Chapter 18 Electrochemistry. 2 GOALS Review: oxidation states oxidation/reduction oxidizing/reducing...

Chapter 18

Electrochemistry

2

GOALS

Review:Review:oxidation statesoxidation statesoxidation/reductionoxidation/reductionoxidizing/reducing agentoxidizing/reducing agentch. 17ch. 17

Balancing redox reactionsBalancing redox reactionsVoltaic cellsVoltaic cellsElectrochemical potentialsElectrochemical potentialsElectrolysisElectrolysis ……& the calculations!!& the calculations!!

Why Study Electrochemistry?Why Study Electrochemistry?Why Study Electrochemistry?Why Study Electrochemistry?

• BatteriesBatteries• CorrosionCorrosion• Industrial production Industrial production

of chemicalsof chemicals such as such as Cl Cl22, NaOH, F, NaOH, F22 and Al and Al

• Biological redox Biological redox reactionsreactions

The heme groupThe heme group

Electron Transfer ReactionsElectron Transfer ReactionsElectron Transfer ReactionsElectron Transfer Reactions• Electron transfer reactions are oxidation-reduction

or redox reactions (i.e. changes in oxidation states).

• Redox reactions can result in the generation of an

electric current (battery), or, may be caused by

applying an electric current (electroplating).

• Therefore, this field of chemistry is often called

ELECTROCHEMISTRY.

ELECTRON TRANSFER REACTIONSELECTRON TRANSFER REACTIONS

0 0 1+ 1-0 0 1+ 1-

2Na(s) + Cl2Na(s) + Cl22(g) (g) 2NaCl(s) 2NaCl(s)

1+ 0 2+ 0 1+ 0 2+ 0

2HCl(aq) + Zn(s) 2HCl(aq) + Zn(s) ZnCl ZnCl22(aq) + H(aq) + H22(g)(g) 0 0 0 0

Cu(s) + 2 AgCu(s) + 2 Ag++(aq) (aq) Cu Cu2+2+(aq) + 2 Ag(s)(aq) + 2 Ag(s)

0 0 3+ 2-0 0 3+ 2-

2Fe(s) + xH2Fe(s) + xH22O(l) + 1½OO(l) + 1½O22(g) (g) Fe Fe22OO33.xH.xH22O(s) O(s) 4+ 2- 0 0 4+ 2- 0 0

COCO22(g) + H(g) + H22O(l) + energy O(l) + energy (CH (CH22O)O)nn + O + O22(g) (g)

Review of Terminology for Redox Reactions• OXIDATION—loss of electron(s) by a species;

increase in oxidation number ; ; e- to the right of arrow.

Na Na+ + e-

• REDUCTION—gain of electron(s); decrease in oxidation number; e- to left of arrow.

½Cl2(g) + e- Cl-

• OXIDIZING AGENT—electron acceptor; it is reduced: ½ Cl2(g) + e- Cl-

• REDUCING AGENT—electron donor; it is oxidized

Na Na+ + e-

• OXIDATION—loss of electron(s) by a species; increase in oxidation number ; ; e- to the right of arrow.

Na Na+ + e-

• REDUCTION—gain of electron(s); decrease in oxidation number; e- to left of arrow.

½Cl2(g) + e- Cl-

• OXIDIZING AGENT—electron acceptor; it is reduced: ½ Cl2(g) + e- Cl-

• REDUCING AGENT—electron donor; it is oxidized

Na Na+ + e-

Electrochemical CellsElectrochemical CellsElectrochemical CellsElectrochemical Cells

• Apparatus for generating an Apparatus for generating an electric current through the use electric current through the use of a product favored reaction of a product favored reaction (spontaneous): (spontaneous): voltaic or galvanic cell.

• An electrolytic cell is used to carry out electrolysis (an (an electric current is used to bring electric current is used to bring about a nonspontaneous about a nonspontaneous chemical reaction).chemical reaction).

Batteries are voltaic Batteries are voltaic cellscells

ElectrochemistryElectrochemistryAlessandro Volta, Alessandro Volta, 1745-1827, Italian 1745-1827, Italian scientist and scientist and inventor.inventor.

Luigi Galvani, 1737-1798, Luigi Galvani, 1737-1798, Italian scientist and Italian scientist and inventor.inventor.

Balancing Equations for Redox Balancing Equations for Redox ReactionsReactions

Some redox reactions have equations that Some redox reactions have equations that must be balanced by special techniques.must be balanced by special techniques.

Mn = +7Mn = +7Fe = +2Fe = +2

Fe = +3Fe = +3Mn = +2Mn = +2

MnOMnO44--(aq) + 5 Fe(aq) + 5 Fe2+2+(aq) + 8 H(aq) + 8 H++(aq) (aq)

MnMn2+ 2+ (aq) + 5 Fe(aq) + 5 Fe3+3+(aq) + 4 H(aq) + 4 H22O(liq)O(liq)

10

Rules for Assigning Oxidation States

• rules are in order of priority1. free elements have an oxidation state = 0

Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)

2. monatomic ions have an oxidation state equal to their charge

Na = +1 and Cl = -1 in NaCl

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

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

11


3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion

N = +5 and O = -2 in NO3–, (+5) + 3(-2) = -1

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

Na = +1 in NaCl

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

Mg = +2 in MgCl2

12


5. in their compounds, nonmetals have oxidation states according to the table below (grp # - 8)

nonmetals higher on the table take priorityNonmetal Oxidation State Example

F -1 CF4

H +1 CH4

O -2 CO2

Group 7A -1 CCl4

Group 6A -2 CS2

Group 5A -3 NH3

Balancing Balancing EquationsEquations

Cu + AgCu + Ag++ givegive Cu Cu2+2+ + Ag + Ag

Balancing EquationsBalancing EquationsStep 1:Step 1: Divide the reaction into half-Divide the reaction into half-

reactions, one for oxidation and the reactions, one for oxidation and the other for reduction.other for reduction.

OxOx Cu Cu Cu Cu2+2+

RedRed Ag Ag++ Ag Ag

Step 2:Step 2: Balance each for mass. Already Balance each for mass. Already done in this case.done in this case.

Step 3:Step 3: Balance each half-reaction for Balance each half-reaction for charge by adding electrons.charge by adding electrons.

OxOx Cu Cu Cu Cu2+2+ + + 2e-2e-

RedRed Ag Ag++ + + e- e- Ag Ag

Balancing EquationsBalancing EquationsStep 4:Step 4: Multiply each half-reaction by a factor so Multiply each half-reaction by a factor so

that the reducing agent supplies as many that the reducing agent supplies as many electrons as the oxidizing agent requires.electrons as the oxidizing agent requires.

Reducing agentReducing agent Cu Cu Cu Cu2+2+ + 2e- + 2e-Oxidizing agentOxidizing agent 22 Ag Ag++ + + 22 e- e- 22 Ag AgStep 5:Step 5: Add half-reactions to give the overall Add half-reactions to give the overall

equation.equation.Cu + 2 AgCu + 2 Ag++ Cu Cu2+2+ + 2Ag + 2Ag

The equation is now balanced for both The equation is now balanced for both charge and mass (charge and mass (the 2ethe 2e-- of the left are of the left are cancelled out with those on the rightcancelled out with those on the right).).

ReductionReduction of VO of VO22++ with with

ZnZn

Balancing EquationsBalancing EquationsBalance the following in Balance the following in acidacid solution— solution—

VOVO22++ + Zn + Zn VO VO2+ 2+ + Zn+ Zn2+2+

Step 1:Step 1: Write the half-reactionsWrite the half-reactions

OxOx Zn Zn Zn Zn2+2+

RedRed VOVO22++ VO VO2+2+

Step 2:Step 2: Balance each half-reaction for Balance each half-reaction for mass.mass.

OxOx Zn Zn Zn Zn2+2+ + + 2e2e--

Red Red VOVO22++ + e-+ e- VO VO2+2+ excess of 2+ on the right

Zn lost two electrons. They are written on the right.Zn lost two electrons. They are written on the right.V is 5+ on the left and 4+ on the right. It gained one V is 5+ on the left and 4+ on the right. It gained one

electron. That is written on the left side.electron. That is written on the left side.

Balancing EquationsBalancing EquationsStep 3:Step 3: Balance half-reactions for chargeBalance half-reactions for charge Reaction is acidic, then we can use HReaction is acidic, then we can use H++..OxOx Zn Zn Zn Zn2+2+ + + 2e- 2e-

RedRed e- e- + 2 H+ 2 H++ + VO + VO22++ VO VO2+2+ + H + H22OO

Step 4:Step 4: Multiply by an appropriate factor.Multiply by an appropriate factor.OxOx Zn Zn Zn Zn2+2+ + + 2e-2e-

RedRed 22e-e- + + 44 H H++ + + 22 VO VO22++

22 VO VO2+2+ + + 22 HH22OO

Step 5:Step 5: Add Add balancedbalanced half-reactions. half-reactions.

Zn + 4 HZn + 4 H++ + 2 VO + 2 VO22++ Zn Zn2+2+ + 2 VO + 2 VO2+2+ + 2 H + 2 H22OO

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.Balance O with OHBalance O with OH-- or H or H22OO..

• Never add HNever add H22 or H atoms to or H atoms to

balance hydrogen.balance hydrogen.

Balance H with HBalance H with H++/H/H22OO in in acid or OH acid or OH--/H/H22O in O in base.base.

{Equations that include oxoanions like SO4

2-, NO3-, ClO- , CrO4

2-, and MnO4-,

also fall into this category}.

Tips on Balancing EquationsTips on Balancing Equations

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

•PRACTICE!!!!!!!!!!!PRACTICE!!!!!!!!!!!

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

21

More Practice - Balance the equations below!I

(aq) + MnO4(aq) I2(aq) + MnO2(s) in basic solution

ClO3- + Cl- Cl2 (in acid)

Cr2O72- + I- IO3

- + Cr3+ (in acid)

MnO4- + H2SO3

SO42- + Mn2+ (in acid)

Cr(OH)4- + H2O2

CrO42- + H2O (in basic soln)

Zn + NO3- Zn(OH)4

- + NH3 (in basic soln)

An alkaline (basic) solution of hypochlorite ions reacts with solid chromium(III) hydroxide to produce chromate and

chloride ions.

VOLTAIC CELLSVOLTAIC CELLS

The Zn|Zn2+ and Cu|Cu2+ CellZn(s) + CuZn(s) + Cu2+2+(aq) (aq) Zn Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

-- use a chemical rxn to use a chemical rxn to produce an electric produce an electric current. current.

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Oxidation: Zn(s) Zn2+(aq) + 2e-Reduction: Cu2+(aq) + 2e- Cu(s)----------------------------------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) Zn Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

Zn metal

Cu2+ ions

Zn metal

Cu2+ ionsElectrons are transferred from Zn to Cu2+, but there is no useful electric current.

Electrons are transferred from Zn to Cu2+, but there is no useful electric current.

With time, Cu plates out onto Zn metal

strip, and Zn strip “disappears.” With time, Cu plates out onto Zn metal

strip, and Zn strip “disappears.”

A CHEMICAL CHANGE PRODUCES AN ELECTRIC CURRENT

•To obtain a useful current, To obtain a useful current, we separate the oxidizing we separate the oxidizing and reducing agents so that and reducing agents so that electron transfer occurs electron transfer occurs through an external wire. through an external wire.

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

This is accomplished in a This is accomplished in a GALVANICGALVANIC or or VOLTAICVOLTAIC cell. cell.

A group of such cells is called a A group of such cells is called a batterybattery..

Zn(s) Zn(s) Zn Zn2+2+(aq) + (aq) + 2e-2e-

A CHEMICAL CHANGE PRODUCES AN ELECTRIC CURRENT

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

••Electrons travel through external wire.Electrons travel through external wire.•Salt bridge Salt bridge allows anions and cations to move allows anions and cations to move between electrode compartments.between electrode compartments.

••Electrons travel through external wire.Electrons travel through external wire.•Salt bridge Salt bridge allows anions and cations to move allows anions and cations to move between electrode compartments.between electrode compartments.

Zn Zn Zn Zn2+2+ + 2e- + 2e- CuCu2+2+ + 2e- + 2e- Cu Cu

AnionsAnionsCations Cations

OxidationOxidationAnodeAnodeNegativeNegative

OxidationOxidationAnodeAnodeNegativeNegative

ReductionReductionCathodeCathodePositivePositive

ReductionReductionCathodeCathodePositivePositive

CELL POTENTIAL, ECELL POTENTIAL, E

• Electrons are “driven” from anode to cathode Electrons are “driven” from anode to cathode by an by an electromotive forceelectromotive force or or emfemf..

• For Zn/Cu cell, this is indicated by a voltage of For Zn/Cu cell, this is indicated by a voltage of 1.10 V at 25 ˚C and when [Zn1.10 V at 25 ˚C and when [Zn2+2+] = [Cu] = [Cu2+2+] = 1.0 M.] = 1.0 M.

Zn and Zn2+,anode

Cu and Cu2+,cathode

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons1.10 V1.10 V

1.0 M1.0 M 1.0 M1.0 M

Need Calculate Cell VoltageNeed Calculate Cell Voltage

• Balanced half-reactions can be added together to get the overall, balanced equation. .

Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e-(aq) + 2e-CuCu2+2+(aq) + 2e- (aq) + 2e- Cu(s) Cu(s)----------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) Zn Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

Zn(s) Zn(s) Zn Zn2+2+(aq) + 2e-(aq) + 2e-CuCu2+2+(aq) + 2e- (aq) + 2e- Cu(s) Cu(s)----------------------------------------------------------------------------------------CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) Zn Zn2+2+(aq) + Cu(s)(aq) + Cu(s)

If we know EIf we know Eoo for each half-reaction, we add for each half-reaction, we add them to get Ethem to get Eoo for overall reaction. for overall reaction.

-need E-need Eoo for Zn & Cu half-cells for Zn & Cu half-cells

28

Standard Reduction Potential

• We can measure ALL other half-cell potentials relative to another half-reaction.

• We select as a standard half-reaction, the reduction of H+ to H2 under standard conditions (1atm, 1 M @ 25 oC) and which we assign a potential difference = 0 V

Standard Hydrogen Electrode, SHE

Zn/Zn2+ half-cell hooked up to a SHE.Eo for the cell = +0.76 V

Zn/Zn2+ half-cell hooked up to a SHE.Eo for the cell = +0.76 V

Volts

ZnH2

Salt Bridge

Zn2+ H+

Zn Zn2+ + 2e- OXIDATION ANODE

2 H+ + 2e- H2REDUCTIONCATHODE

- +

Volts

ZnH2

Salt Bridge

Zn2+ H+

Zn Zn2+ + 2e- OXIDATION ANODE

2 H+ + 2e- H2REDUCTIONCATHODE

- +

Negative Negative electrodeelectrode

Supplier Supplier of of

electronselectrons

Acceptor Acceptor of of

electronselectrons

Positive Positive electrodeelectrode

2 H2 H++ + 2e- + 2e- H H22

ReductionReductionCathodeCathode

Zn Zn Zn Zn2+2+ + 2e- + 2e- OxidationOxidation

AnodeAnode

Zn is a Zn is a betterbetter reducing agent than H reducing agent than H22

Volts

CuH2

Salt Bridge

Cu2+ H+

Cu2+ + 2e- Cu REDUCTION CATHODE

H2 2 H+ + 2e-OXIDATION ANODE

-+

Volts

CuH2

Salt Bridge

Cu2+ H+

Cu2+ + 2e- Cu REDUCTION CATHODE

H2 2 H+ + 2e-OXIDATION ANODE

-+

Acceptor of

electronsSupplier Supplier

of of electronselectrons

CuCu2+2+ + 2e- + 2e- Cu CuReductionReductionCathodeCathode

HH22 2 H 2 H++ + 2e- + 2e-

OxidationOxidationAnodeAnode

PositiveNegative

Cu/Cu2+ half-cell hooked up to a SHE.Eo for the cell = +0.34 V

Cu/Cu2+ half-cell hooked up to a SHE.Eo for the cell = +0.34 V

HH22 is now a is now a betterbetter reducing agent than Cu! reducing agent than Cu!

Zn/Cu Electrochemical Cell

oxid: Zn(s) oxid: Zn(s) Zn Zn2+2+(aq) + 2e- E(aq) + 2e- Eoo = +0.76 V = +0.76 V red: Cured: Cu2+2+(aq) + 2e- (aq) + 2e- Cu(s) Cu(s) E Eoo = +0.34 V = +0.34 V------------------------------------------------------------------------------------------------------------------------------ CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) Zn Zn2+2+(aq) + Cu(s) (aq) + Cu(s)

Eo (calc’d) = +1.10 V

Cathode; positive; sink for electrons

Anode; negative; source of electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons ++

Uses of Eo Values

Organize half-reactions by relative ability to act as oxidizing/reducing agents. Half-rxns are written as reduction rxns!!

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

Zn

Zn2+ ions

Cu

Cu2+ ions

wire

saltbridge

electrons

CuCu2+2+(aq) + 2e- (aq) + 2e- Cu(s) Cu(s) EEoo = +0.34 V = +0.34 VZnZn2+2+(aq) + 2e- (aq) + 2e- Zn(s) Zn(s) EEoo = –0.76 V = –0.76 V

When a reaction is reversed, the sign of E˚ is reversed!When a reaction is reversed, the sign of E˚ is reversed!

34reducing agentsoxidizing agents

Using Standard Potentials, EUsing Standard Potentials, Eoo

Which is the best reducing agent:

Hg (+0.79 V), Al (-1.66 V), or Sn (-0.14 V)?

Which is the best oxidizing agent:O2 (1.23 V); H2O2 (1.77 V) or Cl2 (1.36 V)?

H2O2 (1.77 V)

Al (-1.66 V)


Which element/ion is the best reducing agent?Fe3+

+ e- Fe2+ (+0.77 V) I2 + 2e- 2I- (+0.54 V) Sn4+ + 2e- Sn2+ (+0.15 V)

Which substance is the best oxidizing agent?Cr2O7

2- + 6e- + 14H+ 2Cr3+ + 7H2O (+1.33 V) O2 + 4e- + 4H+ 2H2O (+1.23 V) Fe3+

+ e- Fe2+ (+0.77 V)Cr2O7

2-

Sn2+

Standard Redox Potentials, EStandard Redox Potentials, Eoo

Any substance on the right Any substance on the right will reduce any substance will reduce any substance

HIGHERHIGHER than it on the than it on the

LEFT.LEFT.

• Zn can reduce HZn can reduce H++ and Cu and Cu2+2+..

• HH22 can reduce Cu can reduce Cu2+2+ but not but not

ZnZn2+2+

• Cu cannot reduce HCu cannot reduce H++ or or ZnZn2+2+..

Eo (V)

Cu2+ + 2e- Cu +0.34

2 H+ + 2e- H2 0.00

Zn2+ + 2e- Zn -0.76

oxidizingability of ion

reducing abilityof element

Eo (V)

Cu2+ + 2e- Cu +0.34

2 H+ + 2e- H2 0.00

Zn2+ + 2e- Zn -0.76

oxidizingability of ion

reducing abilityof element


Cu2+ + 2e- --> Cu +0.34

+2 H + 2e- --> H2 0.00

Zn2+ + 2e- --> Zn -0.76

Northwest-southeast rule:Northwest-southeast rule: product-favored product-favored reactions occur between reactions occur between • reducing agent at southeast corner (ANODE) & reducing agent at southeast corner (ANODE) & • oxidizing agent at northwest corner (CATHODE)oxidizing agent at northwest corner (CATHODE)

Any substance on the right will reduce any substance Any substance on the right will reduce any substance higher than it on the left.higher than it on the left.

Ox. agentOx. agent

Red. agentRed. agent


Cu2+ + 2e- --> Cu +0.34 V

2+Ni + 2e- --> Ni -0.25 V

Zn2+ + 2e- --> Zn -0.76 V

Northwest-southeast rule:Northwest-southeast rule: product-favored product-favored reactions occur between reactions occur between • reducing agent at southeast corner (ANODE) & reducing agent at southeast corner (ANODE) & • oxidizing agent at northwest corner (CATHODE)oxidizing agent at northwest corner (CATHODE)

Zn will reduce NiZn will reduce Ni2+2+, Cu, Cu2+2+; Ni will reduce Cu; Ni will reduce Cu2+2+..

Ox. agentOx. agent

Red. agentRed. agent


• In which direction do the following reactions go?In which direction do the following reactions go?

• Cu(s) + 2 AgCu(s) + 2 Ag++(aq) (aq) Cu Cu2+2+(aq) + 2 Ag(s) (aq) + 2 Ag(s) +0.46 V+0.46 V

• CuCu2+2+(aq) + Zn(s) (aq) + Zn(s) Cu(s) + Zn Cu(s) + Zn2+2+(aq) (aq) +1.10 V+1.10 V

Go to the right as writtenGo to the right as written

• FeFe2+2+(aq) +(aq) + Cd(s) Fe(s) + Cd2+(aq)

Goes LEFT, opposite to direction writtenGoes LEFT, opposite to direction written

• What is Eonet for this reverse reaction?

-0.04 V

+0.04 V

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

Cd Cd Cd Cd2+2+ + 2e- + 2e-oror

CdCd2+2+ + 2e- + 2e- Cd Cd (-0.40 V)(-0.40 V)

Fe Fe Fe Fe2+2+ + 2e- + 2e-oror

FeFe2+2+ + 2e- + 2e- Fe Fe (-0.44 V)(-0.44 V)

EEoo for a Voltaic Cell for a Voltaic Cell

Which way does the reaction proceed?Which way does the reaction proceed?In which direction is it spontaneous?In which direction is it spontaneous?

Fe(s) + Cd2+(aq) Cd(s) + Fe2+(aq)

From the From the tabletable, , •• Fe is a better reducing agent Fe is a better reducing agent

than Cd (-0.44 V; anode)than Cd (-0.44 V; anode)•• CdCd2+2+ is a better oxidizing is a better oxidizing

agent than Feagent than Fe2+2+ (-0.40 V; (-0.40 V; cathode)cathode)

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

Volts

Cd Salt Bridge

Cd2+

Fe

Fe2+

EEoo for a Voltaic Cell for a Voltaic Cell

EEoo = E˚ = E˚cathodecathode - E˚ - E˚anode anode (reverse the smaller, (reverse the smaller, more more

negativenegative, then add), then add) cathode: cathode: Cd2+(aq) + 2e- Cd(s) -0.40 V (red) - anode: Fe(s) Fe2+(aq) + 2e- +0.44 V (oxid)Overall :Fe(s) + Cd2+(aq) Cd(s) + Fe2+(aq) +0.04 V

More 0n Cell VoltageMore 0n Cell VoltageWhen two half-rxns (written as reduction) are joined in an electrochemical cell, the one with the larger half-cell potential occurs in the forward direction, and the one with the smaller potential occurs in the reverse direction.

CdCd2+2+ + 2e- + 2e- Cd Cd (-0.40 V) larger(-0.40 V) larger

FeFe2+2+ + 2e- + 2e- Fe Fe (-0.44 V) smaller (-0.44 V) smaller (reverse this rxn)

CdCd2+2+ + 2e- + 2e- Cd Cd (-0.40 V) larger (-0.40 V) larger Fe Fe2+ + 2e- (+0.44 V) overall: CdCd2+2+ + Fe + Fe Cd + Fe Cd + Fe2+2+ (+0.04 V) (+0.04 V)

More 0n Cell VoltageMore 0n Cell VoltageWhen two half-rxns (written as reduction) are joined in an electrochemical cell, the one with the larger half-cell potential occurs in the forward direction, and the one with the smaller potential occurs in the reverse direction.

NiNi2+2+ + 2e- + 2e- Ni Ni (-0.23 V) larger(-0.23 V) larger

MnMn2+2+ + 2e- + 2e- Mn Mn (-1.18 V) smaller (-1.18 V) smaller

(reverse this rxn) NiNi2+2+ + 2e- + 2e- Ni Ni (-0.23 V) (-0.23 V) Mn Mn2+ + 2e- (+1.18 V)overall: NiNi2+2+ + Mn + Mn Ni + Mn Ni + Mn2+2+ (+0.95 V) (+0.95 V)

More 0n Cell VoltageMore 0n Cell VoltageAssumeAssume I I-- ion can reduce water. ion can reduce water.

2 H2O + 2e- H2 + 2 OH- Cathode2 I- I2 + 2e- Anode-------------------------------------------------2 I- + 2 H2O I2 + 2 OH- + H2

2 H2O + 2e- H2 + 2 OH- Cathode2 I- I2 + 2e- Anode-------------------------------------------------2 I- + 2 H2O I2 + 2 OH- + H2

Assuming reaction occurs as written, Assuming reaction occurs as written,

E˚E˚netnet = E˚ = E˚cathodecathode - E˚ - E˚anodeanode (from values in table) (from values in table)

= (-0.828 V) - (+0.535 V) = = (-0.828 V) - (+0.535 V) = -1.363 V-1.363 V

Minus EMinus Enetnet˚ means net rxn. occurs in the ˚ means net rxn. occurs in the

opposite direction (favors Iopposite direction (favors I-- + H + H22O).O).

II2 2 + 2 OH+ 2 OH- - + H + H22 2I 2I-- + H + H22O (+1.363 V)!!!O (+1.363 V)!!!

46

Calculate Ecell for the reaction at 25CAl(s) + NO3

−(aq) + 4 H+

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

(This is the reaction of Al with nitric acid)

Separate the reaction into the oxidation and reduction half-reactions

Find the E for each half-reaction and sum to get Ecell

ox: Al(s) Al3+(aq) + 3 e−

red: NO3−

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

E°red = +0.96 V

Eox = −Ered = +1.66 V

Eox = −(Ered) = +1.66 V

Ered = +0.96 V

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

47

For the reaction Al(s) + NO3

−(aq) + 4 H+

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

ox: Al(s) Al3+(aq) + 3 e−

red: NO3−

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

Eox = −Ered = +1.66 V

Ered = +0.96 V

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

The symbol of the cell is

Al(s)| Al3+(aq) || NO3

−(aq), 4 H+

(aq) , NO(g) |Pt

This is the symbol for the salt bridgeThis is the symbol for the electrode-solxn contact

E at Nonstandard ConditionsE at Nonstandard Conditions

• The NERNST EQUATION

• E = potential under nonstandard conditions

• n = no. of electrons exchanged

• ln = “natural log”

• If [P] and [R] = 1 mol/L, then E = E˚

• If [R] > [P], then E is LARGER than E˚

• If [R] < [P], then E is smaller than E˚

]reactants[

products][ln

0257.0

n

VEE o

]reactants[

products][ln

0257.0

n

VEE o

QnF

RTEE o ln Q

nF

RTEE o ln

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Zn is anode (-0.76 V) in 0.40 M Zn2+(aq) and Cu is cathode (0.34 V) in 4.8 10-3 M Cu2+(aq) .

Eºcell = Eºcathode– Eºanode = (0.34 V) – (–0.76 V)

= 0.34 + 0.76 = +1.10 VSubstituting

cell

2o

2

0.0257 V [Zn ]E E ln

[Cu ]n

Calculate the cell potential.Calculate the cell potential.

Solution: (Solution: (need standard cell potential, Eneed standard cell potential, Eoocellcell))

]108.4[

]40.0[ln

2

V0257.0V10.1E

3

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)Zn is anode (-0.76 V) and Cu is cathode (0.34 V) EEoo

cellcell = 1.10 V = 1.10 V

n = 2 (Cun = 2 (Cu2+2+ + 2e- = Cu + 2e- = Cu00))

Solution:Solution:

E = 1.10 V – (0.01285)(4.42) = 1.04 V = 1.04 V

2Fe3+(aq) + 3Mg(s) 2Fe(s) + 3Mg2+(aq) Fe3+ = 1.0 10-3 M ; Mg2+ = 2.5 M Calculate the cell potential.

Eºcell = Eºcathode– Eºanode = (-0.036 V) – (–2.37 V)

= -0.036 + 2.37 = +2.33 VSubstituting

23

32o

][

][ln

V0257.0EE cell

Fe

Mg

n

Solution: (need standard cell potential, Eocell)

What is n???

2Fe3+(aq) + 3Mg(s) 2Fe(s) + 3Mg2+(aq) Fe3+ = 1.0 10-3 M ; Mg2+ = 2.5 M Calculate the cell potential.

Eºcell = +2.33 V; n = 6

Substituting

23

3

]100.1[

]5.2[ln

6

V0257.0V33.2E

Solution: (need standard cell potential, Eocell)

564.166

V0257.0V33.2E

Ans = 2.26 VV071.0V33.2E

BATTERIESBATTERIESPrimary, Secondary, and Fuel CellsPrimary, Secondary, and Fuel Cells

Dry Cell BatteryDry Cell Battery

Anode (-)Anode (-)

Zn Zn Zn Zn2+2+ + 2e- + 2e-

Cathode (+)Cathode (+)

2 NH2 NH44++ + 2e- + 2e-

2 NH2 NH33 + H + H22

Primary batteryPrimary battery — uses redox — uses redox reactions that cannot be reactions that cannot be restored by recharge.restored by recharge.

Nearly same reactions as in common dry Nearly same reactions as in common dry cell, but under basic conditions.cell, but under basic conditions.

Alkaline BatteryAlkaline Battery

Anode (-): Anode (-): Zn + 2 OHZn + 2 OH-- ZnO + H ZnO + H22O + 2e-O + 2e-

Cathode (+): Cathode (+): 2 MnO2 MnO22 + H + H22O + 2e- O + 2e-

MnMn22OO33 + 2 OH + 2 OH--

Lead Storage Battery Lead Storage Battery

• Secondary batterySecondary battery

• Uses redox reactions Uses redox reactions that can be reversed.that can be reversed.

• Can be restored by Can be restored by rechargingrecharging

Lead Storage BatteryLead Storage Battery

Anode (-) Anode (-) EEoo = +0.36 V = +0.36 V

Pb + HSOPb + HSO44-- PbSO PbSO44 + H + H++ + 2e- + 2e-

Cathode (+)Cathode (+) EEoo = +1.68 V = +1.68 V

PbOPbO22 + HSO + HSO44-- + 3 H + 3 H++ + 2e- + 2e-

PbSO PbSO44 + 2 H + 2 H22OO

Ni-Cad BatteryNi-Cad Battery

Anode (-)Anode (-)

Cd + 2 OHCd + 2 OH-- Cd(OH) Cd(OH)22 + 2e- + 2e-

Cathode (+) Cathode (+)

NiO(OH) + HNiO(OH) + H22O + e- O + e- Ni(OH) Ni(OH)22 + OH + OH--

Fuel Cells: HFuel Cells: H22 as a Fuel as a Fuel•Fuel cellFuel cell - reactants are - reactants are

supplied continuously from supplied continuously from

an external source.an external source.•Cars can use electricity Cars can use electricity

generated by Hgenerated by H22/O/O22 fuel fuel

cells.cells.

•HH22 carried in tanks or carried in tanks or

generated from generated from

hydrocarbons.hydrocarbons.

Storing HStoring H22 as a Fuel as a Fuel

One way to store HOne way to store H22 is to adsorb the gas onto a is to adsorb the gas onto a

metal or metal alloy. metal or metal alloy.

Hydrogen—Air (OHydrogen—Air (O22) Fuel ) Fuel CellCell

Cathode:Cathode: OO22(g) + 2H(g) + 2H22O(liq) + 4e- O(liq) + 4e- 4OH 4OH- -

(aq) (aq)

--------------------------------------------------------------------

Net:Net: OO22(g) + 2H(g) + 2H22(g) (g) 2H 2H22O(liq) O(liq)

Anode:Anode: 2H 2H22(g) (g) 4H 4H++(aq) + 4e- (aq) + 4e-

ElectrolysisElectrolysis

Using electrical energy to produce chemical change.

Sn2+(aq) + 2 Cl-(aq) Sn(s) + Cl2(g)

Electrolysis of water; electroplating; refining metals; production of chemicals.

ElectrolysisElectrolysisElectric Energy Chemical Change

BATTERY

+

Na+Cl-

Anode Cathode

electrons

BATTERY

+

Na+Cl-

Anode Cathode

electrons• • Electrolysis of Electrolysis of

molten NaCl.molten NaCl.

• • Here a battery Here a battery

“pumps” electrons “pumps” electrons

from Clfrom Cl-- to Na to Na++..

• • NOTE:NOTE: Polarity of Polarity of

electrodes is electrodes is

reversed from reversed from

batteries.batteries.

Electrolysis of Molten NaClElectrolysis of Molten NaCl

Anode (+)

2Cl2Cl--((ll) ) Cl Cl22(g) + 2e- (g) + 2e- (-1.36 V)(-1.36 V)

Cathode (-)

NaNa++((ll) + e- ) + e- Na Na (-2.71 V)(-2.71 V)

BATTERY

+

Na+Cl-

Anode Cathode

electrons

BATTERY

+

Na+Cl-

Anode Cathode

electrons

EEoo for cell (in melted NaCl) = E˚ for cell (in melted NaCl) = E˚cc + E˚ + E˚aa

= - 2.71 V + (-1.36 V)= - 2.71 V + (-1.36 V)

= - 4.07 V (in melted NaCl) = - 4.07 V (in melted NaCl) rxn is nonspontaneousrxn is nonspontaneous

External electrical energy needed because EExternal electrical energy needed because Eoo is (-). is (-).

Electrolysis of Aqueous Electrolysis of Aqueous NaOHNaOH

Anode (+) Anode (+) E° = -0.40 VE° = -0.40 V

4 OH4 OH--

OO22(g) + 2 H(g) + 2 H22O + 4e-O + 4e-

Cathode (-)Cathode (-)

4 H4 H22O + 4e- O + 4e-

2 H2 H22 + 4 OH + 4 OH--

EEoo for cell = -1.23 V for cell = -1.23 V

Electric Energy Electric Energy Chemical Change Chemical ChangeAnodeAnode CathodeCathode

HH22O is more easily reduced O is more easily reduced

than Nathan Na++!! !! (E(Eoo -2.71 V)-2.71 V)

NaOH + H2O Na+(aq) + OH-(aq)

Electrolysis of Aqueous NaClElectrolysis of Aqueous NaCl

Anode (+) Anode (+) 2 Cl2 Cl--

ClCl22(g) + 2e(g) + 2e-- E° =-1.36 V E° =-1.36 V

Cathode (-) Cathode (-)

2 H2 H22O + 2e- O + 2e-

HH22 + 2 OH + 2 OH--


Note that HNote that H22O is more O is more

Easily reduced than NaEasily reduced than Na++((E° = -2.71 V)E° = -2.71 V)

2H2H22O(l) O(l) OO22(g) + 4H(g) + 4H++ + 4e + 4e--

BATTERY

+

Na+Cl-

Anode Cathode

H2O

electrons

BATTERY

+

Na+Cl-

Anode Cathode

H2O

electrons

Also, ClAlso, Cl-- is oxidized in preference to H is oxidized in preference to H22O O

because of kinetics (because of kinetics (overvoltageovervoltage))E°E°redred < -1.23 V, may be < -1.23 V, may be down to ~ -2.00 Vdown to ~ -2.00 V

Also, ClAlso, Cl-- is oxidized in preference to H is oxidized in preference to H22O O

because of kinetics (because of kinetics (overvoltageovervoltage))E°E°redred < -1.23 V, may be < -1.23 V, may be down to ~ -2.00 Vdown to ~ -2.00 V

NaCl + H2O Na+(aq) + Cl-(aq)

EEoo and Thermodynamics and Thermodynamics

• EEoo is related to ∆G is related to ∆Goo, the free energy , the free energy change for the reaction.change for the reaction.

• ∆∆G˚ proportional to –nE˚G˚ proportional to –nE˚∆∆GGoo = -nFE = -nFEoo

where F = where F = Faraday constantFaraday constant

= 9.6485 x 10= 9.6485 x 1044 J/V•mol of e J/V•mol of e--

(or 9.6485 (or 9.6485 10 1044 coulombs/mol) coulombs/mol)and n is the number of moles of electrons and n is the number of moles of electrons

transferred.transferred.

Electrolysis of Aqueous CuClElectrolysis of Aqueous CuCl22

Anode (+) Anode (+)

2 Cl2 Cl-- Cl Cl22(g) + 2e-(g) + 2e-

Cathode (-) Cathode (-)

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


Note that Cu is more easily Note that Cu is more easily

reduced than either Hreduced than either H22O or O or

NaNa++ (check redox potentials). (check redox potentials).

BATTERY

+

Cu2+Cl-

Anode Cathode

electrons

H2O

BATTERY

+

Cu2+Cl-

Anode Cathode

electrons

H2O

CuCl2 + H2O Cu2+(aq) + 2Cl-(aq)

Calculate Calculate GGoo for the reaction, for the reaction,ZnZn2+2+(aq) + Ni(s) (aq) + Ni(s) Zn(s) + Ni Zn(s) + Ni2+2+(aq) (aq)

Solution: use Solution: use GGoo = -nFE° = -nFE°no. of electrons, n = 2no. of electrons, n = 2F = 9.6485 F = 9.6485 10 1044 C C

ZnZn2+2+(aq) + 2e- (aq) + 2e- Zn(s), -0.763 V Zn(s), -0.763 V

Ni(s) Ni(s) Ni Ni2+2+(aq) + 2e-, +0.25 V(aq) + 2e-, +0.25 V

cathodecathode

anodeanode

need Eneed Eoocellcell

EEoocellcell = E = Ecathodecathode – (E – (Eanodeanode))

EEoocellcell = -0.763 – (-0.25 V) = -0.51 V = -0.763 – (-0.25 V) = -0.51 V

GGoo = (-2 = (-2 96485 J/V 96485 J/V -0.51 V)-0.51 V) 1 kJ/1000 J1 kJ/1000 J

98 kJ98 kJ∆∆Go > 0, reaction is nonspontaneousGo > 0, reaction is nonspontaneous

Reagents are favoredReagents are favored

EEoo and ∆G and ∆Goo

∆∆GGoo = - n F E = - n F Eoo For a For a product-favoredproduct-favored reaction reaction

Reactants Reactants Products Products

∆∆GGo o < 0 and so E < 0 and so Eo o > 0 > 0

EEoo is positive is positive

For a For a reactant-favoredreactant-favored reaction reaction

Reactants Reactants Products Products

∆∆GGo o > 0 and so E > 0 and so Eo o < 0 < 0

EEoo is negative is negative

72

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

G° < 0 (negative)E° > 0 (positive)K > 1 (large)

QnF

RTEE cell ln0 Q

nF

RTEE cell ln0

When Ecell = 0 (no net rxn), reactants and products are at equilibrium……and Q = K

KnE ocell

ln0257.0

KnE ocell

ln0257.0

Kn

Ecell ln0257.00 Kn

Ecell ln0257.00

Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

Consider electrolysis of aqueous silver ion.Consider electrolysis of aqueous silver ion.

AgAg++ (aq) + e- (aq) + e- Ag(s) Ag(s)

1 mol e-1 mol e- 1 mol Ag 1 mol Ag

If we could measure the moles of e-, we could If we could measure the moles of e-, we could know the quantity of Ag formed.know the quantity of Ag formed.

But do we measure moles of e-?But do we measure moles of e-?

Current = charge passing

timeCurrent =

charge passingtime

I (amps) = coulombsseconds


ButBut , , how is charge related to moles of electrons?how is charge related to moles of electrons?

Current = charge passing

timeCurrent =

charge passingtime



= = 96,500 C/mol e- 96,500 C/mol e- = = 1 Faraday1 Faraday

Michael FaradayMichael Faraday1791-18671791-1867C96,500

emol1

emol1

C96,500 or

Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

Quantitative Aspects of Electrochemistry

1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min.(aq) solution for 15.0 min.

What mass of Ag metal is deposited?What mass of Ag metal is deposited?

SolutionSolution

(a)(a) Calc. chargeCalc. charge

Charge (C) = current (A) x time (t)Charge (C) = current (A) x time (t)

= (1.5 amps)(15.0 min)(60 s/min) = 1350 C= (1.5 amps)(15.0 min)(60 s/min) = 1350 C



Quantitative Aspects of ElectrochemistryQuantitative Aspects of Electrochemistry

SolutionSolution

(a)(a) Charge = 1350 CCharge = 1350 C

(b)(b) Calculate moles of e- usedCalculate moles of e- used



1350 C • 1 mol e -96, 500 C

0.0140 mol e -1350 C • 1 mol e -96, 500 C

0.0140 mol e -

Ag g 1.51or Ag mol 0.0140 -e mol 1

Ag mol 1 • -e mol .01400 Ag g 1.51or Ag mol 0.0140

-e mol 1

Ag mol 1 • -e mol .01400

1.50 amps flow thru a Ag1.50 amps flow thru a Ag++(aq) solution for 15.0 min. What mass of Ag metal is (aq) solution for 15.0 min. What mass of Ag metal is

deposited? deposited? AgAg++ + e- + e- Ag(s) Ag(s)

(c)(c) Calc. quantity of AgCalc. quantity of Ag

Quantitative Aspects of Electrochemistry

The anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(s) + HSOPb(s) + HSO44--(aq) (aq) PbSO PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-

If a battery delivers 1.50 amp, and there is 454 g of Pb, how long will If a battery delivers 1.50 amp, and there is 454 g of Pb, how long will the battery last?the battery last?

Solution Solution

a)a) 454 g Pb = 2.19 mol Pb 454 g Pb = 2.19 mol Pb

b)b) Calculate moles of e-Calculate moles of e-

each Pb atom is loosing 2eeach Pb atom is loosing 2e−−

2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -

c)c) Calculate chargeCalculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C4.38 mol e- • 96,500 C/mol e- = 423,000 C

Quantitative Aspects of ElectrochemistryQuantitative Aspects of ElectrochemistryThe anode reaction in a lead storage battery isThe anode reaction in a lead storage battery is

Pb(s) + HSOPb(s) + HSO44--(aq) (aq) PbSO PbSO44(s) + H(s) + H++(aq) + 2e-(aq) + 2e-

If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery If a battery delivers 1.50 amp, and you have 454 g of Pb, how long will the battery last?last?

SolutionSolution

a)a) 454 g Pb = 2.19 mol Pb454 g Pb = 2.19 mol Pb

b)b) Mol of e- = 4.38 molMol of e- = 4.38 mol

c)c)Charge = 423,000 CCharge = 423,000 C

Time (s) = Charge (C)

I (amps)Time (s) =

Charge (C)I (amps)

Time (s) = 423,000 C1.50 amp

= 282,000 sTime (s) = 423,000 C1.50 amp

= 282,000 s About 78 hoursAbout 78 hours

d)d) Calculate timeCalculate time

Chapter 18 Electrochemistry. 2 GOALS Review: oxidation states oxidation/reduction oxidizing/reducing...

Documents

Transcript of Chapter 18 Electrochemistry. 2 GOALS Review: oxidation states oxidation/reduction oxidizing/reducing...