Chapter 20 B and L notes.ppt

(1)

Electrochemistry

(2)

19.1

Electrochemical processes are oxidation-reduction reactions in which:

• the energy released by a spontaneous reaction is converted to electricity or

• electrical energy is used to cause a nonspontaneous reaction to occur

(3)

Assigning Oxidation Numbers

The charge the atom would have in a molecule (or an

ionic compound) if electrons were completely transferred. These are somewhat hypothetical, keeps track of electrons.

1. Free elements (uncombined state) or natural state have an oxidation number of zero.

Na, Be, K, Pb, H

₂

, O

₂

, P

₄

=

0

2. In monatomic ions, the oxidation number is equal to the charge on the ion.

Li

+

, Li =

+1; Fe

3+

, Fe =

+3; O

2-

, O =

-2

(4)

5. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In

these cases, its oxidation number is –1. LiH = H =-1

6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the

molecule or ion.

HCO

3

-O =

-2

H =

+1

3x(-2) +

1

+

?

= -1

C =

+4

Oxidation numbers of

all the atoms in HCO₃- ?

(5)

Practice Problems:

Figure out the oxidation numbers of each

element in the following substances:

•Ca(OH)

₂

6. H

₂

O

•Cl

₂

7. K

₂

CrO

₄

•Fe

2+

8. PbS

2

•H

₂

C

₂

O

₄

9. H

₂

O

₂

•MnO

₄-

10. OF

(6)

You Try These:

1. Na

2. Cl

-3. I

₂

4. HNO

₃

5. SnO

₂

6. P

₂

O

₅

7. NH

₄+

(7)

Electrochemistry Terminology #1



Oxidation

–

A process in which an element

attains a more positive oxidation state by

losing e

-

(charge increases)

Na(s)



Na

+

+ e

-



Reduction

– A process in which an element

attains a more negative oxidation state by

gaining e

-

(charge decreases)

(8)

G

ain

E

lectrons

=

R

eduction

An old memory device for oxidation

and reduction goes like this…

LEO

says

GER

L

ose

E

lectrons

=

O

xidation

(9)

Identifying Redox Rxns

Given: Cu

(s)

+ AgNO

₃

(aq)



Ag

(s)

+ Cu(NO

₃

)

₂

Identify the reduction: (GER)

Ag+



Ag

*Charge on Ag decreased or was reduced

Identify the oxidation: (LEO)

Cu



Cu

2+

(10)

You Try:

In the following Redox rxns, identify the element

being oxidized and the element being reduced.

(Use the half-reaction notation.)

1.Fe

(s)

+ CuCl

₂

(aq)

 FeCl

₂

(aq)

+ Cu

(s)

Red: Cu

+2

+ 2e

-

 Cu

Ox: Fe  Fe

+2

+ 2e

-•Zn(s) + H

₂

SO

₄

(aq)

 H

₂

(g)

+ ZnSO

₄

(aq)

Red: 2H

+

+ 2e

-

 H

2

(11)

-H

₂

(g) + O

₂

(g)

H

₂

O (g)

H

₂

2H

+

+ 2e

-O

₂

+ 4e

-

2O

2-Oxidation

half-reaction (lose e

-

)

Reduction

half-reaction (gain e

-

)

Balancing Redox Reactions using the half-reaction method: 1. Always start by assigning oxidation states to each

element in the reaction.

2. Look for differences (which were oxidized/reduced) 3. Separate into half reactions and balance atoms and e

-4. Balance e- and sum together (e- should cancel out!)

0 0

(12)

2-Balancing Redox Equations

The oxidation of Fe2+ to Fe3+ by Cr

2O72- in acid solution?

1. Separate the equation into two half-reactions: Red and Ox.

Oxidation:

Cr₂O₇2- Cr3+

+6 +3

Reduction:

Fe+2 2+ Fe+3 3+

2. Balance the atoms other than O and H in each half-reaction in the usual fashion (use coefficients).

Cr₂O₇2- 2Cr3+

Fe2+ + Cr

(13)

Balancing Redox Equations

3. For reactions in acid, add H₂O to balance O atoms, then H+

to balance H atoms.

Cr₂O₇2- 2Cr3+ + 7H 2O

14H+ + Cr

2O72- 2Cr3+ + 7H2O

4. Add electrons to one side of each half-reaction to balance the charges on the half-reaction (this is ion charges, not

oxidation states! _Fe₂₊_Fe₃₊_{+ 1e}

-6e- + 14H+ + Cr

2O72- 2Cr3+ + 7H2O

5. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by

appropriate coefficients to cancel out electrons.

6Fe2+ 6Fe3+ + 6e

-6e- + 14H+ + Cr

(14)

Balancing Redox Equations

6. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel.

6e- + 14H+ + Cr

2O72- 2Cr3+ + 7H2O

6Fe2+ 6Fe3+ + 6e

-Oxidation:

Reduction:

14H+ + Cr

2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O

7. Verify that the number of atoms and the charges are balanced. (You can do this by inspection)

14x1 – 2 + 6x2 = 24 = 6x3 + 2x3

8. For reactions in basic solutions, add OH- to both sides of

the equation for every H+ that appears in the final equation.

(15)

Balance the following Redox rxn

in basic solution:

I

-

(aq)

+ MnO

4-

(aq)



I

2

(aq)

+ MnO

2

(s)

Answer:

6I-(aq)+ 4H

(16)

You Try:

Balance the following redox reaction in

acidic solution using the half reaction

method.

Cu

(s)

+ NO

₃-

(aq)

 Cu

2+

(aq)

+ NO

2

(g)

Answer:

2NO

₃-

+ Cu + 4H

+



Cu

2+

+ 2NO

(17)



Oxidizing agent

The substance that is reduced is

the oxidizing agent



Reducing agent

(18)

Galvanic Cells: http://www.youtube.com/watch?v=C26pH8kC_Wk

spontaneous redox reaction anode

oxidation

(19)



Anode

The electrode

where

oxidation

occurs



Cathode

The electrode

where

reduction

occurs

Memory device:

Red

uction

at the

Cat

hode

(20)

Rxn:

Zn

(s)

+ Cu

2+

(aq)



Cu

(s)

+ Zn

2+

(aq)

1. To draw cell, first figure out which is oxidized and which is reduced.

1. Remember that

oxidation occurs at anode and

(21)

Measuring

Standard

Electrode

Potential

(E°

_cell_cell

)

Potentials are measured against a hydrogen ion

reduction reaction, which is arbitrarily

assigned a potential of

(22)

Table of Table of Reduction Reduction Potentials Potentials Measured Measured against against the the Standard Standard Hydrogen Hydrogen Electrode Electrode

(23)

Galvanic/Voltaic

(Electrochemical) Cells

Spontaneous redox

processes have:

1. A positive

cell potential, E

0

2. A negative free

energy change, (-



G)

*All Voltaic cells are

considered

spontaneous

(24)

Zn - Cu

Galvanic

Cell

Example

Zn

2+

+ 2e

-



Zn E = -0.76V

Cu

2+

+ 2e

-



Cu E = +0.34V

From a table

of reduction

potentials:

(25)

Zn - Cu

Galvanic

Cell

Cu

2+

+ 2e

-



Cu E = +0.34V

The

The less positive less positive

reduction potential reduction potential

becomes the becomes the

oxidation

oxidation and and more more positive or higher

positive or higher

reduction potential reduction potential

becomes

becomes reductionreduction

Zn



Zn

2+

+ 2e

-

E = +0.76V

Zn + Cu

2+



Zn

2+

+ Cu E

0

= + 1.10 V

(26)

Line

Notation

Zn(

s

) | Zn

2+2+

(

aq

) || Cu

2+2+

(

aq

) | Cu(

s

)

An abbreviated

representation of

an electrochemical

cell

Anode Anode solution solution Anode Anode material

(27)

Standard Reduction Potentials

19.3

Pt (s) | H₂ (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)

2e- + Cu2+ (1 M) Cu (s)

H₂ (1 atm) 2H+ (1 M) + 2e

-Anode (oxidation):

Cathode (reduction):

H₂ (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)

E0 = E

cathode - Eanode

cell 0 0

E0 = 0.34 V

cell

E_cell = E_{Cu /Cu}2+ – E_{H /H}+

2 0 0 0

0.34 = E0_{Cu /Cu}₂₊ - 0

(28)

19.3

• E0 is for the reaction as

written

• The more positive E0 the

greater the tendency for the substance to be reduced

• The half-cell reactions are reversible

• The sign of E0 changes

when the reaction is reversed

• Changing the stoichiometric coefficients of a half-cell

(29)

What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO₃)₂ solution and a Cr electrode in a 1.0 M Cr(NO₃)₃ solution?

Cd2+ ₍_aq₎ + 2e- Cd ₍_s₎ E0 = -0.40 V

Cr3+ ₍_aq₎ + 3e- Cr ₍_s₎ E0 = -0.74 V

Cd is the stronger oxidizer Cd will oxidize Cr

2e- + Cd2+ (1 M) Cd (s)

Cr (s) Cr3+ (1 M) + 3e

-Anode (oxidation):

Cathode (reduction):

2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)

x 2

x 3

E0 = E

cathode - Eanode cell 0 0

E0 = -0.40 – (-0.74)

cell

E0 = 0.34 V

cell

(30)

Calculating



G

00

for a Cell

(a measure of spontaneity)

)

10

.

1

)(

500

96

)(

2

(

0

Coulomb

Joules

e

mol

coulombs

e

mol

G





 _



G

00

= -nFE

00

n

= moles of electrons in balanced redox equation

F

= Faraday constant = 96,500 coulombs/mol e

-

-Zn + Cu

2+



Zn

2+

+ Cu

E

0

= + 1.10 V

kJ

Joules

G

0





212267





212



(31)

Spontaneity of Redox Reactions

(32)

(33)

Will the following reaction occur spontaneously at 250C?

Fe2+ ₍_aq₎ + Cd ₍_s₎ Fe ₍_s₎ + Cd2+ ₍_aq₎

2e- + Fe2+ 2Fe

Cd Cd2+ + 2e

-Oxidation:

Reduction:

n = 2

E0 = -0.44

E0 = -0.04 V

G = 7720 J or 7.720 kJ Non-spontaneous

19.5

G

0



nFE

0

E0 = 0.40

(34)

(35)

(36)

The Nernst Equation

(not tested this year)

)

ln(

0

Q

nF

RT

E





Standard potentials assume a concentration of

Standard potentials assume a concentration of

1 M. The Nernst equation allows us to calculate

potential when the two cells are not 1.0 M.

R

= 8.31 J/(mol



K)

T

= Temperature in K

n

= moles of electrons in balanced redox equation

F

(37)

-Equilibrium Constants and Cell Potential

Equilibrium Constants and Cell Potential

At equilibrium, forward and reverse

reactions occur at equal rates, therefore:

1.

The battery is

“

dead

”

2. The cell potential,

E

, is zero volts

Use Le Chatelier’s Principal to predict changes in

voltage:

(38)

Concentration

Cell

Determine which side undergoes oxidation,

and which side undergoes reduction.

Both sides have the same

components but at different concentrations.

???

(39)

Concentration

Cell

Both sides have the same

components but at different concentrations.

The 1.0 M Zn

2+

must decrease in concentration, and

the 0.10 M Zn

2+

must increase in concentration

Zn

2+

(1.0M) + 2e

-



Zn (reduction)

Zn



Zn

2+

(0.10M) + 2e

-

(oxidation)

??? ??? Cathode Cathode Anode Anode

(40)

Use Le Chat to predict:

Given the following voltaic cell:

Ag

+

+ e

-



Ag

E° = +0.80 V

Cu

2+

+ 2e

-



Cu

E° = +0.34 V

•Predict the reaction that will occur.

•If water is added to the Cu

2+

cell, how will

voltage be affected?

•If Cl

-

is added to the Ag

+

cell to create AgCl,

how will voltage be affected?

(41)

You Try: Predict:

If the reduction of mercury (I) is desired:

Hg

₂2+

+ 2e

-



2Hg(s)

E° = +0.80 V

1. Which of the following could be used as

the anode?

Br

₂

+ 2e

-



2Br

-

E° = +1.07 V

Zn

2+

+ 2e

-



Zn

E° = -0.76 V

•Write the reaction that occurs.

•If water is added to the solution of Hg

₂2+

,

how is voltage affected?

(42)

Remember…

Galvanic (Electrochemical) Cells

Spontaneous redox

processes have:

A positive

cell potential, E

0

A negative free

(43)

Electrolytic

Processes

A negative

cell potential, (-E

0

)

A positive free

energy change, (+



G)

Electrolytic

processes are

(44)

(45)

Why Electrolysis?

• Unlike electrochemical cells that

spontaneously supply energy, electrolysis

uses

energy to cause a non-spontaneous

(46)

19.8

Electrolysis is the process in which electrical energy is

used to cause a nonspontaneous chemical reaction to occur.

(47)

How to determine what gets reduced or

oxidized by electrolysis in an aqueous

solution

• Metal ions or water can be reduced.

• When electrolysis occurs in aqueous solutions,

if the metal has a reduction potential more

negative (smaller) than -0.8, then only water

is reduced because water has the larger E

o

red

.

• Negative ions will be oxidized.

• If there is a mixture of metal ions, the

metals will be reduced in order of

largest E

o red

(48)

(49)

Predicting the Products of

Electrolysis

If there is no water present and you

have a pure molten ionic compound,

then:

–

the cation will be reduced

– (gain electrons/go down in charge)

–

the anion will be oxidized

– (lose electrons/go up in charge)

Na

+

+ e- Na (

s

)

2Cl

-

Cl

(50)

(51)

Electrolysis and Mass Changes charge (C) = current (A) x time (s)

1 mole e- = 96,500 C

(52)

How much Ca will be produced in an electrolytic cell of molten CaCl₂ if a current of 0.452 A is passed through the cell for 1.5 hours?

Anode:

Cathode: _Ca2+ _(l) + 2e- Ca _(s)

2Cl- (l) Cl

2 (g) + 2e

-Ca2+ (l) + 2Cl- (l) Ca (s) + Cl

2 (g)

2 mole e- = 1 mole Ca

mol Ca = C

s

1.5 hr x 3600 _hrs

96,500 C 1 mol e

-x

2 mol e

-1 mol Ca x

= 0.0126 mol Ca _{= 0.50 g Ca} x 0.452

(53)

More Practice:

Calculate the number of grams of aluminum

produced in 2.00 hours by electrolysis of molten

AlCl

₃

if a current of 12.0 A is applied

.

8.05 g Al

The half-reaction for the formation of magnesium

metal upon electrolysis of molten MgCl

₂

is: Mg

2+

+

2e

-



Mg. How many seconds would be required

to produce 25.0 g of Mg from MgCl

₂

if the current

is 50.0 A?

(54)

Electroplating of

Silver

Anode reaction

:

Ag



Ag

++

+ e

- -

Electroplating requirements

:

1. Solution of the plating metal

3. Cathode with the object to be plated

2. Anode made of the plating metal

4. Source of current

Cathode reaction

:

Ag

(55)

Solving an Electroplating Problem

Q: How many seconds will it take to plate out

5.0 grams of silver from a solution of AgNO

₃₃

using a 20.0 Ampere current?

5.0 g

Ag

++

+ e

- -



Ag

1 mol Ag

107.87 g

1 mol e

-1 mol Ag

96 500

C

1 mol e

-1 s

20.0 C

= 2.2 x 10

(56)

(57)

http://www.youtube.com/watch?v=C26pH8kC_Wk