4 MATTER ELECTROCHEM

(1)

A) MATTER

B) ELECTROCHEMISTRY

(2)

MATTER

 MATTER HAS _{MATTER HAS}PROPERTIES_PROPERTIES; _;

PROPERTIES DESCRIBE MATTER.

 THERE ARE FOUR GENERAL _{THERE ARE FOUR GENERAL} PROPERTIES OF MATTER.

(3)

MATTER

1) MASS:

1) MASS: THE MASS OF AN OBJECT IS THE MASS OF AN OBJECT IS

ALWAYS CONSTANT.

2)

2) WEIGHT:WEIGHT: WEIGHT IS NOT CONSTANT. WEIGHT IS NOT CONSTANT.

WEIGHT DEPENDS ON LOCATION AND

GRAVITY.

(4)

MATTER

 WEIGHT:WEIGHT: THE AMOUNT OF FORCE THE AMOUNT OF FORCE

EARTHS GRAVITY EXERTS ON AN

OBJECT.

 THE METRIC UNIT USED IS NEWTONS. THE METRIC UNIT USED IS NEWTONS. (N)

(5)

(6)

(7)

MATTER

 (POR) POINT OF REFERENCE: AN _{(POR) POINT OF REFERENCE: AN} OBJECT WITH A MASS OF 1kg IS

OBJECT WITH A MASS OF 1kg IS

PULLED TO THE EARTH WITH A

FORCE OF 9.8(N) NEWTONS

 Example: 200 lbs / 2.2 = 90 kg _{Example: 200 lbs / 2.2 = 90 kg}

(8)

MATTER

 3) VOLUME:_{3) VOLUME:} THE AMOUNT OF SPACE _{THE AMOUNT OF SPACE}

AN OBJECT TAKES UP. THE UNITS

ARE LITERS (L) OR MILLILITERS (ml).

 4) DENSITY:_{4) DENSITY:} THE FOURTH GENERAL _{THE FOURTH GENERAL}

PROPERTY. D= m/v

(9)

MATTER

 MATTER MADE OF THE SAME _{MATTER MADE OF THE SAME}

SUBSTANCE CAN EXIST IN DIFFERENT

STATES; THESE STATES ARE CALLED

STATES; THESE STATES ARE CALLED PHASES.

PHASES.

 THERE ARE FOUR PHASES OF _{THERE ARE FOUR PHASES OF} MATTER.

(10)

MATTER

1.

1. SOLIDSSOLIDS

2.

2. LIQUIDSLIQUIDS

3.

3. GASGAS

4.

(11)

MATTER

 ARCHIMEDES’ PRINCIPLE:_{ARCHIMEDES’ PRINCIPLE:} THE _THE

BUOYANT FORCE ON AN OBJECT IN A

FLUID IS AN UPWARD FORCE EQUAL

TO THE WEIGHT OF THE FLUID THAT

THE OBJECT DISPLACES.

(12)

(13)

MATTER

 PASCAL’S PRINCIPLE:_{PASCAL’S PRINCIPLE:} A CHANGE IN _{A CHANGE IN}

PRESSURE AT ANY POINT IN AN

ENCLOSED FLUID WILL BE

TRANSMITTED EQUALLY TO ALL

PARTS OF THE FLUID.

(14)

(15)

MATTER

 GAS LAWS_{GAS LAWS} DESCRIBE WHY GASES BEHAVE _{DESCRIBE WHY GASES BEHAVE}

THE WAY THEY DO BASED ON THREE THE WAY THEY DO BASED ON THREE

FACTORS: FACTORS:

 TEMPERATURE, PRESSURE AND_{TEMPERATURE, PRESSURE AND} VOLUME._VOLUME.

 THREE PRIMARY GAS LAWS_{THREE PRIMARY GAS LAWS}:_: BOYLES LAW, _{BOYLES LAW,}

(16)

MATTER

 BOYLES LAW:_{BOYLES LAW:} FOR A FIXED AMOUNT _{FOR A FIXED AMOUNT}

OF GAS AT A CONSTANT TEMP. THE

VOLUME OF THE GAS INCREASES AS

ITS PRESSURE DECREASES.

 EXAMPLE: LET A BUBBLE OF AIR OUT _{EXAMPLE: LET A BUBBLE OF AIR OUT} OF YOUR MOUTH 10 m UNDER

OF YOUR MOUTH 10 m UNDER

WATER, THE BUBBLE GETS BIGGER

(17)

(18)

MATTER

 BOYLES LAW FORMULA: P_{BOYLES LAW FORMULA: P} 1

1VV11 = P = P22VV22

 THE MAXIMUM VOLUME A WEATHER THE MAXIMUM VOLUME A WEATHER

BALLOON CAN REACH WITHOUT BALLOON CAN REACH WITHOUT

RUPTURING IS 22,000 LITERS. IT IS RUPTURING IS 22,000 LITERS. IT IS

DESIGNED TO REACH AN ALTITUDE OF 30 DESIGNED TO REACH AN ALTITUDE OF 30

KILOMETERS. AT THIS ALTITUDE, THE KILOMETERS. AT THIS ALTITUDE, THE

ATMOSPHERIC PRESSURE IS 0.0125 atm. ATMOSPHERIC PRESSURE IS 0.0125 atm.

(19)

MATTER

 VV₁₁ = ? V = ? V₂₂ = 22,000 L P = 22,000 L P₁₁ = 1 atm = 1 atm P

P₂₂ = 0.0125 atm = 0.0125 atm

 V1 = (0.0125 atm) (22,000 L)_{V1 = (0.0125 atm) (22,000 L)}

 __________{_________}

 (1 atm)_{(1 atm)}

(20)

MATTER

 TWO LITERS OF AIR AT ATMOSPHERIC _{TWO LITERS OF AIR AT ATMOSPHERIC} PRESSURE ARE COMPRESSED INTO A

PRESSURE ARE COMPRESSED INTO A

0.45 L CANISTER. IF ITS

TEMPERATURE REMAINS CONSTANT,

WHAT IS THE PRESSURE OF THE

COMPRESSED AIR?

(21)

MATTER

 PP₂₂ = ? V = ? V₁₁ = 2 L P = 2 L P₁₁= 1 atm V= 1 atm V₂₂ = 0.045 L = 0.045 L

 PP₂₂ = (2 L) ( 1 atm) = (2 L) ( 1 atm)

 ________{_______}

 (0.45 L) _{(0.45 L)}

(22)

MATTER

 CHARLES LAW:CHARLES LAW: FOR A FIXED AMOUNT FOR A FIXED AMOUNT

OF GAS AT A CONSTANT PRESSURE,

THE VOLUME OF THE GAS INCREASES

AS THE TEMPERATURE INCREASES.

 _example_example: : _{IF YOU HEAT A BALLOON THE VOLUME}_{IF YOU HEAT A BALLOON THE VOLUME} INCREASES AND THE BALLOON POPS. IF YOU

INCREASES AND THE BALLOON POPS. IF YOU

COOL IT, THE VOLUME DECREASES AND THE

BALLOON SHRINKS.

(23)

(24)

MATTER

 GAS VOLUMES AND PRESSURES ARE _{GAS VOLUMES AND PRESSURES ARE}

PROPORTIONAL TO TEMPERATURE ONLY IF PROPORTIONAL TO TEMPERATURE ONLY IF

THE TEMPERATURE IS EXPRESSED IN THE TEMPERATURE IS EXPRESSED IN

KELVINS. WE WILL USE CELSIUS OR KELVINS. WE WILL USE CELSIUS OR

FAHRENHEIT MORE THAN KELVIN. FAHRENHEIT MORE THAN KELVIN.

 TEMPERATURE KELVIN EQUALS _{TEMPERATURE KELVIN EQUALS}

TEMPERATURE CELSIUS PLUS 273 TEMPERATURE CELSIUS PLUS 273

(25)

MATTER

 _{CHARLES FORMULA: V}_{CHARLES FORMULA: V} 1

1/T/T11 = V = V22/T/T22

 A BALLOON IS FILLED WITH 3.0 L OF A BALLOON IS FILLED WITH 3.0 L OF

HELIUM AT 22 DEGREES CELSIUS AT

760 mm Hg. IT IS THEN PLACED

OUTDOORS ON A HOT SUMMER DAY

WHEN THE TEMPERATURE IS 31

DEGREES CELSIUS. IF THE PRESSURE

REMAINS CONSTANT, WHAT WILL THE

(26)

MATTER

 T_T K

K = T = TCC + 273 + 273  T_T

K

K = 22 + 273 = 295 K = 22 + 273 = 295 K

 T_T K

K = 31 + 273 = 304 K = 31 + 273 = 304 K

 V_V 2

2 = (3.0 L) (304 K)= (3.0 L) (304 K)

(27)

MATTER

 GAY-LUSSAC’S LAW:_{GAY-LUSSAC’S LAW:} THE PRESSURE _{THE PRESSURE}

OF A GAS INCREASES AS THE

TEMPERATURE INCREASES IF THE

VOLUME OF THE GAS DOESN’T

CHANGE.

 EXAMPLE: A PRESSURIZED SPRAY CAN _{EXAMPLE: A PRESSURIZED SPRAY CAN}

(28)

(29)

MATTER

 A GAS SAMPLE SOMETIMES _{A GAS SAMPLE SOMETIMES} UNDERGOES CHANGES IN

UNDERGOES CHANGES IN

TEMPERATURE, PRESSURE, AND

VOLUME ALL AT THE SAME TIME.

 WHEN THIS HAPPENS, THREE _{WHEN THIS HAPPENS, THREE}

VARIABLES MUST BE DEALT WITH AT

ONCE.

(30)

MATTER

 THE COMBINED GAS LAW:_{THE COMBINED GAS LAW:}

EXPRESSES THE RELATIONSHIP

BETWEEN PRESSURE, VOLUME, AND

TEMPERATURE OF A FIXED AMOUNT

OF GAS.

(31)

MATTER

 THE COMBINED GAS LAW CAN BE _{THE COMBINED GAS LAW CAN BE}

WRITTEN AS FOLLOWS:

 PP₁₁VV₁₁/T/T₁₁ = P = P₂₂VV₂₂/T/T₂₂

(32)

MATTER

 FROM THIS EXPRESSION, ANY VALUE FROM THIS EXPRESSION, ANY VALUE CAN BE CALCULATED IF THE OTHER

CAN BE CALCULATED IF THE OTHER

FIVE ARE KNOWN.

(33)

MATTER

 A HELIUM FILLED BALLOON HAS A _{A HELIUM FILLED BALLOON HAS A} VOLUME OF 50.0 L AT 25

VOLUME OF 50.0 L AT 2500 C AND 1.08 _{C AND 1.08}

atm. WHAT VOLUME WILL IT HAVE AT

0.855 atm AND 10

0.855 atm AND 100 0 C?_C?

 REARRANGE THE COMBINED GAS _{REARRANGE THE COMBINED GAS} LAW EQUATION AND SOLVE.

(34)

MATTER

 VV₂₂ = P = P₁₁VV₁₁TT₂₂ / P / P₂₂TT₁₁

 VV₂₂ = (1.08 atm)(50 L He)(283 K) = (1.08 atm)(50 L He)(283 K)

 ________________________{_______________________}

(35)

MATTER

 AVOGADROS LAW:_{AVOGADROS LAW:} STATES THAT _{STATES THAT}

EQUAL VOLUMES OF GASES AT THE

SAME TEMPERATURE AND PRESSURE

CONTAIN EQUAL NUMBERS OF

MOLECULES.

(36)

MATTER

 THE VOLUME OF ANY GAS VARIES _{THE VOLUME OF ANY GAS VARIES} DIRECTLY WITH THE NUMBER OF

DIRECTLY WITH THE NUMBER OF

MOLECULES. THE EQUATION FOR

AVOGADROS LAW IS V = kn

 n IS THE AMOUNT OF GAS IN MOLES, _{n IS THE AMOUNT OF GAS IN MOLES,} AND k IS A CONSTANT.

(37)

MATTER

 WHAT VOLUME DOES 0.0685 mol OF GAS _{WHAT VOLUME DOES 0.0685 mol OF GAS}

OCCUPY AT STP? STP IS STANDARD

TEMPERATURE AND PRESSURE, A

CONSTANT, WHICH IS 22.4 L.

(38)

MATTER

 WE HAVE LEARNED EQUATIONS THAT _{WE HAVE LEARNED EQUATIONS THAT} DESCRIBE THE RELATIONSHIPS

DESCRIBE THE RELATIONSHIPS

BETWEEN PRESSURE, VOLUME,

TEMPERATURE, AND MOLES.

 ALL OF THE GAS LAWS WE HAVE _{ALL OF THE GAS LAWS WE HAVE}

LEARNED CAN BE COMBINED INTO A

SINGLE EQUATION.

(39)

MATTER

 THE IDEAL GAS LAW: _{THE IDEAL GAS LAW:}THE _THE

MATHEMATICAL RELATIONSHIP

AMONG PRESSURE, VOLUME,

TEMPERATURE, AND THE NUMBER OF

MOLES OF A GAS.

(40)

MATTER

 PV = nRT_{PV = nRT}

 P pressure, V volume, n number of moles P pressure, V volume, n number of moles or molecules, T temperature, R is a

or molecules, T temperature, R is a

constant.

(41)

MATTER

 THE IDEAL GAS LAW REDUCES _{THE IDEAL GAS LAW REDUCES}

BOYLES, CHARLES, GAY-LUSSACS,

AND AVOGADROS LAW WHEN THE

APPROPRIATE VARIABLES ARE HELD

CONSTANT.

(42)

MATTER

 THE CONSTANT R IS KNOWN AS THE _{THE CONSTANT R IS KNOWN AS THE} IDEAL GAS CONSTANT. ITS VALUE

IDEAL GAS CONSTANT. ITS VALUE

DEPENDS ON THE UNITS CHOSEN

FOR PRESSURE, VOLUME, AND

TEMPERATURE.

 MEASURED VALUES OF P,V,T, AND n _{MEASURED VALUES OF P,V,T, AND n} CAN BE USED TO CALCULATE R.

(43)

MATTER

 R= PV / nT_{R= PV / nT}

 R = (1atm) (22.4 L)R = (1atm) (22.4 L)

 ________________{_______________}

 (1 mol) (273.15 K)_{(1 mol) (273.15 K)}

(44)

MATTER

 THE IDEAL GAS LAW CAN BE APPLIED _{THE IDEAL GAS LAW CAN BE APPLIED} TO DETERMINE THE EXISTING

TO DETERMINE THE EXISTING

CONDITIONS OF A GAS SAMPLE

WHEN THREE OF THE FOUR

VARIABLES, P,V,T, AND n ARE KNOWN.

(45)

MATTER

 WHAT IS THE PRESSURE IN _{WHAT IS THE PRESSURE IN}

ATMOSPHERES EXERTED BY A 0.500

mol SAMPLE OF NITROGEN GAS IN A

10.0 L CONTAINER AT 298 K?

(46)

MATTER

 REARRANGE THE IDEAL GAS LAW AND _{REARRANGE THE IDEAL GAS LAW AND}

SOLVE FOR P.

 P = (0.500 mol) (0.0821) (298 K)_{P = (0.500 mol) (0.0821) (298 K)}

 ________________________{_______________________}

(47)

ELECTROCHEMISTRY

 ELECTROCHEMISTRYELECTROCHEMISTRY: THE STUDY OF : THE STUDY OF

THE RELATIONSHIPS BETWEEN

ELECTRICITY AND CHEMICAL

REACTIONS.

 OXIDATION-REDUCTION REACTION OXIDATION-REDUCTION REACTION (REDOX): REACTIONS OCCUR WHEN

(REDOX): REACTIONS OCCUR WHEN

ELECTRONS ARE TRANSFERRED

FROM AN ATOM THAT IS OXIDIZED TO

AN ATOM THAT IS REDUCED.

(48)

ELECTROCHEMISTRY

 WE DETERMINE WHETHER A GIVEN _{WE DETERMINE WHETHER A GIVEN} CHEMICAL REACTION IS AN

CHEMICAL REACTION IS AN

OXIDATION-REDUCTION REACTION BY

KEEPING TRACK OF THE OXIDATION

NUMBERS (OXIDATION STATES) OF

THE ELEMENTS INVOLVED IN THE

REACTION.

(49)

ELECTROCHEMISTRY

 CONSIDER THE REACTION THAT OCCURS _{CONSIDER THE REACTION THAT OCCURS}

SPONTANEOUSLY WHEN ZINC METAL IS SPONTANEOUSLY WHEN ZINC METAL IS

ADDED TO A STRONG ACID: ADDED TO A STRONG ACID:

 Zn (s) + 2 HZn (s) + 2 H++ (aq) --- (aq) --- Zn Zn2+2+ (aq) + H (aq) + H₂₂ (g) (g)

 THE CHEMICAL REACTION FOR THIS _{THE CHEMICAL REACTION FOR THIS}

(50)

ELECTROCHEMISTRY

 Zn (s) + 2 HZn (s) + 2 H++(aq) (aq)  Zn Zn2+2+ (aq) + H (aq) + H₂₂ (g) (g)  0 +1 +2 0_{0 +1 +2 0}

 ZINC OXIDIZED AND HYDROGEN _{ZINC OXIDIZED AND HYDROGEN} REDUCED.

(51)

ELECTROCHEMISTRY

 THE SUBSTANCE THAT MAKES IT THE SUBSTANCE THAT MAKES IT

POSSIBLE FOR ANOTHER SUBSTANCE

TO BE OXIDIZED IS CALLED THE

OXIDIZING AGENT

OXIDIZING AGENT OR THE OR THE OXIDANTOXIDANT. . THE OXIDIZING AGENT AQUIRES

THE OXIDIZING AGENT AQUIRES

ELECTRONS FROM THE OTHER

SUBSTANCE AND SO IS ITSELF

REDUCED. THE

REDUCED. THE REDUCING AGENT_{REDUCING AGENT} OR _OR

REDUCTANT

(52)

ELECTROCHEMISTRY

 IDENTIFYING OXIDIZING AND _{IDENTIFYING OXIDIZING AND} REDUCING AGENTS:

REDUCING AGENTS:

 THE NICKEL-CADMIUM BATTERY USES THE NICKEL-CADMIUM BATTERY USES THE FOLLOWING REDOX REACTION

THE FOLLOWING REDOX REACTION

TO GENERATE ELECTRICITY:

(53)

ELECTROCHEMISTRY

 FIRST WE ASSIGN OXIDATION STATES FIRST WE ASSIGN OXIDATION STATES OR NUMBERS TO ALL THE ATOMS

OR NUMBERS TO ALL THE ATOMS

AND DETERMINE WHICH ELEMENTS

CHANGE OXIDATION STATE.

(54)

ELECTROCHEMISTRY

 Cd (s) + NiOCd (s) + NiO₂₂ (s) + 2 H (s) + 2 H₂₂O (l) O (l)  Cd(OH) Cd(OH)₂₂ (s) + Ni(OH) (s) + Ni(OH)₂₂ (s) (s)

 0 +4 -2 +1 -2 +2 -2 +1 +2 -2 +1_{0 +4 -2 +1 -2 +2 -2 +1 +2 -2 +1}

 SECOND, WE APPLY THE DEFINITIONS OF _{SECOND, WE APPLY THE DEFINITIONS OF}

OXIDATION AND REDUCTION:

(55)

ELECTROCHEMISTRY

 THE OXIDATION STATE OF Cd THE OXIDATION STATE OF Cd

INCREASES FROM 0 TO +2, AND THAT

OF Ni DECREASES FROM +4 TO +2. THE

Cd ATOM IS OXIDIZED (LOSES

ELECTRONS) AND IS THE REDUCING

AGENT. THE OXIDATION STATE OF Ni

DECREASES AS NiO

DECREASES AS NiO₂₂ IS CONVERTED IS CONVERTED INTO Ni(OH)

INTO Ni(OH)₂₂. NiO. NiO₂₂ IS REDUCED (GAINS IS REDUCED (GAINS ELECTRONS) AND IS THE OXIDIZING

(56)

ELECTROCHEMISTRY

 BALANCING REDOX EQUATIONS:_{BALANCING REDOX EQUATIONS:}

 MUST OBEY THE LAW OF MUST OBEY THE LAW OF CONSERVATION OF MASS

CONSERVATION OF MASS

 AND_AND

(57)

ELECTROCHEMISTRY

 THE METHOD OF _{THE METHOD OF}HALF-REACTIONS_{HALF-REACTIONS} IS _IS A SYSTEMATIC PROCEDURE FOR

A SYSTEMATIC PROCEDURE FOR

BALANCING REDOX REACTIONS.

 HALF REACTIONS SHOW EITHER _{HALF REACTIONS SHOW EITHER}

OXIDATION OR REDUCTION ALONE.

 EXAMPLE: THE OXIDATION OF Sn_{EXAMPLE: THE OXIDATION OF Sn}2+2+ BY _BY

Fe

(58)

ELECTROCHEMISTRY

 SnSn2+2+ (aq) + 2 Fe_{(aq) + 2 Fe}3+3+ (aq) _(aq)_ Sn_Sn4+4+ (aq) + 2 Fe_{(aq) + 2 Fe}2+2+ (aq)_(aq)

 OXIDATION: Sn_{OXIDATION: Sn}2+2+ (aq) _(aq)_ Sn_Sn4+4+ (aq) + 2 e_{(aq) + 2 e}-

- REDUCTION: 2 FeREDUCTION: 2 Fe3+3+ (aq) + 2 e_{(aq) + 2 e}-- _ 2 Fe_{2 Fe}2+2+ (aq)_(aq)

 NOTICE THAT ELECTRONS ARE SHOWN AS _{NOTICE THAT ELECTRONS ARE SHOWN AS}

(59)

ELECTROCHEMISTRY

 BALANCING EQUATIONS BY THE _{BALANCING EQUATIONS BY THE} METHOD OF HALF-REACTIONS:

METHOD OF HALF-REACTIONS:

 BEGIN WITH A SKELETON IONIC _{BEGIN WITH A SKELETON IONIC} EQUATION SHOWING ONLY THE

EQUATION SHOWING ONLY THE

SUBSTANCES UNDERGOING

OXIDATION AND REDUCTION.

(60)

ELECTROCHEMISTRY

 WE WILL FIND THAT H_{WE WILL FIND THAT H}+ + FOR ACIDIC _{FOR ACIDIC}

SOLUTIONS, OH

SOLUTIONS, OH-- FOR BASIC _{FOR BASIC}

SOLUTIONS, AND H

SOLUTIONS, AND H₂₂O IS BEING O IS BEING OXIDIZED OR REDUCED, THESE

OXIDIZED OR REDUCED, THESE

SPECIES DO NOT APPEAR IN THE

SKELETON EQUATION.

(61)

ELECTROCHEMISTRY

 BALANCING A REDOX REACTION THAT _{BALANCING A REDOX REACTION THAT} OCCURS IN ACIDIC AQUEOUS

OCCURS IN ACIDIC AQUEOUS

SOLUTIONS:

 1) DIVIDE THE EQUATION INTO ONE _{1) DIVIDE THE EQUATION INTO ONE}

OXIDATION HALF REACTION AND ONE

REDUCTION HALF REACTION.

(62)

ELECTROCHEMISTRY

 2) BALANCE EACH HALF REACTION.2) BALANCE EACH HALF REACTION.

 A) FIRST, BALANCE ELEMENTS OTHER A) FIRST, BALANCE ELEMENTS OTHER THAN H AND O.

THAN H AND O.

 B) NEXT, BALANCE O ATOMS BY B) NEXT, BALANCE O ATOMS BY ADDING H

ADDING H₂₂O AS NEEDED.O AS NEEDED.

 C) THEN BALANCE H ATOMS BY C) THEN BALANCE H ATOMS BY ADDING H

(63)

ELECTROCHEMISTRY

 3) MULTIPLY HALF REACTIONS BY _{3) MULTIPLY HALF REACTIONS BY}

INTEGERS AS NEEDED TO MAKE THE

NUMBER OF ELECTRONS LOST IN THE

OXIDATION HALF REACTION EQUAL

THE NUMBER OF ELECTRONS GAINED

IN THE REDUCTION HALF REACTION.

(64)

ELECTROCHEMISTRY

 4) ADD HALF REACTIONS AND, IF _{4) ADD HALF REACTIONS AND, IF}

POSSIBLE, SIMPLIFY BY CANCELING

SPECIES APPEARING ON BOTH SIDES

OF THE COMBINED EQUATION.

 5) CHECK TO MAKE SURE THAT _{5) CHECK TO MAKE SURE THAT} ATOMS AND CHARGES ARE

(65)

ELECTROCHEMISTRY

 CONSIDER THE REACTION BETWEEN _{CONSIDER THE REACTION BETWEEN} PERMANGANATE ION (MnO

PERMANGANATE ION (MnO₄₄--) AND _{) AND}

OXALATE ION (C

OXALATE ION (C₂₂OO₄₄2-2-) IN ACIDIC _{) IN ACIDIC}

SOLUTION.

(66)

ELECTROCHEMISTRY

 WHEN MnOWHEN MnO₄₄-- IS ADDED TO AN IS ADDED TO AN ACIDIFIED SOLUTION OF C

ACIDIFIED SOLUTION OF C₂₂OO₄₄2-2-, THE _{, THE}

DEEP PURPLE COLOR OF THE MnO

DEEP PURPLE COLOR OF THE MnO₄₄-

-ION FADES, BUBBLES OF CO

ION FADES, BUBBLES OF CO₂₂ FORM, _FORM,

AND THE SOLUTION TAKES ON A PALE

PINK COLOR OF Mn

(67)

ELECTROCHEMISTRY

 MnOMnO₄₄-- (aq) + C_{(aq) + C} 2

2OO442-2- (aq) (aq)  Mn Mn2+2+ (aq) + CO (aq) + CO22 (aq) (aq)

 1) WRITE THE TWO HALF REACTIONS:_{1) WRITE THE TWO HALF REACTIONS:}

 MnOMnO₄₄- - (aq) (aq)  Mn Mn2+2+ (aq) (aq)

 CC₂₂OO₄₄2-2- (aq) _(aq)_ CO_CO 2

(68)

ELECTROCHEMISTRY

 2) COMPLETE AND BALANCE EACH _{2) COMPLETE AND BALANCE EACH} HALF REACTION.

HALF REACTION.

 MnOMnO₄₄-- (aq) (aq)  Mn Mn2+2+ (aq) (aq)

(69)

ELECTROCHEMISTRY

 3) NEXT WE BALANCE O._{3) NEXT WE BALANCE O.}

 MnOMnO₄₄-- (aq) _(aq)_ Mn_Mn2+2+ (aq) + 4 H_{(aq) + 4 H} 2

2O (l)O (l)

 4) NEXT BALANCE THE H.4) NEXT BALANCE THE H.

(70)

ELECTROCHEMISTRY

 5) BALANCE THE CHARGE._{5) BALANCE THE CHARGE.}

 5 e5 e-- + 8 H_{+ 8 H}++ (aq) + MnO_{(aq) + MnO} 4

4-- (aq) (aq)  Mn Mn2+2+ (aq) + 4 H (aq) + 4 H22OO

 CC₂₂OO₄₄2-2- (aq) _(aq)_ 2 CO_{2 CO} 2

(71)

-ELECTROCHEMISTRY

ELECTROCHEMISTRY

 6) MULTIPLY EACH HALF REACTION _{6) MULTIPLY EACH HALF REACTION} BY AN APPROPRIATE INTEGER SO

BY AN APPROPRIATE INTEGER SO

THAT THE NUMBER OF ELECTRONS

GAINED IN ONE HALF REACTION

EQUALS THE NUMBER OF

ELECTRONS LOST IN THE OTHER:

(72)

ELECTROCHEMISTRY

 10 e10 e-- + 16 H_{+ 16 H}++ (aq) + 2 MnO_{(aq) + 2 MnO} 4

4-- (aq) (aq)  2 Mn 2 Mn2+2+ (aq) (aq)

+ 8 H

+ 8 H₂₂O (l)O (l)

(73)

-ELECTROCHEMISTRY

ELECTROCHEMISTRY

 THE BALANCED EQUATION IS THE _{THE BALANCED EQUATION IS THE} SUM OF THE BALANCED HALF

SUM OF THE BALANCED HALF

REACTIONS:

 16 H16 H++ (aq) + 2 MnO (aq) + 2 MnO₄₄-- (aq) + 5 C (aq) + 5 C₂₂OO₄₄2-2- (aq) (aq)



 2 Mn_{2 Mn}2+2+ (aq) + 8 H_{(aq) + 8 H} 2

(74)

ELECTROCHEMISTRY

 IF A REDOX REACTION OCCURS IN A _{IF A REDOX REACTION OCCURS IN A} BASIC SOLUTION, THE EQUATION

BASIC SOLUTION, THE EQUATION

MUST BE BALANCED BY USING OH

MUST BE BALANCED BY USING OH- -

AND H

AND H₂₂O RATHER THAN H_{O RATHER THAN H}++ AND H_{AND H} 2

(75)

ELETROCHEMISTRY

 BALANCING REDOX EQUATIONS IN _{BALANCING REDOX EQUATIONS IN} BASIC SOLUTION:

BASIC SOLUTION:

 WE GO THROUGH THE FIRST STEP OF _{WE GO THROUGH THE FIRST STEP OF} THE PROCEDURE AS IF THE

THE PROCEDURE AS IF THE

REACTION WERE OCCURING IN

ACIDIC SOLUTION.

(76)

ELECTROCHEMISTRY

 THEN ADD THE APPROPRIATE _{THEN ADD THE APPROPRIATE} NUMBER OF OH

NUMBER OF OH- - IONS TO EACH SIDE _{IONS TO EACH SIDE}

OF THE EQUATION, COMBINING H

OF THE EQUATION, COMBINING H++

WITH OH

WITH OH-- TO FORM H_{TO FORM H} 2

2O.O.

(77)

ELECTROCHEMISTRY

 COMPLETE AND BALANCE THE _{COMPLETE AND BALANCE THE}

FOLLOWING EQUATION FOR A REDOX

REACTION THAT TAKES PLACE IN

BASIC SOLUTION:

 CNCN-- (aq) + MnO_{(aq) + MnO} 4

(78)

ELECTROCHEMISTRY

 1) WRITE THE INCOMPLETE _{1) WRITE THE INCOMPLETE}

UNBALANCED HALF REACTION.

 CN_CN-- (aq) _(aq)_ CNO_CNO-- (aq)_(aq)

 MnOMnO₄₄-- (aq) _(aq)_ MnO_MnO 2

(79)

ELECTROCHEMISTRY

 2) BALANCE EACH HALF REACTION AS _{2) BALANCE EACH HALF REACTION AS} IF IT TOOK PLACE IN ACIDIC

IF IT TOOK PLACE IN ACIDIC

SOLUTION. (BALANCE THE “O”, THE

“H”, AND THE e-)

 _CN_CN-- (aq) + H_{(aq) + H} 2

2O (l) O (l)  CNO CNO-- (aq) + 2 H (aq) + 2 H++ (aq) + 2 e (aq) + 2 e-

- _{3 e}_{3 e}-- + 4 H_{+ 4 H}++ (aq) + MnO_{(aq) + MnO} 4

4-- (aq) (aq)  MnO MnO22 (s) + 2 (s) + 2

H

(80)

ELECTROCHEMISTRY

 3) TAKE INTO ACCOUNT THAT THE _{3) TAKE INTO ACCOUNT THAT THE} REACTION OCCURS IN BASIC

REACTION OCCURS IN BASIC

SOLUTION. ADD OH

SOLUTION. ADD OH- - TO BOTH SIDES _{TO BOTH SIDES}

OF BOTH HALF REACTIONS TO

NEUTRALIZE THE H

(81)

ELECTROCHEMISTRY

 CNCN-- (aq) + H (aq) + H₂₂O (l) + 2 OHO (l) + 2 OH-- (aq) (aq)  CNO CNO- -(aq) + 2 H

(aq) + 2 H++ (aq) + 2 e_{(aq) + 2 e}-- + 2 OH_{+ 2 OH}-- (aq)_(aq)

 3 e3 e-- + 4 H + 4 H++ (aq) + MnO (aq) + MnO₄₄-- (aq) + 4 OH (aq) + 4 OH-- (aq) (aq)



(82)

ELECTROCHEMISTRY

 NEXT, NEUTRALIZE H_{NEXT, NEUTRALIZE H}++ AND OH_{AND OH}-- BY _BY

FORMING WATER WHEN THEY ARE

ON THE SAME SIDE OF EITHER HALF

REACTION.

(83)

ELECTROCHEMISTRY

 CNCN-- (aq) + H_{(aq) + H} 2

2O (l) + 2 OHO (l) + 2 OH-- (aq) (aq)  CNO CNO-- (aq) + 2 H (aq) + 2 H22O (l) O (l)

+ 2 e

+ 2 e-

- 3 e- + 4 H3 e- + 4 H₂₂O (l) + MnOO (l) + MnO₄₄-- (aq) _(aq)_ MnO_MnO 2

2 (s) + 2 H (s) + 2 H22O (l) + 4 O (l) + 4

OH

(84)

-ELECTROCHEMISTRY

ELECTROCHEMISTRY

 NEXT, CANCEL WATER MOLECULES _{NEXT, CANCEL WATER MOLECULES}

THAT APPEAR AS BOTH REACTANTS

AND PRODUCTS.

 CNCN-- (aq) + 2 OH_{(aq) + 2 OH}-- (aq) _(aq)_ CNO_CNO-- (aq) + H_{(aq) + H} 2

20 (l) + 2 e0 (l) + 2 e-

-3 e

(85)

ELECTROCHEMISTRY

 NEXT, CHECK THE ATOMS AND THE NEXT, CHECK THE ATOMS AND THE CHARGES.

CHARGES.

 MULTIPLY THE CYANIDE HALF MULTIPLY THE CYANIDE HALF

REACTION BY 3 WHICH GIVES 6 e- ON

THE PRODUCT SIDE AND MULTIPLY

THE PERMANGATE HALF REACTION

BY 2, WHICH GIVES 6 e- ON THE

REACTANT SIDE.

(86)

ELECTROCHEMISTRY

 3 CN3 CN-- (aq) + 6 OH_{(aq) + 6 OH}-- (aq) _(aq)_ 3 CNO_{3 CNO}-- (aq) + 3 H_{(aq) + 3 H} 2

2O O

(l) + 6 e (l) + 6 e-

- 6 e6 e-- + 4 H + 4 H₂₂O (l) + 2 MnOO (l) + 2 MnO₄₄-- (aq) (aq)  2 MnO 2 MnO₂₂ (s) + 8 (s) + 8

OH

(87)

ELECTROCHEMISTRY

 FINALLY, ADD THE TWO HALF _{FINALLY, ADD THE TWO HALF}

REACTIONS TOGETHER AND SIMPLIFY

BY CANCELING SPECIES THAT

APPEAR AS BOTH REACTANTS AND

PRODUCTS.

(88)

ELECTROCHEMISTRY

 3 CN3 CN-- (aq) + H_{(aq) + H} 2

2O (l) + 2 MnOO (l) + 2 MnO44-- (aq) (aq)  3 CNO 3 CNO- -

(aq) + 2 MnO

(89)

ELECTROCHEMISTRY

 VOLTAIC CELL_{VOLTAIC CELL}: (GALVANIC CELL) A _{: (GALVANIC CELL) A}

DEVICE IN WHICH THE TRANSFER OF

ELECTRONS TAKES PLACE THROUGH

AN EXTERNAL PATHWAY RATHER

THAN DIRECTLY BETWEEN

REACTANTS PRESENT IN THE SAME

REACTION VESSEL.

(90)

ELECTROCHEMISTRY

 TWO SOLID METALS CONNECTED BY _{TWO SOLID METALS CONNECTED BY} THE EXTERNAL CIRCUIT ARE CALLED

THE EXTERNAL CIRCUIT ARE CALLED

ELECTRODES.

 ANODE_ANODE: THE ELECTRODE WHERE _{: THE ELECTRODE WHERE}

OXIDATION OCCURS.

(91)

ELECTROCHEMISTRY

 EACH COMPARTMENT OF A VOLTAIC _{EACH COMPARTMENT OF A VOLTAIC} CELL IS CALLED A HALF-CELL.

CELL IS CALLED A HALF-CELL.

 ONE HALF-CELL IS THE SITE OF THE _{ONE HALF-CELL IS THE SITE OF THE} OXIDATION HALF-REACTION AND THE

OXIDATION HALF-REACTION AND THE

OTHER IS THE SITE OF THE

REDUCTION HALF-REACTION.

(92)

ELECTROCHEMISTRY

 FOR EXAMPLE:_{FOR EXAMPLE:}

 A STRIP OF ZINC (Zn) IS PLACED IN A A STRIP OF ZINC (Zn) IS PLACED IN A 1 M ZnSO

1 M ZnSO₄₄ SOLUTION (Zn SOLUTION (Zn2+2+). _).

(ZINC ELECTRODE)

A PIECE OF COPPER (Cu) IS PLACED

(93)

ELECTROCHEMISTRY

 ANODE_ANODE: OXIDATION HALF REACTION: _{: OXIDATION HALF REACTION:} Zn (s) ----> Zn

Zn (s) ----> Zn2+2+ (aq) + 2 e_{(aq) + 2 e}-

- CATHODE_CATHODE: REDUCTION HALF _{: REDUCTION HALF} REACTION:

REACTION:

(94)

ELECTROCHEMISTRY

 SALT BRIDGE_{SALT BRIDGE}: U-SHAPED TUBE _{: U-SHAPED TUBE} CONTAINING AN ELECTROLYTE

CONTAINING AN ELECTROLYTE

SOLUTION SUCH AS KNO

SOLUTION SUCH AS KNO₃₃ (aq) WHOSE (aq) WHOSE IONS WILL NOT REACT WITH OTHER

IONS WILL NOT REACT WITH OTHER

IONS IN THE VOLTAIC CELL.

 IT ALLOWS IONS TO MIGRATE AND _{IT ALLOWS IONS TO MIGRATE AND}

MAINTAINS ELECTRICAL NEUTRALITY

(95)

ELECTROCHEMISTRY

 IN A VOLTAIC CELL, THE ANODE IS _{IN A VOLTAIC CELL, THE ANODE IS}

LABELED WITH A NEGATIVE SIGN AND

THE CATHODE IS LABELED WITH A

POSITIVE SIGN.

(96)

(97)

ELECTROCHEMISTRY

 DESCRIBING A VOLTAIC CELL:_{DESCRIBING A VOLTAIC CELL:}

 A SOLUTION CONTAINING KA SOLUTION CONTAINING K₂₂CrCr₂₂OO₇₇ AND AND H

H₂₂SOSO₄₄ IS POURED INTO ONE BEAKER. IS POURED INTO ONE BEAKER. A SOLUTION OF KI IS POURED INTO

A SOLUTION OF KI IS POURED INTO

ANOTHER. THEY ARE JOINED BY A

SALT BRIDGE.

(98)

ELECTROCHEMISTRY

 THE OXIDATION REDUCTION _{THE OXIDATION REDUCTION} REACTION:

REACTION:

 CrCr₂₂OO₇₇2-2- (aq) + 14 H (aq) + 14 H++ (aq) + 6 I (aq) + 6 I-- (aq) (aq)

(99)

ELECTROCHEMISTRY

 A) WRITE THE TWO HALF-REACTIONS._{A) WRITE THE TWO HALF-REACTIONS.}

 B) WHICH SOLUTION IS THE ANODE?_{B) WHICH SOLUTION IS THE ANODE?}

 C) WHICH ELEMENT (ION) OXIDIZED, C) WHICH ELEMENT (ION) OXIDIZED, THE SOURCE OF THE ELECTRONS?

THE SOURCE OF THE ELECTRONS?

 D) WHICH SOLUTION IS THE _{D) WHICH SOLUTION IS THE} CATHODE?

CATHODE?

(100)

ELECTROCHEMISTRY

 CrCr₂₂OO₇₇2-2- (aq) + 14 H_{(aq) + 14 H}++ + 6 e_{+ 6 e}-- ---> 2 Cr_{---> 2 Cr}3+3+ (aq) _(aq)

+ 7 H

+ 7 H₂₂O.O.

(101)

-ELECTROCHEMISTRY

ELECTROCHEMISTRY

 THE KI SOLUTION IS THE ANODE AND _{THE KI SOLUTION IS THE ANODE AND} THE 6 I

THE 6 I-- OXIDIZED TO A 3 I_{OXIDIZED TO A 3 I} 2

2..

 THE KTHE K₂₂CrCr₂₂OO₇₇ SOLUTION IS THE SOLUTION IS THE CATHODE AND Cr

CATHODE AND Cr₂₂OO₇₇2-2- IONS _IONS

ACCEPTED THE ELECTRONS.

(102)

ELECTROCHEMISTRY

 CELL POTENTIAL UNDER STANDARD _{CELL POTENTIAL UNDER STANDARD} CONDITIONS:

CONDITIONS:

 ELECTRONS FLOW SPONTANEOUSLY _{ELECTRONS FLOW SPONTANEOUSLY} THROUGH AN EXTERNAL CIRCUIT

THROUGH AN EXTERNAL CIRCUIT

FROM THE ANODE TO THE CATHODE

BECAUSE OF A DIFFERENCE IN

(103)

ELECTROCHEMISTRY

 THE POTENTIAL ENERGY OF _{THE POTENTIAL ENERGY OF}

ELECTRONS IS HIGHER IN THE ANODE

THAN THE CATHODE.

 THE DIFFERENCE IN POTENTIAL _{THE DIFFERENCE IN POTENTIAL} ENERGY (

ENERGY (POTENTIAL DIFFERENCEPOTENTIAL DIFFERENCE) ) BETWEEN TWO ELECTRODES IS

BETWEEN TWO ELECTRODES IS

MEASURED IN

(104)

ELECTROCHEMISTRY

 ONE VOLT (V) IS THE POTENTIAL _{ONE VOLT (V) IS THE POTENTIAL}

DIFFERENCE REQUIRED TO IMPART

1 JOULE (J) OF ENERGY TO A CHARGE

OF 1 COULOMB (C).

(105)

ELECTROCHEMISTRY

 THE POTENTIAL DIFFERENCE _{THE POTENTIAL DIFFERENCE}

BETWEEN THE TWO ELECTRODES OF

A VOLTAIC CELL IS CALLED THE

A VOLTAIC CELL IS CALLED THE CELL CELL

POTENTIAL.

(106)

ELECTROCHEMISTRY

 BECAUSE THE POTENTIAL DIFFERENCE _{BECAUSE THE POTENTIAL DIFFERENCE}

PROVIDES THE DRIVING FORCE THAT PROVIDES THE DRIVING FORCE THAT

PUSHES ELECTRONS THROUGH THE PUSHES ELECTRONS THROUGH THE

EXTERNAL CIRCUIT, IT IS ALSO CALLED THE EXTERNAL CIRCUIT, IT IS ALSO CALLED THE

ELECTROMOTIVE FORCE

(107)

ELECTROCHEMISTRY

 THE MAGNITUDE OF THE CELL _{THE MAGNITUDE OF THE CELL} POTENTIAL DEPENDS ON THE

POTENTIAL DEPENDS ON THE

REACTIONS THAT OCCUR AT THE

ANODE AND CATHODE, THE

CONCENTRATION OF REACTANTS

AND PRODUCTS, AND THE

TEMPERATURE.

(108)

ELECTROCHEMISTRY

 STANDARD CONDITIONS_{STANDARD CONDITIONS} ARE:_ARE:

 1 M CONCENTRATIONS FOR 1 M CONCENTRATIONS FOR

REACTANTS AND PRODUCTS IN

SOLUTION AND 1 ATMOSPHERE

PRESSURE AT 25 DEGREES CELSIUS.

(109)

ELECTROCHEMISTRY

 THE STANDARD CELL POTENTIAL, _{THE STANDARD CELL POTENTIAL,} E

Eoo cell

cell, IS THE STANDARD REDUCTION , IS THE STANDARD REDUCTION

POTENTIAL OF THE CATHODE E

POTENTIAL OF THE CATHODE Eoo

red

CATHODE MINUS THE STANDARD

REDUCTION POTENTIAL OF THE

ANODE E

ANODE Eoo

red

(110)

ELECTROCHEMISTRY

 FOR A Zn-Cu_{FOR A Zn-Cu}2+2+ VOLTAIC CELL, WE _{VOLTAIC CELL, WE}

HAVE:

 Zn (s) + Cu_{Zn (s) + Cu}2+2+ (aq) _(aq)_ Zn_Zn2+2+ + Cu (s)_{+ Cu (s)}

(111)

ELECTROCHEMISTRY

 EEoo_cell_cell = E = Eoo_red_red(cathode) – E(cathode) – Eoo_red_red(anode)(anode)

 1.10 V = E1.10 V = Eoo_red_red(cathode) – (-0.76 V)(cathode) – (-0.76 V)

 EEoo red

(112)

ELECTROCHEMISTRY

 THE STANDARD CELL POTENTIAL IS _{THE STANDARD CELL POTENTIAL IS} 1.46 V FOR A VOLTAIC CELL BASED

1.46 V FOR A VOLTAIC CELL BASED

ON THE FOLLOWING HALF-REACTION:

(113)

-ELECTROCHEMISTRY

ELECTROCHEMISTRY

 USING THE STANDARD REDUCTION _{USING THE STANDARD REDUCTION} POTENTIAL TABLE, CALCULATE E

POTENTIAL TABLE, CALCULATE Eoo

red

FOR THE REDUCTION OF In

FOR THE REDUCTION OF In3+3+ TO In_{TO In}++._.

(114)

ELECTROCHEMISTRY

 A VOLTAIC CELL IS BASED ON THE _{A VOLTAIC CELL IS BASED ON THE} TWO STANDARD HALF-REACTIONS:

TWO STANDARD HALF-REACTIONS:

 Cd_Cd2+2+ (aq) + 2 e_{(aq) + 2 e}-- _ Cd (s)_{Cd (s)}

(115)

ELECTROCHEMISTRY

 USE THE STANDARD REDUCTION _{USE THE STANDARD REDUCTION} POTENTIAL CHART TO DETERMINE:

POTENTIAL CHART TO DETERMINE:

 A) WHICH HALF REACTION OCCURS A) WHICH HALF REACTION OCCURS AT THE CATHODE AND WHICH

AT THE CATHODE AND WHICH

OCCURS AT THE ANODE.

 B) DETERMINE THE STANDARD CELL _{B) DETERMINE THE STANDARD CELL} POTENTIAL.

(116)

ELECTROCHEMISTRY

 THE CATHODE WILL HAVE THE _{THE CATHODE WILL HAVE THE} REDUCTION WITH THE MORE

REDUCTION WITH THE MORE

POSITIVE E

POSITIVE Eoo red

red VALUE AND THE ANODE VALUE AND THE ANODE

WILL HAVE THE LESS POSITIVE E

WILL HAVE THE LESS POSITIVE Eoo

red

(117)

ELECTROCHEMISTRY

 ACCORDING TO OUR CHART:_{ACCORDING TO OUR CHART:}

 EEoo red

red (Cd (Cd2+2+/Cd) = - 0.403 V/Cd) = - 0.403 V

(118)

ELECTROCHEMISTRY

 THE STANDARD REDUCTION _{THE STANDARD REDUCTION} POTENTIAL FOR Sn

POTENTIAL FOR Sn2+2+ IS MORE _{IS MORE}

POSITIVE (LESS NEGATIVE) THAN

THAT FOR Cd

THAT FOR Cd2+2+, THE REDUCTION OF _{, THE REDUCTION OF}

Sn

Sn2+2+ IS THE REACTION THAT OCCURS _{IS THE REACTION THAT OCCURS}

AT THE

(119)

ELECTROCHEMISTRY

 CATHODE:_CATHODE:

 Sn_Sn2+2+ + 2 e_{+ 2 e}-- _ Sn_Sn

 ANODE:_ANODE:

(120)

-ELECTROCHEMISTRY

ELECTROCHEMISTRY

 EEoo cell

cell = E = Eooredred (cathode) – E (cathode) – Eooredred (anode) (anode)

(121)

ELECTROCHEMISTRY

 DETERMINING RELATIVE STRENGTHS _{DETERMINING RELATIVE STRENGTHS} OF OXIDIZING AGENTS:

OF OXIDIZING AGENTS:

 USING OUR CHART, LETS RANK THE _{USING OUR CHART, LETS RANK THE} FOLLOWING IONS IN ORDER OF

FOLLOWING IONS IN ORDER OF

INCREASING STRENGTH AS

OXIDIZING AGENTS:

NO

(122)

ELECTROCHEMISTRY

 THE MORE READILY AN ION IS _{THE MORE READILY AN ION IS}

REDUCED (THE MORE POSITIVE ITS

E

Eoo red

red VALUE) THE STRONGER IT IS AS VALUE) THE STRONGER IT IS AS

AN OXIDIZING AGENT.

(123)

ELECTROCHEMISTRY

 NONO₃₃-- (aq) + 4 H_{(aq) + 4 H}++ (aq) + 3 e_{(aq) + 3 e}-- _ NO (g) + 2 H_{NO (g) + 2 H} 2

2O (l) O (l)

 EEoo red

red = +0.96 V = +0.96 V

 AgAg++ (aq) + e (aq) + e--  Ag (s) E Ag (s) Eoo_red_red = + 0.80 V = + 0.80 V

 CrCr₂₂OO₇₇2-2- (aq) + 14 H_{(aq) + 14 H}++ (aq) + 6 e_{(aq) + 6 e}-- _ 2 Cr_{2 Cr}3+3+ (aq) + 7 H_{(aq) + 7 H} 2

2O (l) O (l)

E

(124)

ELECTROCHEMISTRY

 BECAUSE THE STANDARD _{BECAUSE THE STANDARD}

REDUCTION POTENTIAL OF Ag+ IS

THE MOST POSITIVE, IT IS THE

STRONGEST OXIDIZING AGENT.

 THEY RANK:_{THEY RANK:}

 CrCr₂₂OO₇₇2-2- > NO_{> NO} 3

(125)

ELECTROCHEMISTRY

 IN GENERAL REDOX REACTIONS (NOT _{IN GENERAL REDOX REACTIONS (NOT} JUST VOLTAIC CELLS), DETERMINING

JUST VOLTAIC CELLS), DETERMINING

SPONTANEITY

SPONTANEITY….WE NEED TO MODIFY ….WE NEED TO MODIFY OUR EQUATION:

OUR EQUATION:

 EEoo = E_{= E}oo red

(126)

ELECTROCHEMISTRY

 EEoo = E_{= E}oo red

red ( (REDUCTION PROCESSREDUCTION PROCESS) – E) – Eooredred ( (OXIDATION PROCESSOXIDATION PROCESS))

 WE HAVE DROPPED THE SUBSCRIPT “CELL” AND _{WE HAVE DROPPED THE SUBSCRIPT “CELL” AND}

DROPPED CATHODE AND ANODE.

 WE CAN MAKE A GENERAL STATEMENT ABOUT _{WE CAN MAKE A GENERAL STATEMENT ABOUT}

SPONTANEITY:

 A POSITIVE VALUE OF “E” INDICATES A _{A POSITIVE VALUE OF “E” INDICATES A}

SPONTANEOUS PROCESS; AND A NEGATIVE

(127)

ELECTROCHEMISTRY

 SPONTANEOUS OR NOT?_{SPONTANEOUS OR NOT?}

 USE THE STANDARD REDUCTION _{USE THE STANDARD REDUCTION}

POTENTIAL CHART AND DETERMINE

IF THE FOLLOWING REACTIONS ARE

SPONTANEOUS OR NOT UNDER

STANDARD CONDITIONS:

(128)

ELECTROCHEMISTRY

 1) WRITE HALF REACTIONS_{1) WRITE HALF REACTIONS}

 2) USE THE SRP FROM CHART TO 2) USE THE SRP FROM CHART TO CALCULATE E

CALCULATE Eoo._.

 3) IF THE REACTION IS _{3) IF THE REACTION IS} SPONTANEOUS, E

(129)

ELECTROCHEMISTRY

 2 H2 H++ (aq) + 2 e (aq) + 2 e--  H H₂₂ (g) (reduction) (g) (reduction)

 EEoo_red_red = 0 V = 0 V

 Cu (s) _{Cu (s)}_ Cu_Cu2+2+ (aq) + 2 e_{(aq) + 2 e}-- (oxidation)_(oxidation)

(130)

ELECTROCHEMISTRY

 E_Eoo = (0 V) – (.34 V) = -.34 V_{= (0 V) – (.34 V) = -.34 V}

 BECAUSE EBECAUSE Eoo IS NEGATIVE, THE _{IS NEGATIVE, THE}

REACTION IS NOT SPONTANEOUS.

(131)

ELECTROCHEMISTRY

 ClCl₂₂ (g) + 2 e (g) + 2 e--  2 Cl 2 Cl-- (aq) (reduction) (aq) (reduction)

 EEoo_red_red = + 1.36 V = + 1.36 V

 2 I2 I-- (aq) _(aq)_ I_I 2

2 (s) + 2 e (s) + 2 e-- (oxidation) (oxidation)

E

(132)

ELECTROCHEMISTRY

 E_Eoo = (1.36 V) – (.54 V) = + .82 V_{= (1.36 V) – (.54 V) = + .82 V}

 BECAUSE THE VALUE OF EBECAUSE THE VALUE OF Eoo IS _IS

POSITIVE, THE REACTION IS

SPONTANEOUS.

(133)

ELECTROCHEMISTRY

 GIBBS FREE ENERGY_{GIBBS FREE ENERGY}: THE ENERGY _{: THE ENERGY} ASSOCIATED WITH A CHEMICAL

ASSOCIATED WITH A CHEMICAL

REACTION THAT CAN BE USED TO DO

WORK.

 IT IS A _{IT IS A}MEASURE OF THE _{MEASURE OF THE} SPONTANEITY

SPONTANEITY OF A PROCESS THAT OF A PROCESS THAT OCCURS AT CONSTANT TEMP. AND

OCCURS AT CONSTANT TEMP. AND

PRESSURE.

(134)

ELECTROCHEMISTRY

 GIBBS FREE ENERGY: ^G_{GIBBS FREE ENERGY: ^G}

(135)

ELECTROCHEMISTRY

 n REPRESENTS THE NUMBER OF _{n REPRESENTS THE NUMBER OF}

ELECTRONS TRANSFERRED IN THE

REACTION…. WHEN YOU ASSIGN

OXIDATION NUMBERS TO HALF

REACTIONS.

(136)

ELECTROCHEMISTRY

 F IS FARADAYS CONSTANT, THE _{F IS FARADAYS CONSTANT, THE}

QUANTITY OF ELECTRICAL CHARGE

ON ONE MOLE OF ELECTRONS.

(137)

ELECTROCHEMISTRY

 E IS THE EMF OR E CELL THAT WE _{E IS THE EMF OR E CELL THAT WE}

LEARNED TO CALCULATE EARLIER BY

ASSIGNING SRP NUMBERS.

(138)

ELECTROCHEMISTRY

 ^G IS ALSO RELATED TO THE _{^G IS ALSO RELATED TO THE}

EQUILIBRIUM CONSTANT “K” BY THE

EXPRESSION:

(139)

ELECTROCHEMISTRY

 EQUILIBRIUM CONSTANT K:_{EQUILIBRIUM CONSTANT K:}

 CHEMICAL EQUILIBRIUM IS THE CHEMICAL EQUILIBRIUM IS THE

STATE IN WHICH THE REACTANTS

AND PRODUCTS HAVE NO NET

CHANGE OVER TIME. FORWARD AND

REVERSE REACTIONS OCCUR AT

EQUAL RATES.

(140)

ELECTROCHEMISTRY

 R IS THE IDEAL GAS CONSTANT 8.314_{R IS THE IDEAL GAS CONSTANT 8.314}

 T 1n K MEANS THE 25 DEGREE T 1n K MEANS THE 25 DEGREE

CELSIUS STANDARD TEMP PLUS 273

YIELDING 298 KELVIN.

(141)

ELECTROCHEMISTRY

 ^G = - RT 1 nK_{^G = - RT 1 nK}

 1 nK = ^G / - RT1 nK = ^G / - RT

(142)

ELECTROCHEMISTRY

 USE THE SRP TO CALCULATE THE _{USE THE SRP TO CALCULATE THE} STANDARD FREE ENERGY CHANGE

STANDARD FREE ENERGY CHANGE

^G AND THE EQUILIBRIUM CONSTANT

K AT 298 K FOR THE FOLLOWING

REACTION:

(143)

ELECTROCHEMISTRY

 4 Ag (s) + O4 Ag (s) + O₂₂ (g) + 4 H (g) + 4 H++ (aq) (aq)  4 Ag 4 Ag++ (aq) (aq) + 2 H

+ 2 H₂₂O (l)O (l)

 1) CALCULATE E BY USING SRP _{1) CALCULATE E BY USING SRP} CHART AND BREAKING THE

CHART AND BREAKING THE

EQUATION INTO HALF REACTIONS.

(144)

ELECTROCHEMISTRY

O

O₂₂ (g) + 4 H (g) + 4 H++ (aq) + 4 e_{(aq) + 4 e}-- _ 2 H_{2 H} 2

2O (reduction)O (reduction)

E

Eoo red

red = + 1.23 V = + 1.23 V

4 Ag (s)

(145)

ELECTROCHEMISTRY

(146)

ELECTROCHEMISTRY

 ^G = - nFE_{^G = - nFE}

 n= 4 4 electrons were transferred_{n= 4 4 electrons were transferred}

 F is faradays constant 96,485F is faradays constant 96,485

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ELECTROCHEMISTRY

 ^G = - (4) (96,485) (+.43 V)_{^G = - (4) (96,485) (+.43 V)}

 ^G = -1.7 X 10^G = -1.7 X 1055 JOULES / MOLE_{JOULES / MOLE}

 OR_OR

(148)

ELECTROCHEMISTRY

 FINALLY, CALCULATE “K” USING:_{FINALLY, CALCULATE “K” USING:}

 ^G = - RT 1n K^G = - RT 1n K

 1n K = ((- 1.7 X (1O_{1n K = ((- 1.7 X (1O}55)) / -((8.314) (298))_{)) / -((8.314) (298))}

(149)

ELECTROCHEMISTRY

 AS A VOLTAIC CELL DISCHARGES, _{AS A VOLTAIC CELL DISCHARGES,} REACTANTS OF THE REACTION ARE

REACTANTS OF THE REACTION ARE

CONSUMED AND PRODUCTS ARE

GENERATED, SO THE

CONCENTRATIONS OF THESE

SUBSTANCES CHANGE.

(150)

ELECTROCHEMISTRY

 THE EMF PROGRESSIVELY DROPS _{THE EMF PROGRESSIVELY DROPS}

UNTIL E= 0, AT WHICH POINT WE CALL

THE CELL “DEAD”.

 AT THIS POINT, THE _{AT THIS POINT, THE}

CONCENTRATIONS OF REACTANTS

AND PRODUCTS CEASE TO CHANGE

(151)

ELECTROCHEMISTRY

 THE EMF GENERATED UNDER _{THE EMF GENERATED UNDER}

“NONSTANDARD” CONDITIONS CAN

BE CALCULATED USING THE

BE CALCULATED USING THE NERNST NERNST

EQUATION.

(152)

ELECTROCHEMISTRY

(153)

ELECTROCHEMISTRY

 E_E00 = THE CELL POTENTIAL_{= THE CELL POTENTIAL}

 RT IS A CONSTANT: 0.0592 VOLTS_{RT IS A CONSTANT: 0.0592 VOLTS}

 nF IS THE NUMBER OF ELECTRONSnF IS THE NUMBER OF ELECTRONS

 1nQ BECOMES LOG Q: _{1nQ BECOMES LOG Q:}

 LOG concen of annode moles / concen of _{LOG concen of annode moles / concen of} cathode moles.

(154)

ELECTROCHEMISTRY

 USE THE FOLLOWING EQUATION:USE THE FOLLOWING EQUATION:

 Zn (s) + CuZn (s) + Cu2+2+ (aq) _(aq)_ Zn_Zn2+2+ (aq) + Cu (s)_{(aq) + Cu (s)}

 WRITE HALF REACTIONSWRITE HALF REACTIONS

(155)

ELECTROCHEMISTRY

 ASSIGN CuASSIGN Cu2+2+ MOLARITY OF 5 M_{MOLARITY OF 5 M}

(156)

ELECTROCHEMISTRY

 Zn (s) _{Zn (s)}_ Zn_Zn2+ 2+ + 2 e_{+ 2 e}-

- OXIDIZED ANODE SEP -.763 _{OXIDIZED ANODE SEP -.763}

 Cu_Cu2+2+ + 2 e_{+ 2 e}-- _ Cu_Cu

(157)

ELECTROCHEMISTRY

 E= (1.10 V) – (0.0592 / 2) log (.05/5)_{E= (1.10 V) – (0.0592 / 2) log (.05/5)}

 E = (1.10 V) – (0.0592 / 2) (-2)E = (1.10 V) – (0.0592 / 2) (-2)

(158)

ELECTROCHEMISTRY

 THE AMOUNT (CONCENTRATION) OF _{THE AMOUNT (CONCENTRATION) OF} THE REACTANT Cu

THE REACTANT Cu2+2+ (CATHODE) _(CATHODE)

INCREASED.

 THE AMOUNT (CONCENTRATION) OF _{THE AMOUNT (CONCENTRATION) OF} THE PRODUCT Zn

THE PRODUCT Zn2+2+ (ANODE) _(ANODE)

DECREASED.

(159)

ELECTROCHEMISTRY

 AN INCREASED CONCENTRATION OF _{AN INCREASED CONCENTRATION OF} REACTANT AND DECREASE

REACTANT AND DECREASE

CONCENTRATION OF PRODUCT

YIELDS AN INCREASED EMF.

 A DECREASED CONCENTRATION OF _{A DECREASED CONCENTRATION OF} REACTANT AND AN INCREASED

REACTANT AND AN INCREASED

CONCENTRATION OF PRODUCT WILL

RESULT IN A DECREASED EMF.