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Oxidation-Reduction Reactions

General Chemistry II 2

Setting the Stage

The flow of electrons is commonThe flow of electrons is common It occurs naturally (lightning) and as a result It occurs naturally (lightning) and as a result

of human activity (electricity)of human activity (electricity) Early electrical experiments involved Early electrical experiments involved

chemically generated electricity (batteries)chemically generated electricity (batteries) This electron flow results from one reactant This electron flow results from one reactant

having a greater affinity for electrons than having a greater affinity for electrons than the otherthe other


Formulating some questions

What is involved in the electron exchange What is involved in the electron exchange process and what are the terms used?process and what are the terms used?

How do we keep track of electrons?How do we keep track of electrons? Can we predict favorable and unfavorable Can we predict favorable and unfavorable

electron exchange reactions?electron exchange reactions? How can we make unfavorable reactions How can we make unfavorable reactions

occur?occur?


Nature of Oxidation and Reduction

Na reacts with ClNa reacts with Cl22 quite vigorously, such that the piece of quite vigorously, such that the piece of

Na glows white hot with the heat of the reactionNa glows white hot with the heat of the reaction The process forms ordinary table saltThe process forms ordinary table salt

2 Na2 Na(s)(s) + Cl + Cl2 (g)2 (g) 2 NaCl 2 NaCl(s)(s)

Na + Cl Na+ + Cl-


Half reactions

Electron exchange reactions can be viewed Electron exchange reactions can be viewed as the sum of two as the sum of two half reactionshalf reactions

Half reactions represent either the loss of Half reactions represent either the loss of electrons or the gain of electrons as a electrons or the gain of electrons as a separate balanced equationseparate balanced equationSodium half reactionSodium half reactionNa Na Na Na++ + e + e--

Na loses an electron to form a sodium ionNa loses an electron to form a sodium ion


Half reactions

A substance that loses electrons is said to be A substance that loses electrons is said to be oxidizedoxidized

Chlorine half reactionChlorine half reaction2 e2 e-- + Cl + Cl22 2 Cl 2 Cl--

The neutral chlorine molecule has gained The neutral chlorine molecule has gained two electrons to form chloride ionstwo electrons to form chloride ions

A substance that gains electrons is said to A substance that gains electrons is said to be be reducedreduced


Terminology

Redox or oxidation-reduction reactions are Redox or oxidation-reduction reactions are reactions involving the exchange of reactions involving the exchange of electronselectrons

Oxidizing agent – the species that accepts Oxidizing agent – the species that accepts electronselectrons

Reducing agent – the species that donates Reducing agent – the species that donates electronselectrons


Redox reactions

Electrons are conserved in redox reactionsElectrons are conserved in redox reactions e. g. electrons gained in the reduction e. g. electrons gained in the reduction

process must equal the electrons lost in the process must equal the electrons lost in the oxidation processoxidation process

A common, and unfortunate, redox reaction A common, and unfortunate, redox reaction is the rusting of ironis the rusting of iron4 Fe4 Fe(s)(s) + 3 O + 3 O2 (g)2 (g) 2 Fe 2 Fe22OO3 (s)3 (s)


Oxidation states or numbers

The charge that an atom in a molecule or The charge that an atom in a molecule or ion would have if all atoms were present as ion would have if all atoms were present as monatomic ionsmonatomic ions

Similar to formal charge, but in this case, Similar to formal charge, but in this case, both electrons in the bond are assigned to both electrons in the bond are assigned to the more electronegative atom sharing the the more electronegative atom sharing the electronselectrons


Rules for assigning OS

1.1. OS of an element in its free, natural state is OS of an element in its free, natural state is zero (Cuzero (Cu(s)(s), Cl, Cl2 (g)2 (g)))

2.2. The OS of a monoatomic ion is the same as The OS of a monoatomic ion is the same as the charge of the ionthe charge of the ionalkali metalsalkali metals +1+1 same as group #same as group #alkaline earthsalkaline earths +2+2 same as group #same as group #oxygen oxygen OO2-2-

aluminumaluminum AlAl3+3+


Oxidation state

3.3. Halogens are in a -1 oxidation state in Halogens are in a -1 oxidation state in binary compounds whether ionic or binary compounds whether ionic or covalent when bound to a less covalent when bound to a less electronegative elementelectronegative element

4.4. Oxygen is OOxygen is O2-2- except in peroxides and except in peroxides and superoxides. Oxygen is positive when superoxides. Oxygen is positive when bound to Fbound to F


Oxidation States

5.5. H is usually +1. When bound to a less H is usually +1. When bound to a less electronegative atom (usually a metal) it is electronegative atom (usually a metal) it is -1 (as in LiH)-1 (as in LiH)

6.6. The sum of the OS of all of the atoms in a The sum of the OS of all of the atoms in a compound must equal the charge of the compound must equal the charge of the compound or ioncompound or ione. g. FeO has Fee. g. FeO has Fe2+2+ and O and O2-2-


Oxidation state changes

Since oxidation is the loss of electrons, and Since oxidation is the loss of electrons, and therefore results in an increase in the therefore results in an increase in the oxidation state. Reduction results in the oxidation state. Reduction results in the decrease in the oxidation state.decrease in the oxidation state.

Generally, only one element changes Generally, only one element changes oxidation state in a compound, but this is oxidation state in a compound, but this is not always truenot always true


Balancing redox reactions

The oxidation state method focuses on the The oxidation state method focuses on the atoms of the elements undergoing a atoms of the elements undergoing a change in oxidation statechange in oxidation state

ConsiderConsiderHNOHNO3 (aq)3 (aq) + H + H22SS(aq)(aq) NO NO(g)(g) + S + S(s)(s) + H + H22OO(l)(l)


Oxidation state method

1.1. Identify the atoms Identify the atoms whose OS have whose OS have changedchanged

2.2. Draw a bridge Draw a bridge between the same between the same atoms whose OS has atoms whose OS has changed, indicating changed, indicating the electrons gained the electrons gained or lostor lost

HNO3 + H2S NO + S + H2O+5 -2 +2 0

HNO3 + H2S NO + S + H2O+5 -2 +2 0

+3e-

-2e-



3.3. Multiply the two numbers of electrons (in Multiply the two numbers of electrons (in this case +3 and -2) by whole numbers this case +3 and -2) by whole numbers that produce a common numberthat produce a common number3 3 × 2 = 6× 2 = 6 2 2 × 3 = 6× 3 = 6use these multipliers as coefficients of the use these multipliers as coefficients of the respective compounds or elements respective compounds or elements



4.4. Balance the rest of the elements by inspectionBalance the rest of the elements by inspection2 HNO2 HNO33 + 3 H + 3 H22S S 2 NO + 3 S + 4 H 2 NO + 3 S + 4 H22OO

HNO3 + H2S NO + S + H2O+5 -2 +2 0

+3e- × 2 = 6

-2e- × 3 = 6


Balancing Redox Reactions

Ion-electron or Half reaction methodIon-electron or Half reaction method

1.1. Separate the total reaction into half Separate the total reaction into half reactionsreactions

2.2. Balance the half reactions separatelyBalance the half reactions separately

3.3. Methodology depends on whether the Methodology depends on whether the reactions are done in acidic or basic reactions are done in acidic or basic conditionsconditions


Half reaction method

ConsiderConsiderHH++

(aq)(aq) + Cl + Cl--(aq)(aq) + Cr + Cr22OO77

2-2-(aq)(aq)

Cr Cr3+3+ + Cl + Cl2(aq)2(aq) + H + H22OO(aq)(aq)

1.1. Separate the molecule or ion that contains Separate the molecule or ion that contains atoms of an element that has changed atoms of an element that has changed oxidation state and product containing the oxidation state and product containing the atoms of that element. (you don’t need to atoms of that element. (you don’t need to know the actual oxidation state)know the actual oxidation state)



1.1. Con’t.Con’t.CrCr22OO77

2-2- Cr Cr3+3+

2.2. Balance the atoms other than hydrogen or Balance the atoms other than hydrogen or oxygenoxygenCrCr22OO77

2-2- 2Cr 2Cr3+3+

3.3. Balance the oxygen by adding HBalance the oxygen by adding H22O on the O on the

side missing the oxygenside missing the oxygen



3.3. Con’tCon’t Cr Cr22OO77

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

4.4. Balance the hydrogens by adding HBalance the hydrogens by adding H++ on on the other side of the equation from the the other side of the equation from the HH22OO14 H14 H++ + Cr + Cr22OO77

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

5.5. The atoms in the half reaction should now The atoms in the half reaction should now be balanced.be balanced.



5.5. Add eAdd e-- to balance the charge on both sides to balance the charge on both sides of the equationof the equation6 e6 e-- + 14 H + 14 H++ + Cr + Cr22OO77

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

6.6. Do the same for other half reactionDo the same for other half reaction2 Cl2 Cl-- Cl Cl22 + 2 e + 2 e--

7.7. Multiply each equation by a coefficient so Multiply each equation by a coefficient so that the number of electrons is the same in that the number of electrons is the same in both half reactionsboth half reactions



7.7. Con’tCon’t3 (2 Cl3 (2 Cl-- Cl Cl22 + 2 e + 2 e--))6 Cl6 Cl-- 3 Cl 3 Cl22 + 6 e + 6 e--

8.8. 6 e6 e-- + 14 H + 14 H++ + Cr + Cr22OO772-2- 2Cr 2Cr3+3+

+ 7 H+ 7 H22OO6 Cl6 Cl-- 3 Cl 3 Cl22 + 6 e + 6 e--

__________________________________________________________________ 14 H 14 H++ + Cr + Cr22OO77

2-2- + 6 Cl + 6 Cl-- 3 Cl3 Cl22 + 2Cr + 2Cr3+3+

+ 7 H+ 7 H22OO


Half reaction method – basic sol’n

In basic solution, OHIn basic solution, OH-- is the predominant is the predominant species (besides water)species (besides water)

The simplest way to adjust for basic The simplest way to adjust for basic solution is to balance the reaction as if it solution is to balance the reaction as if it occurred in acid, then neutralize the Hoccurred in acid, then neutralize the H++ by by adding OHadding OH-- to both sides. to both sides.

The HThe H++ will combine with the OH will combine with the OH--


Example

1.1. 2 e2 e-- + 2 H + 2 H++ + ClO + ClO-- Cl Cl-- + H + H22OO

2.2. 2 e2 e-- + (2 H + (2 H++ + 2 OH + 2 OH--) + ClO) + ClO-- ClCl-- + H + H22O + 2 OHO + 2 OH--

3.3. 2 e2 e-- + 2 H + 2 H22O + ClOO + ClO-- ClCl-- + H + H22O + 2 OHO + 2 OH--

4.4. 2 e2 e-- + H + H22O + ClOO + ClO-- ClCl-- + 2 OH + 2 OH--


Spontaneous Redox Reactions

A spontaneous reaction occurs between the A spontaneous reaction occurs between the stronger oxidizing agent and the stronger stronger oxidizing agent and the stronger reducing to form weaker oxidizing and reducing to form weaker oxidizing and reducing agentsreducing agents

By testing reactions between pairs of atoms By testing reactions between pairs of atoms and molecules, we can determine a relative and molecules, we can determine a relative order of strength of oxidizing or reducing order of strength of oxidizing or reducing abilityability


Spontaneous redox reactions

Why does K react spontaneously with Why does K react spontaneously with water? (water? (GGºº

rxnrxn = -407 kJ) = -407 kJ)

2 K2 K(s)(s) + 2 H + 2 H22OO(l)(l)

2 K2 K++(aq)(aq) + 2 OH + 2 OH--

(aq)(aq) + H + H2(g)2(g)

Yet Ag does not (Yet Ag does not (GGººrxnrxn = +314 kJ) = +314 kJ)

2 Ag2 Ag(s)(s) + 2 H + 2 H22OO(l)(l)

2 Ag2 Ag++(aq)(aq) + 2 OH + 2 OH--

(aq)(aq) + H + H2(g)2(g)


Spontaneity

K is easily oxidized while Ag resists K is easily oxidized while Ag resists oxidation.oxidation.

The main factor is the first ionization The main factor is the first ionization energiesenergiesIEIE11(K) = 419 kJ/mol(K) = 419 kJ/mol

IEIE11(Ag) = 731 kJ/mol(Ag) = 731 kJ/mol


Direct and indirect electron transfer

Note that Ag and K represent extremes in Note that Ag and K represent extremes in the spontaneity of electron transferthe spontaneity of electron transfer

Consider Consider CuCu2+2+

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

GGººrxnrxn = -213 kJ = -213 kJ

Re: Re: CuCu2+2+(aq)(aq) + 2 e + 2 e-- Cu Cu(s)(s)

Ox: ZnOx: Zn(s)(s) Zn Zn2+2+(aq)(aq) +2 e +2 e--


Direct electron transfer

This is an example of direct electron This is an example of direct electron transfer (Cutransfer (Cu2+2+ collides with the Zn surface collides with the Zn surface and accepts two electrons)and accepts two electrons)

Scan p 869 Bottom FigureScan p 869 Bottom Figure


Indirect electron transfer

Separate the oxidation and reduction Separate the oxidation and reduction reactions into two compartmentsreactions into two compartments

Arrange for the electrons to flow through Arrange for the electrons to flow through and external circuitand external circuit

Connect the two compartments with a Connect the two compartments with a barrier that only allows inert ions to pass (to barrier that only allows inert ions to pass (to yield charge transfer)yield charge transfer)


The Danielle Cell

Z n C u

C u2 + + 2 e-

Z n2 +-2 e-

e -

aq u eo u s K C l

K+

C l_

M em b ran eP erm eab le toto K C l o n ly


Electrodes

Anode – the electrode at which the Anode – the electrode at which the oxidation takes placeoxidation takes place

Cathode – the electrode at which the Cathode – the electrode at which the oxidation takes placeoxidation takes place

Active electrodes, such as the Zn and Cu Active electrodes, such as the Zn and Cu strips, participate in the redox reactionstrips, participate in the redox reaction

Inert electrodes merely provide a surface at Inert electrodes merely provide a surface at which the reaction occurswhich the reaction occurs


Salt bridge

Since negative charge is flowing through Since negative charge is flowing through the external circuit, a means must be the external circuit, a means must be provided to keep charge balancedprovided to keep charge balanced

The salt bridge allows inert ions to pass The salt bridge allows inert ions to pass between the two compartments while between the two compartments while preventing the species involved in the redox preventing the species involved in the redox reaction from mixingreaction from mixing


Voltaic or galvanic cells

Use a favorable redox reaction to generate Use a favorable redox reaction to generate electrical energy through an external circuitelectrical energy through an external circuit

First practical example was the Danielle cellFirst practical example was the Danielle cellZnZn(s)(s) + Cu + Cu2+2+ Zn Zn2+2+ + Cu + Cu

was used to power telegraphs and other was used to power telegraphs and other early electrical devicesearly electrical devices


The Danielle Cell

A Zn strip is immersed in a ZnA Zn strip is immersed in a Zn2+2+ solution solution and in a separate compartment, a Cu strip is and in a separate compartment, a Cu strip is immersed in a Cuimmersed in a Cu2+2+ solution. solution.

Electrodes are the surfaces in a cell at which Electrodes are the surfaces in a cell at which the reactions take placethe reactions take place

The Zn and Cu strips serve as electrodesThe Zn and Cu strips serve as electrodes


Common voltaic cells

Dry cells (flashlight batteries)Dry cells (flashlight batteries) Lead-acid car batteriesLead-acid car batteries Keep in mind that a battery is a collection Keep in mind that a battery is a collection

of one or more separate cells joined of one or more separate cells joined together in one unittogether in one unit


Cell potentials and current

Compare water flow and electron flowCompare water flow and electron flow Water flows from higher elevation to lower Water flows from higher elevation to lower Electrons flow from species with greater Electrons flow from species with greater

reduction potential to those of lowerreduction potential to those of lower Potential is the force of the flowPotential is the force of the flow Current is the amount of material (water or Current is the amount of material (water or

electrons) that flowselectrons) that flows


Electrode equilibrium

Consider a Zn electrode immersed in ZnSOConsider a Zn electrode immersed in ZnSO44

aqueous solutionaqueous solution

Scan p 876Scan p 876



At the molecular leve, some the of the Zn At the molecular leve, some the of the Zn atoms on the electrode surface are oxidized atoms on the electrode surface are oxidized and move out into solution, leaving and move out into solution, leaving electrons behindelectrons behindZnZn(s)(s) Zn Zn2+2+

(aq)(aq) + 2 e + 2 e--(metal surface)(metal surface)

An excess of electrons remains on the An excess of electrons remains on the surface of the Zn electrodesurface of the Zn electrode



A ZnA Zn2+2+ ion collides with the surface and captures ion collides with the surface and captures two electrons to become a neutral Zn atomtwo electrons to become a neutral Zn atomZnZn(aq)(aq) + 2 e + 2 e--

(metal surface)(metal surface) Zn Zn(s)(s)

This dynamic equilibrium results in ca one excess This dynamic equilibrium results in ca one excess surface electron for every 10surface electron for every 101414 Zn atoms Zn atoms

The same situation applies to the Cu electrodeThe same situation applies to the Cu electrode



A slight charge imbalance is generated on each A slight charge imbalance is generated on each surface, but the imbalance is greater for the Zn surface, but the imbalance is greater for the Zn electrode since Zn is easier to oxidize (e.g. the electrode since Zn is easier to oxidize (e.g. the concentration of excess electrons is greater on the concentration of excess electrons is greater on the Zn electrode)Zn electrode)

When two electrodes with different amounts of When two electrodes with different amounts of excess charge are connected, there is a difference excess charge are connected, there is a difference in electrical potential between themin electrical potential between them



Scan Fig 18-13Scan Fig 18-13 Electrons flow “downhill” from the Zn to Electrons flow “downhill” from the Zn to

the Cu electrodesthe Cu electrodes


Electrode potential

The potential difference is measured in The potential difference is measured in volts (V), given the symbol E and termed volts (V), given the symbol E and termed the cell potentialthe cell potential

In a galvanic cell, the electrode with the In a galvanic cell, the electrode with the higher potential is termed the negative higher potential is termed the negative electrode. The other is the positive electrode. The other is the positive electrodeelectrode


Standard cell potentials

Recall standard conditions for Recall standard conditions for thermodynamics (1 M concentration, 1 atm thermodynamics (1 M concentration, 1 atm pressure, 25 pressure, 25 ºC)ºC)

Standard reduction potential = EºStandard reduction potential = Eº The Zn-Cu galvanic cell generates a 1.10 V The Zn-Cu galvanic cell generates a 1.10 V

potential difference under standard potential difference under standard conditionsconditions


The reference half-cell

In order to compare redox reactions against In order to compare redox reactions against each other, it is convenient to measure cell each other, it is convenient to measure cell potentials against a common reactionpotentials against a common reaction

Conventionally, the standard hydrogen Conventionally, the standard hydrogen electrode is usedelectrode is used(SHE)2 H(SHE)2 H33OO++

(aq, 1.00 M)(aq, 1.00 M) + 2 e + 2 e-- ↔↔

HH2 (g, 1.00 atm)2 (g, 1.00 atm) + 2 H + 2 H22OO(l)(l) Eº = 0.00 V Eº = 0.00 V


Standard reduction potentials

All reactions are written as reductionsAll reactions are written as reductions The Cu reactionThe Cu reaction

CuCu2+2+ + 2 e + 2 e-- ↔ Cu↔ Cu(s)(s) Eº = 0.34 V vs SHEEº = 0.34 V vs SHE


Calculating standard cell potentials

EEººcellcell = = EEººcathodecathode - - EEººanodeanode

The half reaction with the more positive The half reaction with the more positive reduction potential occurs at the cathode as reduction potential occurs at the cathode as a reductiona reduction

The half reaction with the more negative The half reaction with the more negative reduction potential occurs at the anode as an reduction potential occurs at the anode as an oxidationoxidation


Example

CuCu2+2+(aq, 1.00 M)(aq, 1.00 M) + 2 e + 2 e-- ↔ Cu↔ Cu(s)(s) Eº = +0.34V Eº = +0.34V

ZnZn2+2+(aq, 1.00 M)(aq, 1.00 M) + 2 e + 2 e-- ↔ Zn↔ Zn(s)(s) Eº = -0.76VEº = -0.76V

EEººcellcell = = EEººcathodecathode - - EEººanodeanode

EEººcellcell = = EEººCuCu - - EEººZnZn = +0.34 – (-0.76) = 1.10V = +0.34 – (-0.76) = 1.10V It is important to note that cell potentials do It is important to note that cell potentials do

not depend on the amount of material not depend on the amount of material involved in the reaction (an intensive involved in the reaction (an intensive property)property)


Free energy and electron transfer

Recall Recall G < 0 for spontaneous reactionsG < 0 for spontaneous reactions EEº > 0 for spontaneous reactionsº > 0 for spontaneous reactions The total energy change depends on the The total energy change depends on the

potential difference (E) and the amount of potential difference (E) and the amount of charge that flows through the cell (n, the charge that flows through the cell (n, the number of moles of electrons)number of moles of electrons)


Free energy and electron transfer

G = -nFE, where F = 96,485.34 C molG = -nFE, where F = 96,485.34 C mol-1-1

GGºº = -nFE = -nFEºº GGºº = -nFE = -nFEº and º and GGºº = -RT ln K = -RT ln Keqeq

After some mathAfter some mathEEº = -(RT/nF) ln Kº = -(RT/nF) ln Keqeq

At STPAt STP EEº = -(0.0592 V/n) log Kº = -(0.0592 V/n) log Keqeq


The Nernst Equation

Relates amounts of material and driving Relates amounts of material and driving forceforce

E = EE = Eº -(RT/nF) ln Qº -(RT/nF) ln Q Q = (products)/(reactants)Q = (products)/(reactants) pH meter is a common device that exploit pH meter is a common device that exploit

the Nernst equation to allow us to measure the Nernst equation to allow us to measure concentration electricallyconcentration electrically


Electrochemical stoichiometry

Measurement of electricityMeasurement of electricity Charge – CoulombsCharge – Coulombs Current – rate of electric current flow Current – rate of electric current flow

(measured in amperes)(measured in amperes) 1 A = 1 C s1 A = 1 C s-1-1 = I (symbol for current) = I (symbol for current) Charge = I Charge = I × × tt Moles of electrons = n = (I Moles of electrons = n = (I × × t)/Ft)/F


Fuel Cells

Uses the direct reaction of chemicals such as Uses the direct reaction of chemicals such as hydrogen and oxygen to produce electricityhydrogen and oxygen to produce electricity

Must be used in applications where the electrical Must be used in applications where the electrical flow must be continuous (no recharging)flow must be continuous (no recharging)

Can also be turned on and off by controlling the Can also be turned on and off by controlling the flow of reagentsflow of reagents

Very expensive, but much research is going into Very expensive, but much research is going into changing thatchanging that


Electrolytic cells

Convert electrical energy into chemical Convert electrical energy into chemical energyenergy

Electrolysis of HElectrolysis of H22O (using KO (using K22SOSO44) to yield ) to yield HH22 and O and O22

2 H2 H22OO(l)(l) 2 H 2 H2 (g)2 (g) + O + O2 (g)2 (g)

Electroplating is another electrolytic cell Electroplating is another electrolytic cell that uses a base metal object (a spoon) as that uses a base metal object (a spoon) as the cathode and a Ag bar as a cathodethe cathode and a Ag bar as a cathode


Dry cells

Not rechargeable, but cheap and portableNot rechargeable, but cheap and portable Zn casing serves as the anode and a Zn casing serves as the anode and a

graphite rod serves as an inert cathode.graphite rod serves as an inert cathode. In between the two is an aqueous paste In between the two is an aqueous paste

containing NHcontaining NH44Cl, MnOCl, MnO22 and carbon and carbonanode: anode: ZnZn(s)(s) Zn Zn2+2+

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

cathode:cathode: NHNH44++ + 2 MnO + 2 MnO2 (s)2 (s) + 2 e + 2 e--

MnMn22OO3 (s)3 (s) + 2 NH + 2 NH3 (aq)3 (aq) + H + H22OO(l)(l)


The Dry Cell

The disadvantage of the dry cell creates an acidic The disadvantage of the dry cell creates an acidic solution.solution.

This acidic solution slowly dissolves the zinc, so This acidic solution slowly dissolves the zinc, so the shelf life of the battery is only a few monthsthe shelf life of the battery is only a few months

Alkaline batteries use NaOH or KOH in place of Alkaline batteries use NaOH or KOH in place of NHNH44Cl, which is more expensive but the shelf life Cl, which is more expensive but the shelf life

is much greater. The anode becomes:is much greater. The anode becomes:ZnZn(s)(s) + 2 OH + 2 OH--

(aq)(aq) ZnO ZnO(s)(s) + H + H22OO(l)(l) + 2 e + 2 e--


The lead acid car battery

Anode:Anode:PbPb(s)(s) + H + H22SOSO4 (aq)4 (aq) PbSO PbSO4 (s)4 (s) + 2H + 2H++

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

Cathode:Cathode:2 e2 e-- + 2 H + 2 H++

(aq)(aq) + PbO + PbO2 (s)2 (s) + H + H22SOSO4 (aq)4 (aq) PbSOPbSO4 (s)4 (s) + 2 H + 2 H22OO(l)(l)

Total reactionTotal reaction Pb Pb(s)(s) + PbO + PbO2 (s)2 (s) + 2 H + 2 H22SOSO4 (aq)4 (aq)

2 PbSO2 PbSO4 (s)4 (s) + 2 H + 2 H22OO


The Lead Storage Battery

The anode is a Pb plateThe anode is a Pb plate The cathode is Pb impregnated with PbOThe cathode is Pb impregnated with PbO22

Both electrodes become coated with PbSOBoth electrodes become coated with PbSO44 during the course of the reactionduring the course of the reaction

The solution is 1 M HThe solution is 1 M H22SOSO44

No porous separator is needed since both No porous separator is needed since both starting materials and products are insoluble starting materials and products are insoluble in sulfuric acidin sulfuric acid


Other batteries

Silver:Silver:ZnZn(s)(s) + Ag + Ag22OO(s)(s) ZnO ZnO(s)(s) + 2 Ag + 2 Ag(s)(s)

Mercury:Mercury:ZnZn(s)(s) + HgO + HgO(s)(s) ZnO ZnO(s)(s) + Hg + Hg(l)(l)

NiCd:NiCd:CdCd(s)(s) + NiO + NiO2 (s)2 (s) + 2 H + 2 H22OO(l)(l) Ni(OH) Ni(OH)2 (s)2 (s) + +

Cd(OH)Cd(OH)2 (s)2 (s) (rechargeable) (rechargeable)


Corrosion

Scan p 895Scan p 895 Process is the destructive oxidation of Process is the destructive oxidation of

metalsmetals Turns refined metals into more stable metal Turns refined metals into more stable metal

oxidesoxides


Oxygen

OO22 is a potent oxidizer, especially in the is a potent oxidizer, especially in the

presence of aqueous acidspresence of aqueous acids OO2 (g)2 (g) + 2 H + 2 H22OO(l)(l) + 4 e + 4 e--

↔ 4 OH↔ 4 OH--(aq)(aq) E = 0.401 VE = 0.401 V

OO2 (g)2 (g) + 4 H + 4 H33OO++(aq)(aq) + 4 e + 4 e--

↔ 6 H↔ 6 H22OO(l)(l) E = 0.401 VE = 0.401 V


Atmospheric oxygen

P(OP(O22) in atmosphere is 0.20 atm) in atmosphere is 0.20 atm Atmospheric water vapor is saturated with Atmospheric water vapor is saturated with

carbon dioxide, which results in a hydrogen carbon dioxide, which results in a hydrogen ion concentration of 2 ion concentration of 2 × 10× 10-6-6 M M

Under these conditions, E(OUnder these conditions, E(O22) = 0.88 V) = 0.88 V Hence anything with a standard potential Hence anything with a standard potential

reduction potential less than 0.88 V reduction potential less than 0.88 V oxidizes spontaneouslyoxidizes spontaneously


Structural material oxidation

4 Fe4 Fe(s)(s) + 3O + 3O2 (g)2 (g) + 12 H + 12 H33OO++(aq)(aq)

4 Fe 4 Fe3+3+(aq)(aq) + 18 H + 18 H22OO(l)(l)

E = 0.88 V – (-0.037 V) = +0.92 VE = 0.88 V – (-0.037 V) = +0.92 V 4 Al4 Al(s)(s) + 3 O + 3 O2 (g)2 (g) + 12 H + 12 H33OO++

(aq)(aq)

4 Al 4 Al3+3+(aq)(aq) + 18 H + 18 H22OO(l)(l)

E = 0.88 V – (-1.662 V) = 2.54 VE = 0.88 V – (-1.662 V) = 2.54 V


Overall reaction

4 Fe4 Fe(s)(s) + 3 O + 3 O2 (g)2 (g) 2 Fe 2 Fe22OO3 (s)3 (s)

4 Al4 Al(s)(s) + 3 O + 3 O2 (g)2 (g) 2 Al 2 Al22OO3 (s)3 (s)

Ions in aqueous solution (such as rainwater) Ions in aqueous solution (such as rainwater) accelerate the process by facilitating charge accelerate the process by facilitating charge transporttransport


Preventing corrosion

Paint prevents oxygen from getting to the Paint prevents oxygen from getting to the metalmetal

Galvanization – coating Fe with a thin film Galvanization – coating Fe with a thin film of a more easily oxidized metal like Zn. of a more easily oxidized metal like Zn. The Zn protects the Fe even if the Zn is The Zn protects the Fe even if the Zn is scratchedscratched


Electrosynthesis

Electrical energy is used to synthesize a Electrical energy is used to synthesize a wide range of commercial productswide range of commercial products

Na and Mg are commonNa and Mg are common Al and ClAl and Cl22 are probably the most important are probably the most important


Electrolysis

Electrolytic cells use electric current to Electrolytic cells use electric current to drive redox reactions (like charging a drive redox reactions (like charging a battery)battery)

Apply an external voltage of opposite sign Apply an external voltage of opposite sign and greater than the redox potential for celland greater than the redox potential for cell


Electrolysis of water

Anode: 6 HAnode: 6 H22OO(l)(l) O O2 (g)2 (g) + 4 H + 4 H33OO++(aq)(aq) + 4e + 4e--

Cathode: 4 HCathode: 4 H33OO++(aq)(aq) + 4 e + 4 e--

2 H 2 H2 (g)2 (g) + 4 H + 4 H22OO(l)(l)

Net reaction: 2 HNet reaction: 2 H22OO(l)(l) 2 H 2 H2 (g)2 (g) + O + O2 (g)2 (g)


Example

What volume of HWhat volume of H22 and O and O22 is evolved if is evolved if

0.775 A is applied for 45 minutes0.775 A is applied for 45 minutes n = [(0.775 C/s)(45 min)(60 s/min)]/Fn = [(0.775 C/s)(45 min)(60 s/min)]/F = 2.169 = 2.169 × 10× 10-2-2 mol of electrons mol of electrons 4 mol of electrons produces 2 mol of H4 mol of electrons produces 2 mol of H22 and and

1 mol of O1 mol of O22

reaksi redoks

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