l)) E° = +1.34 v

Qn

logV 0592.0EE cellcell −= o


Concentration Cells• it is possible to get a spontaneous reaction when the oxidation

and reduction reactions are the same, as long as the electrolyteconcentrations are different

• the difference in energy is due to the entropic difference in the solutions

the more concentrated solution has lower entropy than the less concentrated

• electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution

oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anodereduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode


when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow

Concentration Cell

when the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)

Cu(s) Cu2+(aq) (0.010 M) Cu2+

(aq) (2.0 M) Cu(s)


LeClanche’ Acidic Dry Cell• electrolyte in paste form

ZnCl2 + NH4Clor MgBr2

• anode = Zn (or Mg)Zn(s) → Zn2+(aq) + 2 e-

• cathode = graphite rod• MnO2 is reduced

2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-

→ 2 NH4OH(aq) + 2 Mn(O)OH(s)

• cell voltage = 1.5 v• expensive, nonrechargeable, heavy,

easily corroded


Alkaline Dry Cell• same basic cell as acidic dry cell, except

electrolyte is alkaline KOH paste• anode = Zn (or Mg)

Zn(s) → Zn2+(aq) + 2 e-

• cathode = brass rod• MnO2 is reduced

2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-

→ 2 NH4OH(aq) + 2 Mn(O)OH(s)

• cell voltage = 1.54 v• longer shelf life than acidic dry cells and

rechargeable, little corrosion of zinc


Lead Storage Battery• 6 cells in series• electrolyte = 30% H2SO4

• anode = PbPb(s) + SO4

2-(aq) → PbSO4(s) + 2 e-

• cathode = Pb coated with PbO2

• PbO2 is reducedPbO2(s) + 4 H+(aq) + SO4

2-(aq) + 2 e-

→ PbSO4(s) + 2 H2O(l)

• cell voltage = 2.09 v• rechargeable, heavy


NiCad Battery• electrolyte is concentrated KOH solution• anode = CdCd(s) + 2 OH-1(aq) → Cd(OH)2(s) + 2 e-1 E0 = 0.81 v• cathode = Ni coated with NiO2• NiO2 is reducedNiO2(s) + 2 H2O(l) + 2 e-1 → Ni(OH)2(s) + 2OH-1 E0 = 0.49 v• cell voltage = 1.30 v• rechargeable, long life, light – however

recharging incorrectly can lead to battery breakdown


Ni-MH Battery• electrolyte is concentrated KOH solution• anode = metal alloy with dissolved hydrogen

oxidation of H from H0 to H+1

M·H(s) + OH-1(aq) → M(s) + H2O(l) + e-1 E° = 0.89 v• cathode = Ni coated with NiO2• NiO2 is reduced

NiO2(s) + 2 H2O(l) + 2 e-1 → Ni(OH)2(s) + 2OH-1 E0 = 0.49 v

• cell voltage = 1.30 v• rechargeable, long life, light, more environmentally

friendly than NiCad, greater energy density than NiCad


Lithium Ion Battery• electrolyte is concentrated KOH

solution• anode = graphite impregnated with Li

ions• cathode = Li - transition metal oxide

reduction of transition metal• work on Li ion migration from anode

to cathode causing a corresponding migration of electrons from anode to cathode

• rechargeable, long life, very light, more environmentally friendly, greater energy density



Fuel Cells• like batteries in which

reactants are constantly being added

so it never runs down!• Anode and Cathode

both Pt coated metal• Electrolyte is OH–

solution• Anode Reaction:

2 H2 + 4 OH–

→ 4 H2O(l) + 4 e-

• Cathode Reaction: O2 + 4 H2O + 4 e-

→ 4 OH–


Electrolytic Cell• uses electrical energy to overcome the energy barrier

and cause a non-spontaneous reactionmust be DC source

• the + terminal of the battery = anode• the - terminal of the battery = cathode• cations attracted to the cathode, anions to the anode• cations pick up electrons from the cathode and are

reduced, anions release electrons to the anode and are oxidized

• some electrolysis reactions require more voltage than Etot, called the overvoltage



electroplating In electroplating, the work piece is the cathode.

Cations are reduced at cathode and plate to the surface of the work piece.The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution


Electrochemical Cells• in all electrochemical cells, oxidation occurs at the

anode, reduction occurs at the cathode• in voltaic cells,

anode is the source of electrons and has a (−) chargecathode draws electrons and has a (+) charge

• in electrolytic cellselectrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the batteryelectrons are forced toward the anode, so it must have a source of electrons, the − terminal of the battery


Electrolysis• electrolysis is the process of using

electricity to break a compound apart

• electrolysis is done in an electrolytic cell

• electrolytic cells can be used to separate elements from their compounds

generate H2 from water for fuel cellsrecover metals from their ores


Electrolysis of Water


Electrolysis of Pure Compounds• must be in molten (liquid) state• electrodes normally graphite• cations are reduced at the cathode to metal

element• anions oxidized at anode to nonmetal element


Electrolysis of NaCl(l)


Mixtures of Ions

• when more than one cation is present, the cationthat is easiest to reduce will be reduced first at the cathode

least negative or most positive E°red

• when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode

least negative or most positive E°ox


Electrolysis of Aqueous Solutions• Complicated by more than one possible oxidation and reduction• possible cathode reactions

reduction of cation to metalreduction of water to H2

2 H2O + 2 e-1 → H2 + 2 OH-1 E° = -0.83 v @ stand. cond.E° = -0.41 v @ pH 7

• possible anode reactionsoxidation of anion to elementoxidation of H2O to O2

2 H2O → O2 + 4e-1 + 4H+1 E° = -1.23 v @ stand. cond.E° = -0.82 v @ pH 7

oxidation of electrodeparticularly Cugraphite doesn’t oxidize

• half-reactions that lead to least negative Etot will occurunless overvoltage changes the conditions


Electrolysis of NaI(aq) with Inert Electrodes

possible oxidations2 I-1 → I2 + 2 e-1 E° = −0.54 v2 H2O → O2 + 4e-1 + 4H+1 E° = −0.82 v

possible reductionsNa+1 + 1e-1 → Na0 E° = −2.71 v2 H2O + 2 e-1 → H2 + 2 OH-1 E° = −0.41 v

possible oxidations2 I-1 → I2 + 2 e-1 E° = −0.54 v2 H2O → O2 + 4e-1 + 4H+1 E° = −0.82 v

possible reductionsNa+1 + 1e-1 → Na0 E° = −2.71 v2 H2O + 2 e-1 → H2 + 2 OH-1 E° = −0.41 v

overall reaction2 I−(aq) + 2 H2O(l) → I2(aq) + H2(g) + 2 OH-1

(aq)


Faraday’s Law

• the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs

charge that flows through the cell = current x time


Example 18.10- Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction

Au3+(aq) + 3 e−→ Au(s)

units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−

Check:

Solve:

Concept Plan:

Relationships:

3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min

mass Au, gGiven:

Find:

s 1C 5.5

Au g 6.5Au mol 1

g 196.97 mol 3Au mol 1

C 96,485 mol 1

s 1C 5.5

min 1s 60min 25

=

××××× −

−

ee

t(s), amp charge (C) mol e− mol Au g Au

C 6,4859 mol 1 −e

−e mol 3Au mol 1

Au mol 1g 196.97


Corrosion• corrosion is the spontaneous oxidation of a

metal by chemicals in the environment• since many materials we use are active

metals, corrosion can be a very big problem


Rusting• rust is hydrated iron(III) oxide• moisture must be present

water is a reactantrequired for flow between cathode and anode

• electrolytes promote rustingenhances current flow

• acids promote rustinglower pH = lower E°red



Preventing Corrosion• one way to reduce or slow corrosion is to coat

the metal surface to keep it from contacting corrosive chemicals in the environment

paintsome metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding

• another method to protect one metal is to attach it to a more reactive metal that is cheap

sacrificial electrodegalvanized nails


Sacrificial Anode

Documents

Chapter 18 Electrochemistry - Center for Nonlinear Sciencesflow.chem.pdx.edu/web_data/spring08/chem_223/Chapter18_LEC.pdf · Chapter 18 Electrochemistry 2007, Prentice Hall ... A