Chem chapter 18 LEC

8/13/2019 Chem chapter 18 LEC

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Chapter 18

Electrochemistry

2007, Prentice Hall

Chemistry: A Molecular Approach, 1stEd.

Nivaldo Tro

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA


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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 reactions

half-reactions include electrons

oxidizing agentis reactant molecule that causes oxidationcontains element reduced reducing agentis reactant molecule that causes reduction

contains the element oxidized


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Oxidation & Reduction

oxidationis the process that occurs whenoxidation number of an element increases

element loses electrons

compound adds oxygen

compound loses hydrogenhalf-reaction has electrons as products

reductionis the process that occurs whenoxidation number of an element decreases

element gains electronscompound loses oxygen

compound gains hydrogen

half-reactions have electrons as reactants


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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 equalto their charge

Na = +1 and Cl = -1 in NaCl

3. (a) the sum of the oxidation states of all theatoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0


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Rules for Assigning Oxidation States

5. in their compounds, nonmetals have oxidationstates according to the table below nonmetals higher on the table take priority

Nonmetal Oxidation State Example

F -1 CF4

H +1 CH4

O -2 CO2

Group 7A -1 CCl4

Group 6A -2 CS2

Group 5A -3 NH3


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Oxidation and Reduction

oxidation occurs when an atoms oxidation stateincreases during a reaction

reduction occurs when an atoms oxidation statedecreases during a reaction

CH4 + 2 O2 CO2+ 2 H2O-4+1 0 +42 +1 -2

oxidation

reduction
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OxidationReduction oxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them

the reactant that reduces an element in another reactantis called the reducing agent

the reducing agent contains the element that is oxidized

the reactant that oxidizes an element in another reactantis 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, Cl2is the oxidizing agent


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Identify the Oxidizing and Reducing Agents

in Each of the Following3 H2S + 2 NO3

+ 2 H+ 3S + 2 NO + 4 H2O

MnO2+ 4 HBr MnBr2+ Br2+ 2 H2O


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Identify the Oxidizing and Reducing Agents

in Each of the Following3 H2S + 2 NO3

+ 2 H+ 3S + 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

oxidation

reduction

red agox ag


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12

Common Reducing AgentsReducing Agent Product when Oxidized

H2 H+1

H2O2 O2

I-1

I2

NH3, N2H4 N2

S

-2

, H2S SSO3

-2SO4

-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


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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 arrow

3) balance half-reactions by massa) first balance elements other than H and O

b) add H2O where need O

c) add H+1where need H

d) 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


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Ex 18.3Balance the equation:

I(aq)+ MnO4

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

Assign

Oxidation

States

I(aq)+ MnO4

(aq)I2(aq)+ MnO2(s)

Separate

into half-

reactions

ox:

red:

Assign

Oxidation

States

Separate

into half-

reactions

ox: I(aq)I2(aq)red: MnO4

(aq)MnO2(s)


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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)

Balance half-

reactions by

mass

then O byadding H2O

ox: 2 I(aq)I2(aq)red: MnO4



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Balancehalf-

reactions

by mass

in base,

neutralize

the H+

with OH-

ox: 2 I

(aq)I2(aq)red: 4 H+(aq)+ MnO4


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)


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I(aq)+ MnO4


Balance

Half-

reactions

bycharge

ox: 2 I(aq)I2(aq)+ 2 e

red: MnO4

(aq)+ 2 H2O(l) + 3 eMnO2(s) + 4 OH(aq)

Balance

electrons

between

half-

reactions

ox: 2 I(aq)I2(aq)+ 2 e}x3red: MnO

4

(aq)

+ 2 H2

O(l)

+ 3 eMnO2(s)

+ 4 OH(aq)

}x2

ox: 6 I(aq)3 I2(aq)+ 6 ered: 2 MnO4

(aq)+ 4 H2O(l) + 6 e

2 MnO2(s) + 8 OH(aq)


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I(aq)+ MnO4


Add the

Half-

reactions

ox: 6 I(aq)3 I2(aq)+ 6 ered: 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)

Check ReactantCount Element

ProductCount

6 I 6

2 Mn 2

12 O 12

8 H 8

2 charge 2


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Practice - Balance the Equation

H2O2+ KI + H2SO4K2SO4+ I2+ H2O


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H2O2+ KI + H2SO4K2SO4+ 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 H2O

2+ 2 KI + H

2SO

4K

2SO

4+ 1 I

2+ 2 H

2O


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ClO3-1+ Cl-1 Cl2 (in acid)


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ClO3-1+ 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+ 2ClO3

-1+ 12 H+ 6 Cl2 + 6 H2O

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


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Electrical Current

when we talk about the currentof a liquid in a stream, we arediscussing the amount of waterthat passes by in a given periodof time

when we discuss electriccurrent, we are discussing theamount of electric charge that

passes a point in a given period

of timewhether as electrons flowingthrough a wire or ions flowingthrough a solution


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Redox Reactions & Current

redox reactions involve the transfer of electronsfrom one substance to another

therefore, redox reactions have the potential togenerate an electric current

in order to use that current, we need to separate

the place where oxidation is occurring from theplace that reduction is occurring


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Electric Current Flowing

Indirectly Between Atoms


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Electrochemical Cells oxidation and reduction reactions kept separatehalf-cells

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

requires a conductive solid (metal or graphite)electrodeto allow the transfer of electrons

through external circuit

ion exchange between the two halves of the system electrolyte


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Electrodes Anode

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

cell

loses weight in electrolytic cell Cathodeelectrode where reduction occurscations attracted to it

connected to negative end of battery in electrolyticcellgains weight in electrolytic cellelectrode where plating takes place in electroplating


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Voltaic Cell

the salt bridge is

required to complete

the circuit and

maintain charge

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

second is the current unit = Ampere

1 A of current = 1 Coulomb of charge flowing by each second

1 A = 6.242 x 1018electrons/second

Electrode surface area dictates the number of electrons that canflow

the difference in potential energy between the reactantsand products is the potential difference unit = Volt

1 V of force = 1 J of energy/Coulomb of charge the voltage needed to drive electrons through the external circuit

amount of force pushing the electrons through the wire is calledthe electromotive force, emf


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Cell Potential the difference in potential energy between the

anode the cathode in a voltaic cell is called thecell potential

the cell potential depends on the relative ease

with which the oxidizing agent is reduced at thecathode and the reducing agent is oxidized at theanode

the cell potential under standard conditions iscalled the standard emf, Ecell25C, 1 atm for gases, 1 M concentration of solution

sum of the cell potentials for the half-reactions


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Cell Notation

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

the right single | = phase barrierif multiple electrolytes in same phase, a comma is

used rather than |

often use an inert electrode

double line || = salt bridge


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


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

occur has a large + half-cell potential when two half-cells are connected, theelectrons will flow so that the half-reactionwith the stronger tendency will occur

we cannot measure the absolute tendencyof a half-reaction, we can only measure itrelative to another half-reaction

we select as a standard half-reaction thereduction of H+to H2under standard

conditions, which we assign a potentialdifference = 0 v standard hydrogen electrode, SHE


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Half-Cell Potentials

SHE reduction potential is defined to be exactly 0 v half-reactions with a stronger tendency towardreduction than the SHE have a + value for Ered

half-reactions with a stronger tendency towardoxidation than the SHE have a value for Ered

Ecell= Eoxidation+ EreductionEoxidation= Ereduction

when adding Evalues for the half-cells, do not multiplythehalf-cell E values, even if you need to multiply the half-

reactions to balance the equation


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Ex 18.4Calculate Ecellfor the reaction at 25CAl(s)+ NO3

(aq) + 4 H

+(aq)Al3+(aq)+ NO(g)+ 2 H2O(l)

Separate thereaction into

the oxidation

and reduction

half-reactions

ox: Al(s)Al3+(aq)+ 3 e

red: NO3

(aq)+ 4 H+

(aq) + 3 eNO(g) + 2 H2O(l)

find the Eforeach half-

reaction andsum to get

Ecell

Eox= Ered= +1.66 v

Ered= +0.96 v

Ecell= (+1.66 v) + (+0.96 v) = +2.62 v

E 18 4 P edi t if the f ll i e ti i


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Ex 18.4aPredict if the following reaction is

spontaneous under standard conditions

Fe(s)+ Mg2+

(aq)Fe2+(aq)+ Mg(s)

Separate thereaction into

the oxidation

and reduction

half-reactions

ox: Fe(s)Fe2+(aq)+ 2 e

red: Mg2+(aq)+ 2 eMg(s)

look up the

relative

positions of thereduction half-

reactions

red: Mg2+(aq)+ 2 eMg(s)

red: Fe2+(aq)+ 2 eFe(s)

since Mg2+reduction is below Fe2+reduction, the reaction is NOT spontaneous

as written


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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 eFe(s)

sketch the cell and

label the parts

oxidation occurs atthe anode; electrons

flow from anode to

cathode

P ti Sk t h d L b l th V lt i C ll


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Practice - Sketch and Label the Voltaic Cell

Fe(s) Fe2+(aq) Pb2+(aq) Pb(s) , Write theHalf-Reactions and Overall Reaction, and Determine

the Cell Potential under Standard Conditions.


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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


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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 ofH+(aq)

metals whose ion reduction

reaction lies below H+

reductionon the table will dissolve in acid


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Ecell,DG and K

for a spontaneous reactionone the proceeds in the forward direction with the

chemicals in their standard states

DG < 1 (negative)

E > 1 (positive)

K > 1

DG = RTlnK = nFEcell

nis the number of electrons

F = Faradays Constant = 96,485 C/mol e


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E l 18 7 C l l t K t 25C f th ti


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125.11 102.310

5.11

V0592.0

mol2V34.0log

K

eK

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

+(aq)H2(g)+ Cu

2+(aq)

since K< 1, the position of equilibrium lies far tothe left under standard conditions

Answer:

Solve:

Concept Plan:

Relationships:

Cu(s)+ 2 H+

(aq)H2(g)+ Cu2+

(aq)K

Given:Find:

Eox, Ered Ecell K

redoxcell

EEE Kn

logV0592.0

Ecell

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

logV0592.0

Ecell


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Nonstandard Conditions -

the Nernst Equation DG = DG + RT ln Q

E = E - (0.0592/n) log Q at 25C

when Q = K, E = 0

use to calculate E when concentrations not 1 M


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l l l f h i


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

(aq)+ 8 H

+(aq)2 MnO2(s)+ Cu

2+(aq)+ 4 H2O(l)

units are correct, Ecell> Ecell as expected because

[MnO4] > 1 M and [Cu2+] < 1 M

Check:

Solve:

Concept Plan:

Relationships:

3 Cu(s)+ 2 MnO4

(aq)+ 8 H

+

(aq)2 MnO2(s)+ Cu

2+

(aq)+ 4 H2O(l)[Cu2+] = 0.010 M, [MnO4

] = 2.0 M, [H+] = 1.0 M

Ecell

Given:

Find:

Eox, Ered Ecell Ecell

redoxcell EEE Qn

logV0592.0EE cellcell

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

]C[V05920 32


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V41.1E

.0]1[.0]2[

]010.0[log

6

V0592.0V34.1E

][H][MnO

]Cu[log

V0592.0EE

cell

83

3

cell

834

32

cellcell

n

ox: Cu(s) Cu2+

(aq)+ 2 e}x3E= 0.34 v

Qn

logV0592.0

EE cellcell


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when the cell

concentrations

are equal there isno difference in

energy between

the half-cells and

no electrons flow

Concentration Cell

Cu(s)Cu2+(aq)(0.010 M) Cu2+(aq)(2.0 M)Cu(s)

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)

L Cl h A idi D C ll
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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

Alk li D C ll
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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 reduced2 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 andrechargeable, little corrosion of zinc

d S


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Lead Storage Battery

6 cells in series electrolyte = 30% H2SO4

anode = PbPb(s) + SO

4

2-(aq) PbSO4(s) + 2 e-

cathode = Pb coated with PbO2

PbO2 is reducedPbO2(s) + 4 H

+(aq) + SO42-(aq) + 2 e-

PbSO4(s) + 2 H2O(l)


rechargeable, heavy

NiC d B tt


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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

-1Ni(OH)2(s) + 2OH-1 E0= 0.49 v

cell voltage = 1.30 v rechargeable, long life, lighthowever

recharging incorrectly can lead to batterybreakdown

Ni MH B tt


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Ni-MH Battery

electrolyte is concentrated KOH solution anode = metal alloy with dissolved hydrogen oxidation of H from H0to H+1

MH(s) + OH-1(aq) M(s) + H2O(l) + e-1 E = 0.89 v

cathode = Ni coated with NiO2 NiO2 is reducedNiO2(s) + 2 H2O(l) + 2 e

-1Ni(OH)2(s) + 2OH-1 E0= 0.49 v


rechargeable, long life, light, more environmentallyfriendly than NiCad, greater energy density than

NiCad


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Lithium Ion Battery electrolyte is concentrated KOH

solution

anode = graphite impregnated with Liions

cathode = Li - transition metal oxide reduction of transition metal

work on Li ion migration from anodeto cathode causing a correspondingmigration of electrons from anode tocathode

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


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F l C ll


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Fuel Cells like batteries in which

reactants are constantlybeing added so it never runs down!

Anode and Cathodeboth Pt coated metal

Electrolyte is OHsolution

Anode Reaction:2 H2+ 4 OH

4 H2O(l) + 4 e

-

Cathode Reaction:O2+ 4 H2O + 4 e

- 4 OH


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electroplating


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electroplating

In electroplating, the work

piece is the cathode.

Cations are reduced at

cathode and plate to

the surface of the workpiece.

The anode is made of

the plate metal. The

anode oxidizes andreplaces the metal

cations in the solution


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Electrochemical Cells

in all electrochemical cells, oxidation occurs at theanode, reduction occurs at the cathode

in voltaic cells, anode is the source of electrons and has a () charge

cathode draws electrons and has a (+) charge

in electrolytic cells electrons are drawn off the anode, so it must have a place to

release the electrons, the + terminal of the battery

electrons are forced toward the anode, so it must have a

source of electrons, the terminal of the battery


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Electrolysis of Water


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Electrolysis of Water


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Electrolysis of Pure Compounds

must be in molten (liquid) state

electrodes normally graphite

cations are reduced at the cathode to metalelement

anions oxidized at anode to nonmetal element


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Mixtures of Ions

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

least negative or most positive Ered

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

the anodeleast negative or most positive Eox

Electrolysis of Aqueous Solutions


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Electrolysis of Aqueous Solutions Complicated by more than one possible oxidation and reduction possible cathode reactions reduction of cation to metal reduction of water to H2

2 H2O + 2 e-1H2+ 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 doesnt oxidize

half-reactions that lead to least negative Etotwill occur unless overvoltage changes the conditions

Electrolysis of NaI


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Electrolysis of NaI(aq)with Inert Electrodes

possible oxidations

2 I-1 I2+ 2 e-1 E = 0.54 v2 H2O O2+ 4e-1+ 4H+1 E = 0.82 v

possible reductionsNa+1+ 1e-1Na0 E = 2.71 v2 H2O + 2 e

-1H2+ 2 OH-1 E = 0.41 v

possible oxidations

2 I-1 I2+ 2 e-1 E = 0.54 v2 H2O O2+ 4e-1+ 4H+1 E = 0.82 v

possible reductionsNa+1+ 1e-1Na0 E = 2.71 v2 H2O + 2 e

-1H2+ 2 OH-1 E = 0.41 v

overall reaction

2 I(aq)+ 2 H2O(l)I2(aq)+H2(g) + 2 OH-1(aq)


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Faradays Law

the amount of metal deposited duringelectrolysis is directly proportional to the charge

on the cation, the current, and the length of timethe cell runs

charge that flows through the cell = current xtime

Example 18.10- Calculate the mass of Au that can be plated in

25 i i 5 5 A f h h lf i


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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 minmass Au, g

Given:

Find:

s1

C5.5

Aug6.5

Aumol1

g196.97

mol3

Aumol1

C96,485

mol1

s1

C5.5

min1

s60min25

e

e

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

C6,4859mol1

e emol3

Aumol1

Aumol1

g196.97


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Corrosion

corrosionis the spontaneous oxidation of ametal by chemicals in the environment

since many materials we use are activemetals, corrosion can be a very big problem
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Rusting

rust is hydrated iron(III) oxide moisture must be presentwater is a reactant

required for flow between cathode and anode electrolytes promote rustingenhances current flow

acids promote rustinglower pH = lower Ered


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P i C i


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Preventing Corrosion

one way to reduce or slow corrosion is to coatthe metal surface to keep it from contactingcorrosive chemicals in the environment

paint

some metals, like Al, form an oxide that stronglyattaches to the metal surface, preventing the restfrom corroding

another method to protect one metal is to attachit to a more reactive metal that is cheap

sacrificial electrode

galvanized nails


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Sacrificial Anode

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Chem chapter 18 LEC