Redox dan Electrochemistry (Kimia)

1

REDOX REACTION AND REDOX REACTION AND ELECTROCHEMISTRYELECTROCHEMISTRY

22Why Study Redox & Why Study Redox & Electrochemistry?Electrochemistry?

Why Study Redox & Why Study Redox & Electrochemistry?Electrochemistry?

• Electroplating of metal

• Batteries

• Corrosion

• Industrial production of chemicals such as Cl2, NaOH, F2 and Al

• Biological redox reactions

• Combustion of Fuel

Figure 2.1Figure 2.1

33The Concept Map

electrochemistryelectrochemistry

voltaic cellvoltaic cell electrolytic cellelectrolytic cellredox reactionredox reaction

Oxidation number Oxidation number method method

Half reaction Half reaction methodmethod

Ba

lan

ce b

yB

ala

nce

by

includesincludes

44

Terminology for Redox Terminology for Redox ReactionsReactions

Terminology for Redox Terminology for Redox ReactionsReactions

• OXIDATION—loss of electrons by a species; increase in oxidation number; gain of oxygen.

• REDUCTION—gain of electrons; decrease in oxidation number; loss of oxygen;

• OXIDIZING AGENT—electron acceptor; species is reduced.

• REDUCING AGENT—electron donor; species is oxidized.

• OXIDATION—loss of electrons by a species; increase in oxidation number; gain of oxygen.

• REDUCTION—gain of electrons; decrease in oxidation number; loss of oxygen;

• OXIDIZING AGENT—electron acceptor; species is reduced.

• REDUCING AGENT—electron donor; species is oxidized.

55

Another way to remember

L

E

O

oss of

lectrons is

xidation

G

E

R

ain of

lectrons is

eduction

LEO the lion says GER!!

GER!GER!

66Balancing Equations of

Redox Reaction

Oxidation Number Method

E.g. : KMnOE.g. : KMnO4 4 + H + H22CC22OO44 + H + H22SOSO44 K K22SOSO44 + MnSO + MnSO44 + CO + CO22 + H + H22OO

StepsSteps

1. 1. Determine whether it is oxidation and reduction reaction based on the Determine whether it is oxidation and reduction reaction based on the elements undergoing oxidation number change elements undergoing oxidation number change

KMnOKMnO4 4 + H + H22CC22OO44 + H + H22SOSO44 K K22SOSO44 + MnSO + MnSO44 + CO + CO22 + H + H22OO+7+7 +2+2

+3+3+4+4

oxidationoxidation

reductionreduction

Molecular reaction

77

22.. Balance the total atom in elements undergoing oxidation and reduction Balance the total atom in elements undergoing oxidation and reduction by giving coefficient of reactionby giving coefficient of reaction

KKMnMnOO4 4 + H + H22CC22OO44 + H + H22SOSO44 K K22SOSO44 + + MnMnSOSO44 + + CCOO22 + H + H22OO22

3. Determine the total change of oxidation number for each oxidation and 3. Determine the total change of oxidation number for each oxidation and reduction reaction and balance the total change of oxidation number by reduction reaction and balance the total change of oxidation number by fixed coefficientfixed coefficient

KKMnMnOO4 4 + H + H22CC22OO44 + H + H22SOSO44 K K22SOSO44 + + MnMnSOSO44 + + 2 2 CCOO22 + H + H22OO

+7+7+2+2

2(+4) = +82(+4) = +82(+3) = +62(+3) = +6 2e2e

5e5e x 2x 2

x 5x 5

KMnOKMnO4 4 + H + H22CC22OO44 + H + H22SOSO44 K K22SOSO44 + MnSO + MnSO44 + + COCO22 + H + H22OO22 55 22 1010

88

4. Balance the other elements from metal, non-metal, H and O4. Balance the other elements from metal, non-metal, H and O

2 K2 KMnMnOO4 4 + 5 H + 5 H22CC22OO44 + H + H22SOSO44 K K22SOSO44 + 2 MnSO + 2 MnSO44 + 10 + 10 COCO22 + + H H22OO33 88

Elements Left Right

KK 22 22

SS 11 33

HH 1616 22

5.5. Redox reaction will be balance if the total of O on left equals with the total Redox reaction will be balance if the total of O on left equals with the total of O on rightof O on right

2 K2 KMnMnOO4 4 + 5 H + 5 H22CC22OO44 + H + H22SOSO44 K K22SOSO44 + 2 MnSO + 2 MnSO44 + 10 + 10 COCO22 + + H H22OO3 3 88

99

Ionic reaction

E.g. : MnOE.g. : MnO44-- + C + C22OO44

2-2- Mn Mn2+2+ + CO + CO22 ( in acidic solution ) ( in acidic solution )

StepsSteps

1. Determine whether it is oxidation and reduction reaction based on the 1. Determine whether it is oxidation and reduction reaction based on the elements undergoing oxidation number change elements undergoing oxidation number change

MnOMnO44-- + C + C22OO44

2-2- Mn Mn2+2+ + CO + CO22+7+7

+3+3+2+2

+4+4

2.2. Balance the total atom in elements undergoing oxidation and reduction Balance the total atom in elements undergoing oxidation and reduction by giving coefficient of reactionby giving coefficient of reaction

MnMnOO44-- + + CC22OO44

2-2- MnMn2+2+ + + CCOO2222

oxidationoxidation

reductionreduction

10103. Determine the total change of oxidation number for each oxidation and 3. Determine the total change of oxidation number for each oxidation and

reduction reaction and balance the total change of oxidation number by reduction reaction and balance the total change of oxidation number by fixed coefficientfixed coefficient

MnMnOO44-- + + CC22OO44

2-2- MnMn2+2+ + + 2 C 2 COO22

+7+7 +2+25e5e

2(+3) = +62(+3) = +6 2(+4) = +82(+4) = +82e2e

x 2x 2

x 5x 5

MnOMnO44-- + C + C22OO44

2-2- Mn Mn2+2+ + CO + CO2222 55 22 1010

4. Balance the charge on the left side and on the right side in the redox 4. Balance the charge on the left side and on the right side in the redox reaction by :reaction by :

Adding HAdding H++ if in acidic solution if in acidic solution

Adding OHAdding OH- - if in alkaline solutionif in alkaline solution

1111

4. Balance H4. Balance H+ + by adding of water (Hby adding of water (H22O)O)

2 MnO2 MnO44-- + 5 C + 5 C22OO44

2-2- + 16 H + 16 H+ + 2 Mn 2 Mn2+2+ + 10 CO + 10 CO22 + 8 H+ 8 H22OO

2 MnO2 MnO44-- + 5 C + 5 C22OO44

2-2- 2 Mn 2 Mn2+2+ + 10 CO + 10 CO22

the total of charge on left = -12the total of charge on left = -12the total of charge on right = +4the total of charge on right = +4

2 2 MnOMnO44-- + 5 C + 5 C22OO44

2-2- 2 Mn 2 Mn2+2+ + 10 CO + 10 CO22+ 16 H+ 16 H++

Cause of Cause of in acidic solutionin acidic solution : :

1212

Practice Exercise

Balance the following redox reaction by oxidation number methodBalance the following redox reaction by oxidation number method

1.1. KMnOKMnO44 + SnF + SnF22 + HF + HF MnF MnF22 + SnF + SnF44 + KF + H + KF + H22OO

2.2. CrCr22OO33 + Na + Na22COCO33 + KNO + KNO33 Na Na22CrOCrO44 + CO + CO22 + KNO + KNO22

3.3. IOIO33-- + I + I- - I I22

+ H+ H22OO (in acidic solution)(in acidic solution)

4.4. ClCl-- + MnO + MnO44-- MnO MnO22 + Cl + Cl2 2 (in alkali solution) (in alkali solution)

1313Balancing Equations of

Redox Reaction

Half Reaction Method(Ion-electron)

E.g. : CrE.g. : Cr22OO772-2- + Cl + Cl- - Cr Cr3+3+ + Cl + Cl22 ( in acidic solution ) ( in acidic solution )

StepsSteps

1. Divide the reaction into half-reactions, one for oxidation and the other for 1. Divide the reaction into half-reactions, one for oxidation and the other for reduction.reduction.

CrCr22OO772-2- Cr Cr3+3+

ClCl-- Cl Cl22

( reduction )( reduction )

( oxidation )( oxidation )

1414

2.2. Balance the total atom in each reaction (except O & H) Balance the total atom in each reaction (except O & H)

CrCr22OO772-2- Cr Cr3+3+

ClCl-- Cl Cl22

22

22

3. If in 3. If in acidicacidic solution, so add solution, so add HH22O on O-deficientO on O-deficient side and add H side and add H++ on other on other

side for H-balance.side for H-balance.

If in If in alkaline alkaline solution, so add solution, so add HH22O on O-over O on O-over side and add OHside and add OH- - on other side on other side

for H-balancefor H-balance

CrCr22OO772-2- 2 Cr 2 Cr3+3+ + 7 H+ 7 H22OO+ 14 H+ 14 H++

2 Cl2 Cl-- Cl Cl22

4. Balance the charge of each reaction by adding electrons4. Balance the charge of each reaction by adding electrons

CrCr22OO772-2- + 14 H + 14 H++ 2 Cr 2 Cr3+ 3+ + 7 H+ 7 H22O O

2 Cl2 Cl-- Cl Cl22

+ 6 e+ 6 e

+ 2 e+ 2 e

1515

Practice Exercise

Balance the following redox reaction by ion-electron methodBalance the following redox reaction by ion-electron method

1.1. II-- + NO + NO22-- I I22 + NO (in acidic solution) + NO (in acidic solution)

2.2. MnOMnO44-- + AsO + AsO33

3-3- MnO MnO22 + AsO + AsO443-3- (in alkaline solution) (in alkaline solution)

3.3. MnOMnO44-- + I + I-- Mn Mn2+2+ + I + I22 (in acidic solution) (in acidic solution)

4.4. BrBr22 + Zn + Zn2+2+ BrO BrO33-- + Zn (in alkaline solution) + Zn (in alkaline solution)

5.5. CrCr22OO772-2- + H + H22SOSO33 Cr Cr3+3+ + HSO + HSO44

-- (in acidic solution) (in acidic solution)

CrCr22OO772-2- + 14 H + 14 H+ + + 6 e + 6 e 2 Cr 2 Cr3+ 3+ + 7 H+ 7 H22O O

2 Cl2 Cl-- Cl Cl22 + 2 e + 2 e

5. Balance the total electrons and add both half reactions5. Balance the total electrons and add both half reactions

x 1x 1

x 3x 3

CrCr22OO772-2- + 14 H + 14 H+ + + 6 Cl+ 6 Cl-- 2 Cr 2 Cr3+ 3+ + 3 Cl+ 3 Cl2 2 + 7 H+ 7 H22O O

1616

Practice Exercise

Balance the following redox reaction by ion-electron or oxidation number Balance the following redox reaction by ion-electron or oxidation number methodmethod

1.1. CrCr22OO772-2- + Fe + Fe2+2+ Cr Cr3+3+ + Fe + Fe3+3+ (in acidic solution) (in acidic solution)

2.2. MnOMnO44-- + SO + SO33

2-2- MnO MnO442-2- + SO + SO44

2-2- (in alkaline solution) (in alkaline solution)

3.3. SnSn2+2+ + IO + IO44-- Sn Sn4+4+ + I + I-- (in acidic solution) (in acidic solution)

4.4. SS22OO332-2- + I + I22 S S44OO66

2-2- + I + I-- (in alkaline solution) (in alkaline solution)

5.5. MnMn2+ 2+ + H+ H22OO22 MnO MnO22 + H + H22O (in alkaline solution)O (in alkaline solution)

6.6. MnOMnO44-- + IO + IO33

-- MnO MnO22 + IO + IO44-- ((in alkaline solution) ((in alkaline solution)

1717

Tips on Balancing Tips on Balancing EquationsEquations

• Never add ONever add O22, O atoms, or O, O atoms, or O2-2- to balance oxygen.to balance oxygen.

• Never add HNever add H22 or H atoms or H atoms to balance hydrogen. to balance hydrogen.

• Be sure to write the correct Be sure to write the correct charges on all the ions.charges on all the ions.

• Check your work at the end Check your work at the end to make sure mass and to make sure mass and charge are balanced.charge are balanced.

• PRACTICE!PRACTICE!

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OXIDATION-REDUCTION OXIDATION-REDUCTION REACTIONSREACTIONS

OXIDATION-REDUCTION OXIDATION-REDUCTION REACTIONSREACTIONS

Direct Redox ReactionDirect Redox Reaction

Oxidizing and reducing agents in direct contact.Oxidizing and reducing agents in direct contact.

CuCu(s)(s) + 2 Ag + 2 Ag++(aq)(aq) ---> Cu ---> Cu22++

(aq)(aq) + 2 Ag + 2 Ag(s)(s) ZnZn(s)(s) + Cu + Cu22++ (aq)(aq) ---> Cu ---> Cu(aq)(aq) + Zn + Zn22++(s)(s)

Figure 2.2Figure 2.2 Figure 2.3Figure 2.3

1919

OXIDATION-REDUCTION REACTIONS

OXIDATION-REDUCTION REACTIONS

Indirect Redox ReactionIndirect Redox Reaction

A battery functions by transferring electrons A battery functions by transferring electrons through an external wire from the reducing through an external wire from the reducing

agent to the oxidizing agent.agent to the oxidizing agent.


2020

ELECTROCHEMISTRYELECTROCHEMISTRYElectrochemistry isElectrochemistry is field of chemistry that studies about relationship field of chemistry that studies about relationship between electric energy and chemical energy or reverse.between electric energy and chemical energy or reverse.

Electrochemical process occurs in an electrochemical cell Electrochemical process occurs in an electrochemical cell

Electrochemical CellElectrochemical Cell

ElectrodeElectrode Electrolyte Electrolyte solutionsolution

AnodeAnode( a place of ( a place of

oxidation occurs)oxidation occurs)

CathodeCathode( a place of ( a place of

reduction occurs)reduction occurs)

AnoxAnox

CaredCared

componentscomponents

2121

Kinds of Electrochemical cell Kinds of Electrochemical cell

Electrochemical Cell

Voltaic Cell Electrolytic Cell

chemical energy electric energy

KPAN { Cathode ( + ) , Anode ( - ) }

spontaneous redox reaction

has salt bridge

electric energy chemical energy

KNAP { Cathode ( - ) , Anode ( + ) }

not spontaneous redox reaction

has no salt bridge

2222

VOLTAIC VOLTAIC (GALVANIC) (GALVANIC)

CELLCELLZn Zn Zn Zn2+2+ + 2 e + 2 e

CuCu2+2+ + 2 e + 2 e Cu Cu

(oxidation)(oxidation)

(reduction)(reduction)

Zn + CuZn + Cu2+2+ Zn Zn2+2+ + Cu + Cu

Electrons move from Zn to Cu directly.Electrons move from Zn to Cu directly.

It is an indication that these redox reaction obtains It is an indication that these redox reaction obtains electric current. But the electric current can’t be electric current. But the electric current can’t be detected (measured).detected (measured).

At the figure 2.2At the figure 2.2

2323

• • Electrons move through external wire.Electrons move through external wire.• The function of The function of salt bridgesalt bridge is to neutralize ions is to neutralize ions

• • Electrons move through external wire.Electrons move through external wire.• The function of The function of salt bridgesalt bridge is to neutralize ions is to neutralize ions


2424

DIAGRAM OF VOLTAIC CELLDIAGRAM OF VOLTAIC CELL

Oxidation Oxidation ║ Reduction ║ Reduction

anodeanode electrolyte electrolyte ║ electrolyte║ electrolyte cathode cathode

Salt bridgeSalt bridge

From figure 2.6, we can write down diagram of voltaic cell :From figure 2.6, we can write down diagram of voltaic cell :

ZnZn Zn Zn2+2+(aq)(aq) ║ Cu║ Cu2+2+

(aq)(aq) Cu Cu

2525

ELECTRODE POTENTIAL ELECTRODE POTENTIAL and CELL POTENTIALand CELL POTENTIAL

• Reduction reaction can cause electric potential which is called reduction potential or electrode potential (E).

• Value of E from reduction reaction cannot be measured because there is no reduction reaction without oxidation reaction

• Electrode potential (E) can be measured by using standard electrode (H2)

• So, the value of E is called E0 (standard reduction potential or standard electrode potential)

• The difference of reduction potential of two reduction reaction in different substance is called cell potential

• At figure 2.5, the potential is +1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 M.

• +1.10 V is difference of potential for Zn/Cu cell or it is called as cell potential

2626

CuSO4

0,34 ve e

CuCu

HCl

HH22

PtPt

Salt bridgeSalt bridge


HH2(g)2(g) 2 H 2 H++ + 2 e (Ox) + 2 e (Ox)

CuCu2+2+ + 2e + 2e Cu (Red) Cu (Red)

Electrode potential of HElectrode potential of H22 = 0,00 v, so +0,34 v at the figure 2.6 is = 0,00 v, so +0,34 v at the figure 2.6 is

standard reduction potential of Custandard reduction potential of Cu

2727

NotesNotes ::

If standard reduction potential for an electrode = + , If standard reduction potential for an electrode = + , so the electrode is easier reduced than electrode of Hso the electrode is easier reduced than electrode of H22

If standard reduction potential for an electrode = If standard reduction potential for an electrode = - - , , so the electrode is more difficult reduced than electrode of Hso the electrode is more difficult reduced than electrode of H22

CuCu2+2+ + 2 e + 2 e Cu Cu EE00 = +0.34 v = +0.34 v

2H2H++ + 2 e + 2 e H H22 EE00 = 0,00 v = 0,00 v

ZnZn2+2+ + 2 e + 2 e Zn Zn EE00 = -0,76 v = -0,76 v

increasing strength of reducing agentincreasing strength of reducing agent

2828

Standard Reduction Potential for Several Electrodes in Aqueous Solution , E0 (volt )

Reduction Half-Reaction E0 (v)

Li+(aq) + e- → Li -3,04

K+(aq) + e → K -2,92

Ba2+(aq) + 2e- → Ba -2,90

Ca2+(aq) + 2e- → Ca -2,87

Na+(aq) + e- → Na -2,71

Mg2+(aq) + 2e- → Mg -2,37

Al3+(aq) + 3e- → Al -1,66

Mn2+(aq) + 2e- → Mn -1,18

2H2O(l)+ 2e- → H2(g)+ 2OH-(aq ) -0,83

Zn2+(aq) + 2e- → Zn -0,76

Cr3+(aq) + 3e- → Cr -0,74

Fe2+(aq) + 2e- → Fe -0,44

Cd2+(aq) + 2e → Cd -0,28

Co2+(aq) + 2e- → Co -0,28

Reduction Half-Reaction E0 (v)

Ni2+(aq) + 2e‑ → Ni -0,23

Sn2+(aq) + 2e- → Sn -0,14

Pb2+(aq) + 2e‑ → Pb -0,13

2H+(aq) + 2e → H2 0,00

Cu2+(aq) + 2e → Cu +0,34

O2(g) + 2H2O(l) + 4e- → 4OH-(aq) +0,40

I2(s) + 2e → 2I-(aq) +0.56

Ag+(aq) + e- → Ag +0,80

Hg2+(aq) + 2e- → Hg +0,85

Br2(l) + 2e- → 2Br-(aq) +1,06

O2(g) + 4H+(aq)

+ 4e- → 2H2O(l) +1,23

Cl2(g) + 2e- → 2Cl-(aq) +1,36

Au3+(aq) + 3e- → Au +1,52

F2(g) + 2e- → 2F- +2,87

2929

3030

From value of standard reduction potential for several electrodes in From value of standard reduction potential for several electrodes in the table above, we can write down reactivity series of metal OR the table above, we can write down reactivity series of metal OR voltaic series (deret volta) voltaic series (deret volta)

Li K Ba Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H Cu Hg Ag Pt AuLi K Ba Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H Cu Hg Ag Pt Au

0,00 v0,00 v

EE00 <<<<

Strong reducing agentStrong reducing agent

Weak oxidizing agentWeak oxidizing agent

Be easiest oxidizedBe easiest oxidized

EE00 Weak reducing agentWeak reducing agent

Strong oxidizing agentStrong oxidizing agent

Be easiest reducedBe easiest reduced

3131

Cell Potential (ECell Potential (E00cellcell) is ) is difference of electrode potentialdifference of electrode potential

EE00cellcell = E = E00

highhigh - E - E00lowlow


cathodecathode - E - E00anodeanode


redred - E - E00oxox

3232ExampleExample

Given 2 electrodes :Given 2 electrodes :

MgMg2+2+(aq)(aq) + 2e + 2e Mg Mg(s)(s) EE00 = = -2,37 v-2,37 v

ZnZn2+2+(aq)(aq) + 2e + 2e Zn Zns)s) EE00 = = -0,76 v-0,76 v

Calculate ECalculate E00cellcell of both electrodes ! of both electrodes !

SolutionSolution

oxidationoxidation

reductionreduction

/ anode/ anode

/ cathode/ cathode

EE00cellcell = E= E00

highhigh - E - E00lowlow

= -0,76 v – (-237 v)= -0,76 v – (-237 v)

= +1,61 v= +1,61 v

Step 1Step 1 Step 2Step 2

Cell potential can be calculated by :Cell potential can be calculated by :

CathodeCathode :: ZnZn2+2+ + 2e + 2e Zn Zn EE00 = -0,76 V = -0,76 V

AnodeAnode :: Mg Mg Mg Mg2+2+ + 2e + 2e EE00 = 2,37 V = 2,37 V

Mg + ZnMg + Zn2+2+ Mg Mg2+2+ + Zn + Zn EE00 = +1,61 V = +1,61 V

3333

Redox reaction will be spontaneous, when

EE00cellcell = + = + A metal is left other metal ionA metal is left other metal ion

in voltaic seriesin voltaic series

Li K Ba Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H Cu Hg Ag Pt AuLi K Ba Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H Cu Hg Ag Pt Au

MnMn2+2+ FeFe++

KK++ ++ NaNa

MnMn FeFe2+2+++

3434

ExampleExample

Predict whether the following reactions will be spontaneous or not !Predict whether the following reactions will be spontaneous or not !

a.a. ZnZn(s)(s) + Mg + Mg2+2+(aq)(aq) Zn Zn2+2+

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

b.b. MgMg(s)(s) + Ag + Ag++(aq) (aq) Mg Mg2+2+

(aq)(aq) + Ag + Ag(s)(s)

c.c. ZnZn2+2+(aq)(aq) + Ag + Ag(s)(s) Zn Zn(s)(s) + Ag + Ag++

(aq)(aq)

SolutionSolution

a.a. Zn is oxidized and Mg is reducedZn is oxidized and Mg is reduced


redred – E – E00oxox = -2,37 – (-0,76) = = -2,37 – (-0,76) = - 1,61- 1,61

EE00 Zn = -0,76 vZn = -0,76 v

EE00 Mg = -2,37 v Mg = -2,37 v

EE00 Ag = +0,80 v Ag = +0,80 v

not spontaneousnot spontaneous

b. Mg is oxidized and Ag is reduced b. Mg is oxidized and Ag is reduced


red red – E– E00oxox = +0,80 – (-2,37) = = +0,80 – (-2,37) = + 3,17 v+ 3,17 v spontaneousspontaneous

c. Ag is oxidized and Zn is reduced c. Ag is oxidized and Zn is reduced


red red – E– E00oxox = -0,76 – (+0,80) = = -0,76 – (+0,80) = - 1,56 v- 1,56 v not spontaneousnot spontaneous

3535

Other SolutionOther Solution

Rearrange the metals based on ERearrange the metals based on E00 Mg Zn AgMg Zn Ag

From the arrangement can be concluded :From the arrangement can be concluded :

1.1. Mg + ZnMg + Zn2+2+ Mg Mg2+2+ + Zn + Zn

2.2. Zn + MgZn + Mg2+2+

3.3. AgAg++ + Mg + Mg Ag + Mg Ag + Mg2+2+

4.4. ZnZn2+ 2+ + Ag + Ag

Only reaction of B can occursOnly reaction of B can occurs

B'wina 36

Practice Exercise

1.1. Given :Given :

NiNi2+2+ + 2e + 2e Ni Ni EE00 = = -0,23 v-0,23 vAgAg++ + e + e Ag Ag EE00 = = +0,80 v+0,80 v

From the electrode potential above :From the electrode potential above :a.a. Draw the voltaic cell !Draw the voltaic cell !b.b. Write down the reduction reaction !Write down the reduction reaction !c.c. Write down the oxidation reaction !Write down the oxidation reaction !d.d. Write down the redox reaction !Write down the redox reaction !e.e. Write down the cell diagram !Write down the cell diagram !f.f. Calculate ECalculate E00

cell cell !!

2.2. Predict whether the following reaction will be spontaneous or not ! Predict whether the following reaction will be spontaneous or not ! a.a. SnSn2+2+ + Mn + Mn Sn + Mn Sn + Mn2+2+ d.d. Fe + CuFe + Cu2+2+ Fe Fe2+2+ + Cu + Cub.b. NiNi2+2+ + 2Ag + 2Ag Ni + 2Ag Ni + 2Ag++ e.e. Al + 3PbAl + 3Pb2+2+ 2Al 2Al3+3+ + Pb + Pb

3737

3. Below is list standard reduction potential of some elements3. Below is list standard reduction potential of some elements

AgAg++ + e + e Ag Ag EE00 = +0,80 v = +0,80 v

SnSn2+2+ + 2e + 2e Sn Sn EE00 = = -0,14 v-0,14 v

CuCu2+2+ + 2e + 2e Cu Cu EE00 = = +0,34 v+0,34 v

CrCr3+3+ + 3e + 3e Cr Cr EE00 = = -0,74 v-0,74 v

FeFe2+2+ + 2e + 2e Fe Fe EE00 = = -0,44 v-0,44 v

a.a. Choose pair of 2 half-cells which has the highest EChoose pair of 2 half-cells which has the highest E00cell cell

and the lowest Eand the lowest E00cellcell

b.b. Write down cell diagram of the reactionWrite down cell diagram of the reaction

3838

The Kinds & Applications ofThe Kinds & Applications of Voltaic Cell Voltaic Cell

Kinds ofKinds of Voltaic CellVoltaic Cell

Primary Primary Voltaic CellVoltaic Cell

SecondarySecondaryVoltaic CellVoltaic Cell Fuel CellFuel Cell

UnrenewableUnrenewableIrreversibleIrreversible

RenewableRenewableReversibleReversible

Unrenewable but Unrenewable but can’t be exhausedcan’t be exhaused

Dry batteryDry batteryAlkaline batteryAlkaline battery

Mercury oxide batteryMercury oxide battery

AccumulatorAccumulatorNi-Cd batteryNi-Cd battery

Lithium ion batteryLithium ion battery

Cell containing of Cell containing of The mixture of The mixture of

HH22 & O & O22

3939

Dry Cell BatteryDry Cell Battery

A (-) : ZnA (-) : Zn(s)(s) Zn Zn2+2+(aq)(aq) + 2e- + 2e-

C (+) : 2 MnOC (+) : 2 MnO2(s) 2(s) + 2NH+ 2NH44++

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

2Mn2Mn22OO3 (s) 3 (s) + 2 NH+ 2 NH3(aq)3(aq) + H + H22O O (l)(l)


Dry cell consists of zinc cylinder filled with paste from MnODry cell consists of zinc cylinder filled with paste from MnO22 mixture, mixture,

NHNH44Cl, carbon and some waterCl, carbon and some water

Overall Reaction :Overall Reaction :

4040

Alkaline BatteryAlkaline Battery

Nearly same reactions as in Nearly same reactions as in common dry cell, but under common dry cell, but under basic conditions (e.g. KOH)basic conditions (e.g. KOH)


Anode Anode (Zn)(Zn)

Cathode Cathode (MnO(MnO22))


4141

Mercury BatteryMercury Battery


4242


AnodeAnode

CathodeCathode


4343

AccumulatorAccumulator

Accumulator consists of :Accumulator consists of :• Pb (anode)Pb (anode)• PbOPbO22 (cathode) (cathode)• HH22SOSO44 ( electrolyte solution ) ( electrolyte solution )

The reaction in accumulator at time produce electricity :The reaction in accumulator at time produce electricity :

A (-) : PbA (-) : Pb(s)(s) + SO + SO442-2-

(aq)(aq) PbSO PbSO4(s)4(s) + 2e + 2e- -

C (+) : PbOC (+) : PbO2(s)2(s) + 4H + 4H++(aq)(aq) + SO + SO44

2-2-(aq)(aq) + 2e + 2e- - PbSOPbSO4(s)4(s) + 2 H + 2 H22OO(l)(l)


Cathode Cathode (PbO(PbO22))Anode Anode

(Pb)(Pb)

HH22SOSO44

UsageUsageRechargeRecharge

4444

Accumulator can be Accumulator can be recharged recharged by reversing the electron flow if by reversing the electron flow if the potential exhausted. Electrons are flowed from cathode to the potential exhausted. Electrons are flowed from cathode to anode. anode. PbPb electrode electrode is connected to the negative terminalis connected to the negative terminal of the of the current sourcecurrent source

A ( + / PbOA ( + / PbO22 ) : PbSO ) : PbSO4(s)4(s) + 2 H + 2 H22OO(l) (l) PbOPbO2(s)2(s) + 4H + 4H++(aq)(aq) + SO + SO44

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

C ( - / Pb) : PbSOC ( - / Pb) : PbSO4(s)4(s) + 2e + 2e- - PbPb(s)(s) + SO + SO442-2-

(aq)(aq)


4545

Ni-Cad BatteryNi-Cad Battery

Anode : CdAnode : Cd(s)(s) + 2 OH + 2 OH-- (aq)(aq) ---> Cd(OH) ---> Cd(OH)2(s)2(s) + 2e- + 2e-

Cathode : NiOCathode : NiO2(s)2(s) + 2 H + 2 H22OO(l)(l) + 2e- ---> Ni(OH) + 2e- ---> Ni(OH)2(s)2(s) + 2 OH + 2 OH--(aq)(aq)


Ni-Cd battery consists of :Ni-Cd battery consists of :• Cadmium, Cd (anode)Cadmium, Cd (anode)• Nickel oxide solid (NiONickel oxide solid (NiO22) (cathode)) (cathode)• Electrolyte solution (KOH)Electrolyte solution (KOH)


4646

Fuel CellFuel Cell

Cars can use electricity generated by H2/O2 fuel cells.H2 carried in tanks or generated from hydrocarbons


4747• Fuel cell consists of a mixture of hydrogen and oxygen, electrode

with porous carbon containing platinum metal and alkali solution (KOH) which is between the two electrodes

• O2 gas is flowed to cathode through electrode

• H2 gas is flowed to anode

A : 2HA : 2H2(g)2(g) + 4OH + 4OH--(aq) (aq) 4H 4H22OO(l)(l) + 4e + 4e--

C : OC : O2(g)2(g) + 2H + 2H22OO(l)(l) + 4e + 4e-- 4OH 4OH--

(aq)(aq)

2H2H2(g)2(g) + O + O2(g)2(g) 2H 2H22OO(l)(l)

4848

CorrosionCorrosion

Corrosion is electrochemical reaction between a Corrosion is electrochemical reaction between a metal and its environmentmetal and its environment

The corrosion of iron, Fe is called rusting. The corrosion of iron, Fe is called rusting.

The process of rustingThe process of rusting

• Iron, Fe will rust when both Iron, Fe will rust when both oxygen, Ooxygen, O22

and water, Hand water, H22OO are present. are present.

• Other factors cause rusting of iron are ; Other factors cause rusting of iron are ; acid, electrolyte solution and carbon.acid, electrolyte solution and carbon.

• The rusting of iron, Fe is redox reaction. The rusting of iron, Fe is redox reaction. Oxidation and reduction occur at separate Oxidation and reduction occur at separate parts of metal surface.parts of metal surface.

4949The process of rusting involved the following steps :The process of rusting involved the following steps :

AnodeAnode

Fe Fe Fe Fe2+ 2+ + 2e+ 2e

CathodeCathode

OO22 + 2H + 2H22O + 4e O + 4e 4 OH 4 OH--

waterwater

OO22

Rust Rust (Fe(Fe22OO33.3H.3H22O)O)

OO22

ee

FeFe2+2+

FeFe3+3+

Overall Reaction : 2Fe + 2HOverall Reaction : 2Fe + 2H22OO + O+ O2 2 2 Fe 2 Fe2+2+ + 4 OH+ 4 OH--

2 Fe(OH)2 Fe(OH)22

FeFe2+2+ will be oxidized by oxygen will be oxidized by oxygen

4 Fe(OH)4 Fe(OH)2(s)2(s) + O + O2(g)2(g) + 2 H + 2 H22OO(l)(l) 4 Fe(OH) 4 Fe(OH)3(s)3(s)

FeFe22OO33.3H.3H22OO(s)(s)

5050

Preventing rusting of ironPreventing rusting of iron

1.1. Iron, Fe are Iron, Fe are coated on its surface with on its surface with paint, oil and plastic, oil and plastic

2.2. Coating Coating iron, Fe with iron, Fe with other metalother metal through electroplating through electroplating

Iron, Fe is electroplated by metals which has smaller EIron, Fe is electroplated by metals which has smaller E00 than than EE00 of Fe, such as Cr, Zn, Al and Mg of Fe, such as Cr, Zn, Al and Mg

HH22O + OO + O22

FeFe

Layer of ZnLayer of Zn

layer of Zn layer of Zn which flawwhich flaw

Figure 2.17Figure 2.17 Coating Fe with Zn Coating Fe with Zn

Anode Zn (-) : 2 ZnAnode Zn (-) : 2 Zn(s)(s) 2 Zn 2 Zn2+2+(aq)(aq) + 4e + 4e

Cathode C (+) : 2HCathode C (+) : 2H22OO(l)(l) + O + O2(g)2(g) + 4e + 4e 4OH 4OH--(aq)(aq)

Redox : 2ZnRedox : 2Zn(s)(s) + 2H + 2H22OO(l)(l) + O + O2(g) 2(g) 2Zn(OH)2Zn(OH)2(s)2(s)

5151

3.3. Make alloyMake alloy, such as stainless steel (mixture of Fe, Cr & Ni). , such as stainless steel (mixture of Fe, Cr & Ni). Contents of Cr in the alloys is 12 % - 18 %. Cr will be oxidized to Contents of Cr in the alloys is 12 % - 18 %. Cr will be oxidized to form strong and transparency layer is layer of Crform strong and transparency layer is layer of Cr22OO33..

4.4. Give protection of cathodeGive protection of cathode..

This way is used to cover pipelines in earth from corrosion, This way is used to cover pipelines in earth from corrosion, where metal which more reactive (e.g. Mg) than iron, Fe (based where metal which more reactive (e.g. Mg) than iron, Fe (based on reactivity series of metal or volta series) is planted in earth on reactivity series of metal or volta series) is planted in earth and is connected with Fe.and is connected with Fe.

Iron, Fe acts as cathode and magnesium, Mg acts as anodeIron, Fe acts as cathode and magnesium, Mg acts as anode

5252

ELECTROLYTIELECTROLYTIC CELLC CELL

Inert Inert electrodeelectrode

Inert Inert electrodeelectrode


5353

From figure 2.18, can be explained the process of electrolysisFrom figure 2.18, can be explained the process of electrolysis

1.1. Electron moves from pole (-) of an electric current source Electron moves from pole (-) of an electric current source (battery) to cathode. (battery) to cathode.

2.2. The electron from the cathode moves to solution. Ion (+), NaThe electron from the cathode moves to solution. Ion (+), Na++ is is reduced to be neutral atom (Na solid) at the cathode . Ion (-), reduced to be neutral atom (Na solid) at the cathode . Ion (-), ClCl-- will be pulled to anode and will be oxidized to be neutral atom will be pulled to anode and will be oxidized to be neutral atom (Cl(Cl2 2 gas)gas)

3.3. Electron which produced on the anode moves to electric current Electron which produced on the anode moves to electric current source source

Electrolysis of the molten NaCl can be written :Electrolysis of the molten NaCl can be written :

NaClNaCl(l)(l) Na Na++(l)(l) + Cl + Cl--(l)(l)

Cathode :Cathode : NaNa+ + + e + e Na Na

Anode :Anode : 2 Cl2 Cl-- Cl Cl22 + 2e + 2e ++

x 2x 2

x 2x 2

Reaction : 2 NaClReaction : 2 NaCl Na + Cl Na + Cl22

5454ELECTROLYSISReaction on cathode(Cation is reduced)

Reaction on anode ( Anion is oxidized )

1. Ions of IA, IIA, Al+3, Mn2+

( aqueous )

1. Ions from remain of acid which contain oxygen (SO4

2-, NO3- etc)

2. Other metal ions 2. Halide ions (F-, Cl-, Br-, I-)

3. H+ ion ( acid )

3. OH- ion ( base) and O2- ion

4. All metal ions ( molten ) • 1 , 2 & 3 occurs when use inert electrode (Pt, C, Au)

• If use non-inert electrode, so the anode will undergo oxidation

2H2O + 2e 2OH- + H2

Mn+ + ne M

2H+ + 2e H2

2 H2O 4H+ +O2 + 4e

2 X- X2 + 2e

4 OH- 2 H2O + O2 + 4e

M Mn+ + ne

Mn+ + ne M

2 O2- O2 + 4 e

Reaction on cathode(Cation is reduced)

Reaction on anode ( Anion is oxidized )

1. Ions of IA, IIA, Al+3, Mn2+

( aqueous )

1. Ions from remain of acid which contain oxygen (SO4

2-, NO3- etc)

2. Other metal ions 2. Halide ions (F-, Cl-, Br-, I-)

3. H+ ion ( acid )

3. OH- ion ( base) and O2- ion

4. All metal ions ( molten ) • 1 , 2 & 3 occurs when use inert electrode (Pt, C, Au)

• If use non-inert electrode, so the anode will undergo oxidation

2H2O + 2e 2OH- + H2

Mn+ + ne M

2H+ + 2e H2

2 H2O 4H+ +O2 + 4e

2 X- X2 + 2e

M Mn+ + ne

Mn+ + ne M

2 O2- 2 + 4 e

5555

Examples

1. Electrolysis of KI solution by carbon electrode

KI (aq) → K+ (aq) + I- (aq)

Cathode : 2H2O + 2e → 2OH- + H2

Anode : 2I- → I2 + 2e

Reaction : 2KI + 2H2O → 2 K+ + 2 OH- + H2 + I2

2KI + 2H2O → 2 KOH + H2 + I2

2. Electrolysis of molten MgCl2 by carbon electrode

MgCl2(l) → Mg2+(l) + 2 Cl-(l)

Cathode : Mg2+ + 2e → Mg

Anode : 2 Cl- → Cl2 + 2e

Reaction : MgCl2 → Mg + Cl2

++

x 2x 2

++

5656

3. Electrolysis of AgNO3 solution by carbon electrode

AgNO3 (aq) → Ag+ (aq) + NO3- (aq)

Cathode : Ag+ + e → Ag

Anode : 2H2O → 4H+ + O2 + 4e

Reaction : 4AgNO3 + 2H2O → 4Ag + 4H+ + 4NO3- + O2

4. Electrolysis of NaOH solution by iron electrode

NaOH(aq) → Na+(aq) + OH-

(aq)

Cathode : 2H2O + 2e → 2OH- + H2 Anode : Fe → Fe3+ + 3e

Reaction : 6 H2O + 2 Fe → 6 OH- + 2 Fe3+ + 3 H2

++

x 4x 4

++

x 4x 4

x 3x 3x 2x 2

5757

5. Electrolysis of molten MgSO4 by carbon electrode

MgSO4(l) → Mg2+(l) + SO4

2-(l)

Cathode : Mg2+ + 2e → Mg

Anode : SO42- → SO2 + O2 + 2e

Reaction : MgSO4 → Mg + SO2 + O2

++

6. Electrolysis of CaBr2 solution by iron cathode and carbon anode

CaBr2(aq) → Ca2+(aq) + 2 Br-

(aq)

Cathode : 2 H2O + 2 e → 2 OH- + H2

Anode : 2 Br- → Br2 + 2e

Reaction : CaBr2 + 2 H2O → Ca2+ + 2 OH- + H2 + Br2

++

Point to note : Electrolysis of a molten, the cation will be reduced and the anion will be oxidized.

B'wina 58

Practice Exercise

Write down complete reaction of :

• NaCl solution by copper electrode

• Molten AlCl3 by carbon electrode

• CdSO4 solution by iron electrode

• Ba(NO3)2 solution by platinum electrode

• AgNO3 solution by electrode

5959FARADAY’S LAWFARADAY’S LAW

Faraday IFaraday I

““The quantity of substance produced on electrode is equal to The quantity of substance produced on electrode is equal to the quantity of electricity passing through the electrolysis”the quantity of electricity passing through the electrolysis”

m = e Fm = e F

Fxno

Arm

.

1 Faraday = 1 mole of electron1 Faraday = 1 mole of electron

9650096500

tiQF

WhereWhere : :

mm == mass of substance which produced for electrolysis (gram)mass of substance which produced for electrolysis (gram)o.no.n == oxidation numberoxidation numberFF == the total electricity in Faradaythe total electricity in FaradayQQ == the total electricity in Coulombthe total electricity in Coulombii == the total electricity in Amperethe total electricity in Amperett == time of electrolysis (second)time of electrolysis (second)

6060

Faraday IIFaraday II

““If the same quantities of electricity are passed trough If the same quantities of electricity are passed trough several electrolytic cells, substances produced by each cell several electrolytic cells, substances produced by each cell equals to the equivalent weight”equals to the equivalent weight”

nFFF 21

n

n

e

m

e

m

e

m...........

2

2

1

1

(1)(1) = 1= 1stst cell cell

(2)(2) = 2= 2ndnd cell cell

6161

ExamplesExamples

1.1. A molten KCl is electrolyzed using a current of 193 A for 1 minutes. (Ar A molten KCl is electrolyzed using a current of 193 A for 1 minutes. (Ar K = 39 ; Cl = 35,5)K = 39 ; Cl = 35,5)

a. Write down complete reaction!a. Write down complete reaction!b.b. Calculate mass of substance plated on the cathode ! Calculate mass of substance plated on the cathode ! c.c. Calculate volume of gas which formed on the anode at STP !Calculate volume of gas which formed on the anode at STP !

SolutionSolutionKClKCl(l)(l) K K++

(l)(l) + Cl + Cl--(l)(l)

Cathode :Cathode : KK+ + + e + e K K

Anode :Anode : 2 Cl2 Cl-- Cl Cl22 + 2e + 2e ++

x 2x 2

x 2x 2

a.a. Reaction : 2 KClReaction : 2 KCl 2 K + Cl 2 K + Cl22

b.b.

Fxti

F 12,096500

60193

96500

0,12 F = 0,12 moles of electron = 0,12 moles of K0,12 F = 0,12 moles of electron = 0,12 moles of K

11stst way way

6262m of K = mole x Ar = 0,12 x 39 = 4,68 gm of K = mole x Ar = 0,12 x 39 = 4,68 g

22ndnd way way

gx

xti

xno

Arm 68,4

96500

60193

1

39

96500.

c.c. Based on reaction 0,12 F = 0,12 moles of electron Based on reaction 0,12 F = 0,12 moles of electron

1 mole of Cl1 mole of Cl22 = 2 moles of electron = 2 moles of electron

mole of Clmole of Cl22 = ½ x mole of electron = ½ x 0,12 = 0,06 moles = ½ x mole of electron = ½ x 0,12 = 0,06 moles

V ClV Cl22 (STP) = n x 22,4 L = 0,06 x 22,4 = 1,344 L (STP) = n x 22,4 L = 0,06 x 22,4 = 1,344 L

2.2. An electrolysis of AgNOAn electrolysis of AgNO33 solution produced 54 g of Ag. Calculate solution produced 54 g of Ag. Calculate

mass of Cu which produced when the same electric current is passed mass of Cu which produced when the same electric current is passed through CuSOthrough CuSO44 solution ! (Ar Ag = 108 ; Cu = 63,5) solution ! (Ar Ag = 108 ; Cu = 63,5)

SolutionSolution

2/5.631/108

54 Cu

Cu

Cu

Ag

Ag m

e

m

e

m m Cu = 15.9 gm Cu = 15.9 g

6363

Practice Exercise

1. How many minutes will it take an electric current of 5 A to precipitate all the copper from 500 mL of 0,250 M CuSO4(aq) ? (Ar Cu = 63,5)

2. A current of 0,5 F is passed through 250 L of NaI solution with inert electrodes. When volume of the solution is assumed constant, a. write down complete reaction!b. calculate the volume of gas which produced on cathode at STP?c. what is pH of the solution after electrolysis?

3. Electrolysis cells with inert electrodes were arranged in series. The cells contained solution of CuSO4, AgNO3 and AlCl3. 21,6 g of silver in cell 2 was formed for electrolysis. Calculatea. How many gram of copper was plated out in cell 1?b. What is the volume of Cl2 was produced in cell 3 at RTP?

4. An electrolysis of MSO4 solution produced 0,28 g of M on the cathode. The product of electrolysis can be neutralized by 50 mL of 0,2 M NaOH solution. What is the relative atomic mass of M ?

6464Electrolysis in Electrolysis in IndustriesIndustriesThere are many industrial applications of electrolysis. The most common There are many industrial applications of electrolysis. The most common

application include :application include :

1.1. Extraction of metalsExtraction of metals

Reactive metals such as aluminium and magnesium can be extracted Reactive metals such as aluminium and magnesium can be extracted from their ores by electrolysis. In this process, molten compound, from their ores by electrolysis. In this process, molten compound, concentrated solutions of salts or hydroxide solutions are electrolyzed.concentrated solutions of salts or hydroxide solutions are electrolyzed.

2.2. Purification of metalsPurification of metals

Pure copper and silver can be obtained through the process Pure copper and silver can be obtained through the process electrolysis.electrolysis.

6565

3.3. Electroplating of metalsElectroplating of metals

In electroplating of metals, electrolysis is used to coat one metal onto In electroplating of metals, electrolysis is used to coat one metal onto another metal. another metal.

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Redox dan Electrochemistry (Kimia)