electrochemistry - dr. marten’s ap chemistry cell galvanic cell voltaic cell by separating the...

Electrochemistry

Thursday, April 2, 2009

All electrochemical processes involve the transfer of e-’s from one substance to another

and are therefore redox reactions.

(the reverse) use electrical current to drive chemical reactions that would otherwise be

nonspontaneous (eg. electroplating)

use spontaneous chemical reactions to generate electric current (eg. battery)

or

Electrochemistry


Redox Reactions

key terms:oxidizing agentreducing agentoxidation number


Consider the oxidation-reduction reaction

CuCu2+ (aq) + Zn(s) + Zn2+ (aq) (s)

Cu2+ electronsoxidation number ; is

Zn0 electronsoxidation number ; is

is the agent

is the agent

gains 2decreases reduced

oxidizing

loses 2increases oxidized

reducing

______

________

_____________

______________

_____________

________

________


In studying a redox reaction we often think of it as two half reactions.

oxidation-reduction reactions

Cu2+(aq) + Zn(s) + Zn2+(aq)Cu (s)

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

Cu2+(aq) Cu (s)+ 2e- reduction


Balancing Oxidation-Reduction Equations

Half-Reaction Method


Half-Reaction Method in Acid

1. Write the unbalanced equation in ionic form.

2. Separate the equation into two half-reactions.

3. Balance each half reaction (except for O and H).

5. Balance the charges by adding electrons.

4. In acid solution, balance O by adding H2O and H by adding H+

6. Add the half reactions

7. Check to make sure atoms and charges are balanced


Example Balance the following equation for the reaction in

acid solution. All species are (aq).MnO4

- + Fe2+ Mn2+ + Fe3+

1. Write the unbalanced equation in ionic form.

MnO4- + Fe2+ Mn2+ + Fe3+

2. Separate the equation into two half-reactionsFe2+ Fe3+

MnO4- Mn2+

3. Balance each half reaction (except for O and H).4. Balance O by adding H2O and H by adding H+

Fe2+ Fe3+

MnO4- Mn2+ 4H2O+8H++


5. Balance the charges by adding electrons.

Fe2+ Fe3+ + 1e-

MnO4- Mn2+8H++ 4H2O++ 5e-

5 x1 x

You need the same number of electrons on both sides of the equation. (

( ))

MnO4-

Mn2+

8H++

4H2O ++

+ 5Fe2+

5Fe3+

5Fe2+ 5Fe3+ + 5e-

MnO4- Mn2+8H++ 4H2O++ 5e-

6. Add the half reactions

This is the balanced equationThursday, April 2, 2009

Half-Reaction Method in Base

1. Use the half-reaction method as specified for acidic solutions to obtain the final balanced equation as if H+

ions were present.

2. Add the number of OH- ions to both sides of the equation to turn the remaining H+ ions to H2O

3. Eliminate waters that appear on both sides of the equation.


Example

Balance the following equation for the reaction in basic solution. All species are (aq).

+ H+

2H+ + + H2O

+ 2e-

2e- +

HS- + NO3- S + NO2

-

HS- S

NO3- NO2

-


Example

HS- S

NO3- NO2

-H+ + + H2O

HS- + NO3- S + NO2

-

NO3-H+ +HS- + NO2- + H2O+ SOH- + + OH-



Example

HS- S

NO3- NO2

-H+ + + H2O

HS- + NO3- S + NO2

-

NO3-HS- + NO2- + H2O+ S + OH-H2O +

NO3-HS- + NO2

- + S + OH-



Electrochemical Cells


Electrochemical cell Galvanic cell

Voltaic cell

by separating the oxidizing agent from the reducing agent electrons are transferred via an

external conducting medium

a device in which electricity is produced by a spontaneous redox reaction


Construct a galvanic cell based on the oxidation-reduction reaction


goal is to generate an electrical current

demo: CuSO4 (aq) + Zn (s)


Zn

Cu2+

Cu2+

Cu2+

Cu2+

SO42-

SO42-

SO42-

SO42-


Zn

SO42-

SO42-

SO42-

Cu2+ Cu2+SO42-

Zn2+

Zn2+

electrons flow but no useful current is generated

Cu


SO42-

SO42-

Cu2+

Zn

Consider the same two species in separate vessels

- - -

-

-

Cu2+

SO42-

Cu2+


SO42-

SO42-

Cu2+

Cu

Zn

Zn2+

Consider the same two species in separate vessels

- - -

SO42-

Cu2+


SO42-

SO42-

Cu2+

Cu

Zn

Zn2+

Consider the same two species in separate vesselselectrons soon stop flowing

SO42-

Cu

Zn2+


K

SO42-

SO42-

Cu2+

Cu

Zn

Zn2+

Connect the vessels by a salt bridge

SO42-Zn2+

Cu


SO42-

SO42-

Cu2+

Cu

Zn

Zn2+

or a porous disk

SO42-Zn2+

Cu


a salt bridgecompletes the electric circuit

composed of an inert electrolyte (KCl, NH4NO3) whose ions will not react with the other ions or

electrodes in the cell

K+


K+

a salt bridgecompletes the electric circuit

composed of an inert electrolyte (KCl, NH4NO3) whose ions will not react with the other ions or

electrodes in the cell

Cl-K+

K+

Cl-Cl-


SO42-

SO42-

Cu2+

Cu

Zn

Zn2+

a salt bridge

SO42-Zn2+

Cu

K

Cl- K+ Cl-K+

Cl-

K+

Cl-

K+


Ion flow balances charges

SO42-

Cu2+

Cu

Zn

Zn2+Cl-

K+

K+

Cl-

SO42-Zn2+

Cu

K

Cl- K+


Ion flow balances charges and permits electron flow

Cu2+

Zn

Zn2+

- - - --

Cl-K+

Cl-

Zn2+ K+ SO42-

SO42-

Cu

Cu

K

Cl- K+


Zn

- - - --

Galvanic cellAnode Cathodewhere

__________ occurs

where ___________

occurs

SO42-

Cu2+

CuZn2+

Cl-K+

Cu

Cl-

Zn2+ K+ SO42-

oxidation reduction

K

Cl- K+


Cell diagram

Zn(s) Cu2+(aq)

Anode Cathode

Zn2+(aq) Cu (s)

phase boundarysalt bridge

phase boundary

By convention the anode is written first. The other components appear as you would encounter them moving to

the cathode.


Cell potential

also called electromotive force (emf)

results from difference in energy of an electron at the two electrodes

symbol is script E (εcell)

unit is volt (V)

1 V = 1 joule/coulomb


SO42- SO4

2-Cu2+

Cu

Zn

Cell potential = 1.10 V

Recall the galvanic

cell

Zn2+

when all species are in their standard state

K


Standard Electrode Potentials (ε º)

just as the overall cell reaction can be thought of as the sum of two half-cell

reactions

It is impossible however to measure the potential of a single electrode.

the measured emf of the cell can be treated as the sum of the electrical

potentials.

ε ºcell ε ºox ε ºred+=


The standard hydrogen electrode*

*H2 pressure = 1atm; [HCl] = 1M

reference all reduction half-reactions to:2H+(aq ) + 2e- H2(g )

is assigned the arbitrary potential value of 0 voltsε º = 0.0 V


represented by the cell diagram:

Zn(s) H+(aq, 1 M)Zn2+(aq, 1 M) H2(g, 1 atm ) Pt(s)ε º = 0.76 Vhas a standard cell potential

2H+(aq, 1 M ) + 2e- H2(g, 1 atm ) ε º = 0.0 Vby convention:

Zn(s) Zn 2+(aq, 1 M) + 2e- ε º = 0.76 V

0.76 V is the standard oxidation potential for Zn/Zn2+

ε ºox for Zn→ Zn2+

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


therefore: 0.76V ε ºox 0 V+=


the standard oxidation potential for the half-reaction:

Zn(s) Zn 2+(aq, 1 M ) + 2e- ε º = 0.76 V

the custom is to tabulate values for half-reactions written as reductions

ε º

the standard reduction potential for the half-reaction:

Zn(s) Zn 2+(aq, 1 M ) + 2e- ε º = - 0.76 V


SO42- SO4

2-Cu2+

Cu

Zn

Cell potential = 1.10 V

Recall the galvanic

cell

Zn2+

when all species are in their standard state Thursday, April 2, 2009

use the measured standard cell potential and the standard reduction

potential of Zn2+ to calculate the standard reduction potential of Cu2+



Zn(s) Cu2+(aq)Zn2+(aq) Cu (s)

Cu2+(aq) + Zn(s) + Zn2+(aq)Cu (s)

ε º = 1.10 V

Zn(s) Zn 2+(aq) + 2e- ε º = 0.76 V

Zn(s) Zn 2+(aq) + 2e- ε º = -0.76 V

Cu2+(aq) + 2e- Cu (s)Therefore:

ε º = 1.10 V - 0.76 V = + 0.34 V


Standard Reduction Potentials

H2(g) 2H+(aq) + 2e- 0.00 VZn(s) Zn2+(aq) + 2e- -0.76 V

Cu(s) Cu2+(aq) + 2e- +0.34 V

Li(s) Li +(aq) + 1e- -3.05 V

2F-(aq) F2(g) + 2e- +2.87 V

Better reducing agent. Recall Li metal reacts violently in water to give up an e-. So it likes to do the opposite of the reaction shown.

Better oxidizing agent. Recall F is most electronegative element and “likes” to be “-”. ε º


What are the half- reactions?

Can Zn2+ (aq ) oxidize Sn under standard state conditions?

Example

Zn2+(aq) + Sn(s) + Sn2+(aq)Zn (s)

Zn(s) Zn2+(aq) + 2e- -0.76 Vε ºSn(s) Sn2+(aq) + 2e- -0.14 Vε º

Zn2+ is a worse oxidizing agent than Sn2+;Zn2+ will not oxidize Sn

recall: the more positive the value of , the better the oxidizing agent

ε º


Zn(s) Zn2+(aq) + 2e- -0.76 V

ε º

Standard Reduction Potentials

Better reducing agent

Li(s) Li +(aq) + 2e- -3.05 V

H2(g) 2H+(aq) + 2e- 0.00 V

Sn(s) Sn2+(aq) + 2e- - 0.14 V

Better oxidizing agent


If the sign is negative; the reaction is not spontaneous in the direction written.

Example: what is the cell potential?

Zn(s) Zn2+(aq) + 2e- -0.76 V

Sn(s) Sn2+(aq) + 2e- +0.14 V

ε º

Zn2+(aq) + Sn(s) + Sn2+(aq)Zn (s)

-0.62 V


Example: What is the the standard emf of a galvanic 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?

Cr Cd

1M Cr(NO3)3 1M Cd(NO3)2


Example: What is the the standard emf of a galvanic 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?

ε º =Cr(s) Cr3+(aq) + 3e- -0.74 VCd(s) Cd2+(aq) + 2e- -0.40 Vε º =

2. ε ºcell must be (+) for the reaction to go. So reverse the equation with more negative εº, multiply the half-reactions so e-’s cancel (but not the εº’s) and add

ε º =2(Cr(s) Cr3+(aq) + 3e-) +0.74 VCd(s)) 3(Cd2+(aq) + 2e- -0.40 Vε º =

0.34 Vε º =3Cd2+(aq)2Cr(s) + 2Cr3+(aq) + 3Cd(s)

1. Write down the half-reactions as reductions (we don’t know which will be a reduction and which an oxidation yet)


0.34 Vεº =

3Cd2+(aq)2Cr(s) + 2Cr3+(aq) + 3Cd(s)

εº is an intensive property. (Doesn’t matter how much of each substance you have. The

voltage is the same.)

so multiplying the reduction of Cd2+ by 3 and the oxidation of Cr by 2 has no

effect on the value of the voltage! Thursday, April 2, 2009

Cr

emf = 0.34 V

Cd

e-

1M Cr(NO3)3 1M Cd(NO3)2

anode cathode

Which way do the e-’s go and which metal is anode, cathode?


Spontaneity of Redox Reactions

relationship between thermodynamics and electrochemistry


Cell potential, Electric work, Free energy

the work that can be accomplished when electrons are transferred through a wire depending on the “push”

volts = 1 Joule 1 coulomb

- wqε = w = - q εthus

(volts i.e. EMF)



the work that can be accomplished when electrons are transferred through a wire depends on the “push”

- wqε = w = qεthus

(volts i.e. EMF)

the minus sign represents electrical energy leaving the system



1 mol of electrons has a charge of 96485 Coulombs

w = ΔG = -qε ΔG = -qε = -n F ε

and ΔG º = -n F ε º

the cell potential is directly related to the Free Energy difference between reactants and products

= 1 Faraday (F )Cmol96485


ΔG is negative when ε is positive; the reaction is spontaneous

ΔG º = -n F ε º

at 25 º C(298 K)

0.0592n

log Kε º =

= -RTln K



The Effect of Concentration on Cell EMF


Cu2+Cu

Zn

Zn2+


Cell potential and concentration

What would Le Chatelier say should happen if you did the following?

Add more Cu2+ ion?

Add more Zn2+ ion?

Reaction shifts to form more products.

Reaction shifts to form more reactants.


the dependence of the cell potential on concentration results directly from the

dependence of free energy on concentration

ΔG = ΔG° + RT ln(Q )

ΔG = ΔG° + RT ln [products][reactants]

Cell potential and concentration


The Nernst Equation

ΔG = ΔG º + RTln (Q)

at 25 º C:

-n F ε = -n F ε º + RTln (Q)

ε = ε ºn F

RTln (Q) -

ε = ε ºn

0.0592log (Q) -

ΔG º = -n F ε º

ΔG = -n F ε


The Nernst Equation

at equilibrium

ε = ε ºn

0.0592log (Q) -

ε = 0

0.0592n

log Kε º =

Q = K


Practice Exercise

Fe2+(aq ) + 2Ag(s) Fe(s ) + 2Ag+(aq)

What is K for the reaction shown at 25 ºC?

Fe2+(aq ) + 2e- Fe(s ) ε º = -0.44 V

2Ag(s) 2Ag+(aq) + 2e- ε º = -0.80 V

Fe2+(aq ) + 2Ag(s) Fe(s ) + 2Ag+(aq)

ε º = -1.24 V


Fe2+(aq ) + 2Ag(s) Fe(s ) + 2Ag+(aq)

ε º = -1.24 V

0.0592n

log Qε º -ε =

0.05922 x 1.24

log K = - = - 41.9 K = 1.3 x 10-42

0.05922

log K- 1.24 -0 =

at equilibrium:


Is the reaction spontaneous at 25 ºC?ExampleFe2+(aq) + Cd(s) + Cd2+(aq)Fe (s)0.60 M 0.01 M

= -0.04 V - (-0.05 V) = +0.01 V

n0.0592

log (Q) ε = ε º--0.04 Vε º =

20.0592 logε = - 0.04 V - 0.60

0.01

ε º =Fe(s) Fe2+(aq) + 2e- -0.44 V

Cd(s) Cd2+(aq) + 2e- +0.40 Vε º =

> 0 (barely) so yes!

spontaneous under standard (1 M)

conditions?< 0 so no!


Batteries


Batteries

the voltage is the sum of individual cell voltages for group of galvanic cells

connected in series

a galvanic cell


C, MnO2 cathodeZn anode e-

NH4ClNH4Cl

Dry Cell Battery

Zn

Zn 2 ++ 2e- Mn3 +

MnO2 + e-

ZnO Mn2O3


Pb, PbO2 cathodePb anode e-

H2SO4

Lead Storage Battery

Pb

Pb2 ++ 2e-Pb2 +

PbO2 + 2e-

PbSO4

H2SO4

PbSO4


Fuel Cells

a galvanic cells for which the reactants are continuously supplied


C cathodeC anode e-

Hydrogen-Oxygen Fuel Cell

4H2 + 4HO-

4H2O + 4e- 4HO-

O2 + H2O + 4e-


C cathodeC anode e-


4H2 + 4HO-

4H2O + 4e- 4HO-

O2 + H2O + 4e-KOH

H2Osteam

H2 O2


http://www.utcpower.com/fs/com/Videos/UTCPower.wmv





http://www.apple.com/

http://www.apple.com/

Concentration Cells

A galvanic cell in which both compartments contain the same

components but differ in concentrations.


Zn anode Zn cathode

1M ZnSO4

1M ZnSO4

The EMF of the cell is Zero.


e-

0.1M ZnSO4

1M ZnSO4

What is the EMF of the cell ?

Zn anode Zn cathode


What is the emf of the cell ?

0.0592

nlog Qεº -ε =

0.0592

nlogεº -ε =

[Zn2+]

[Zn2+]

Zn2+(aq, 1.0 M ) Zn2+(aq, 0.1 M ) ε˚ = 0

Zn2+(aq, 1.0 M ) + 2e- Zn(s )

Zn(s) Zn2+(aq, 0.1 M ) + 2e-

cathode

anode


What is the emf of the cell ?

0.0592

nlog Qεº -ε =

0.0592

2log0 -ε =

0.1

1.0

ε = 0.0296 V

Zn2+(aq, 1.0 M ) Zn2+(aq, 0.1 M ) ε˚ = 0

Zn2+(aq, 1.0 M ) + 2e- Zn(s )

Zn(s) Zn2+(aq, 0.1 M ) + 2e-

cathode

anode


Electrolysis


Electrolysis

the process in which electrical energy is used to cause a nonspontaneous chemical reaction to

occur

two examples

has many important applications in industry, mainly in the extraction and purification of

metals


Molten Sodium Chloride

2Cl-

Cl2 + 2e- 2 Na

2Na+ + 2e-

Battery

Molten Sodium Chloride

e-

e- cathodeanode

the battery serves as an electron pump• driving electrons to the cathode• withdrawing electrons from the anode


cathodeanode

water

2H2O

O2 + 4H+ + 4e- 1/2 H2

H+ + e-

Battery

e-

e-

H2O

2H2O O22H2 + ΔG° = 474.4 kJ

PtPt

H2 SO4 (dilute)


cathodeanode

water

Battery

e-

e-

H2O

2H2O O22H2 + ΔG° = 474.4 kJ

PtPt

H2 SO4 (dilute)

2H2O O2 + 4H+ + 4e-

1/2 H2 )4( H+ + e-

2H2O O22H2 +


Quantitative Aspects of Electrolysis


Faraday observed that the mass of product formed (or reactant consumed) at an electrode is

proportional to both the amount of electricity transferred and the molar mass of the substance

in question

Quantitative Aspects of Electrolysis

We generally measure the amount of electricity transferred (current) in amperes, A

amperes = coulomb

secondThursday, April 2, 2009

current in amperes(1 Amp = 1 C/s)

The steps involved in calculating amounts of substances reduced or oxidized in electrolysis

charge in coulombs

Faraday’s constant

grams of substance reduced or oxidized

moles of substance reduced or oxidized

time

molarmass


A current of 0.452 A is passed through an electrolytic cell containing molten CaCl2 for 1.50 hours. Write the electrode reactions and calculate the quantity of products formed at the electrodes.

Example


cathodeanode

2Cl-

Cl2 + 2e- Ca

Ca2+ + 2e-

Battery

e-

e-

CaCl2(molten)

Ca2+ + 2Cl- + Cl2 ( g )Ca (l )

Example


3600s

1 hx

0.452 C

s1.50 h

1 mol e-

96485 Cx

1 mol Ca2+

2 mol e-x

40 g

1 mol Ca2+

x = 0.506g Ca

0.452 C 3600s

1 hsx1.50 h

1 mol e-

96485 Cx

1 mol Cl2

2 mol e-x

70 g

1 mol Cl2

x = 0.896g Cl2


electrochemistry - dr. marten’s ap chemistry cell galvanic cell voltaic cell by separating the...

Documents