Determination of Electrode Potentials Oral Report

8/7/2019 Determination of Electrode Potentials Oral Report

1/65

Annjaneth Briones and Nathalie Dagmang


2/65


3/65

an electron is transferred from one

reactant to another

Zn(s)

+ Cu2+ Cu(s)

+ Zn2+


4/65

An electron-exchange

reaction/oxidation or reduction

at an electrodeqa spontaneous redox reaction

produces energy or

a non-spontaneous redox

reaction uses up energy


5/65

Direct Electrochemical Cell

No work done Work done/needed


6/65

Spontaneous

reactionGenerates or

stores energy

ex. batteries

Non-spontaneous

reactionUses up electrical

ex. Metal plating,

recovery of metal

from ores

Voltaic/GalvanicCell ElectrolyticCell


7/65

Voltaic/GalvanicCell ElectrolyticCell

load Power

supply

energy

energy

+- + -

Work done by

system to

surroundings

Work done on

system by

surroundings

cathodecathode anodeanode


8/65

ANODE CATHODE

Rxn Ion movingtowards it

Sign Rxn Ion movingtowards it

Sign

Galvanic

Cell

oxidation anions _ reduction cations +

Electrolytic

celloxidation anions + reduction cations _

difference b/w the anode and cathode of galvanic and

electrolytic cells is their polarity

Flow of electrons: ANODE CATHODE (ALWAYS)


9/65

Voltaic/Galvanic Cell

Volt-

meter

+

Cu

CuSO4

-

Zn

ZnSO4

Salt bridge

e- e-

2K+

SO42-

Zn(s)Zn2+

(aq) + 2e- Cu2+ (aq) + 2e

- Cu(s)

Cathode = ReductionAnode = Oxidation

Zn|Zn2+

||Cu2+

|Cu

1

2

3


10/65

Has infinite resistance = wont consumeelectricity

Measures the potential difference of the 2half cells (Voltage)

parameter for the tendency of the reaction toproceed to an equilibrium state (same Vs)

potential continues to decrease until it reaches0.00 V.

Measures the Electromotive Force (EMF)

Electric work it can do/ can be extracted


11/65

Prevents the direct reaction but stillmaintains electrical contact b/w the halfcells.

Provides ions to neutralize its charges

Negative ions positive half-cell (Zinc)

Positive ions negative half-cell (Copper)

Some SO42-ions (from the cathode vessel) and some

Zn2+ ions (from the anode vessel) also migrate intothe salt bridge.


12/65

Solid conductor through which an electric

current enters or leaves

Collector or emitter of electric charge


13/65

-

Oxidation takes place

Zn(s)Zn2+

(aq) + 2e-

ZnSO4electrolyte

Zn2+

SO42-

Zn(s)

e-

e-

Reducing agent causes the cathode

to reduce

less e-

affinity


14/65

+

Reduction takes place

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

CuSO4electrolyte

Cu2+

SO42- Cu(s)

e-

e-

Oxidizing agent causes the anode

to oxidize

more e-

affinity


15/65

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

Zn(s) Zn2+

(aq) + 2e-

Zn(s) + Cu2+ Cu(s) + Zn

2+

(reduction)

(oxidation)

(Over-all reaction)


16/65

Electric current - transfer of charge throughelectrolytic and metallic conduction

Electrolytic or ionic conduction- charge is carried

by ions: cations move toward the cathode and anionsmove toward the anode

Metallic conduction- charge is carried by flow ofelectrons from the negative pole to the positive pole

the electrical conductors in an electrochemical cellare the electrodes (via metallic conduction), and theelectrolyte solution (via electrolytic conduction)


17/65


18/65

Potential at standard conditions: 1.00 M

1 atm

25 C

Independent on number of moles Cu2+

(aq)+ 2e- Cu

(s)E = 0.34 V

2Cu2+ (aq) + 4e- 2Cu(s) E = 0.34 V


19/65

Measure of the tendency for a reduction

to occur (when SHE is the anode) (+) V = spontaneous

(-) V = non-spontaneous

When reaction is reversed (turned into

oxidation), V changes sign

Standard reduction potential Potential when the SHE (Standard Hydrogen

Electrode) is the anode


20/65

Potential assigned as 0.00 V

2H+ + 2e- H2

H2

2H+ + 2e-

E = 0.00 V

E = 0.00 V


21/65

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

Zn(s) Zn2+

(aq) + 2e-

Zn(s) + Cu2+ Cu(s) + Zn

2+ Ecell = 1.103 V

E cathode = 0.340 V

E anode = -0.763 V

E cathode - E anode = E cell*reduction potentials are used


22/65

Factors affecting cell potential: Concentration

Temperature

Nature of reactant

E = E - RT ln QnF

where:

R (gas constant) = 8.314 J/molK,

T = temperature in Kelvin,

n= number of moles e- in rxn

F (Faraday ) = 96, 485 Coulombs

Q = reaction quotient


23/65

(Substituting the constants and assuming 25C

temperature)

E = E - 0.0592 log Qn


24/65

At equilibrium, G = 0, Ecell = 0, and Q = Keq

thus,

becomes

Solving for Keq, we get:

Note that Keq can be Ksp, Ka, Kb, Kf, etc.

Qlog

n

0592.0EE cell

ocell !

eqcello

Klogn

0592.0E !

!

0592.0

)E)(n(logantiK

cello

eq


25/65

Thus, the reaction is spontaneous only ifG isnegative and Ecell is positive

Example: Calculate the Go in J/mol for

3 Sn4+

+ 2 Cr

3 Sn2+

+ 2 Cr3+

Cathode: 3(Sn4+ + 2e- Sn2+) Eo = + 0.15 VAnode: 2(Cr Cr3+ +3e-) Eo = - 0.74 V

Eocell = (+0.15 V) (-0.74 V) = + 0.89 V

The very negative value ofG indicates that thereaction is product-favored. This is consistent with the

positive value of Ecell

cellnFEG !(

where F = 96485

C/mol e-, n = no. of

e

-

transferred

mol/J230,515V89.0emol

C485,96

rxnmol

emol6nFEG cell

oo !

!!(

Note: 1 J = 1 CV


26/65

Let us now calculate for the Keq of theprevious example

Recall: E

o

cell = (+0.15 V) (-0.74 V) = + 0.89 VThus,

the very large value of Keq reinforces ourprevious conclusion that the reaction isproduct-favored

90eq 10x59.1

0592.0

)89.0)(6(loganti !

!


27/65

C = AtWhereC = amount of electricity passing

A (ampere) = current

T= seconds

1 F = 96,500 Coulomb/mole e-

the mass of the product formed or the reactant

used is proportional to the amount of electricity

that passes through the system


28/65


29/65

1 Set-up the following half-cells:

Zn

1 M ZnSO4

Zn|Zn2+(1 M)

C

2 M FeSO4+2 M FeCl3

Fe3+(1 M), Fe2+|Fe

Cu

1M CuSO4

Cu|Cu2+(1 M)


30/65

C

2 M FeSO4+2 M FeCl2

Fe3+(1 M), Fe2+|Fe

* Why graphite and not Iron nail?

Desired reaction is

Fe3+ Fe2+ not Fe3+ Fe

When iron nail is used:

Fe3+ Fe2+ FeOFe3+ is oxidized to Fe2+ which will cause rusting

Graphite = semi-conductor

- inert; wont participate, will only conduct


31/65

2Connect the Cu2+/Cu half-cell with avoltmeter and a KNO3 salt bridge

Volt-meter

X

CuSO4

Salt bridge

Cu

X-


32/65

3Record Voltmeter reading

Volt-meter

X

CuSO4

Salt bridge

Cu

X-


33/65

1 Set-up the following half-cells:

C

1 M KCl

Cl- (1 M), Cl2|C

1 M KI1M KBr

Br-(1 M), Br2|C I-(1 M), I2|C

C C C C C

Carbon = inert wont participate in the rxn


34/65

2 Connect to 1.5 V dry cells for 1 minto generate X2

C

1 M KCl

Cl-(1 M), Cl2|C

1 M KX1M KBr

Br-(1 M), Br2|C I-(1 M), I2|C

C C C C C

batt batt batt


35/65

* What happened:

C

KI + I2

I-(1 M), I2|C

C

batt

Bubbles

formed from

H20

Formed brownish

substance

(Iodine / I2)

2I- I2 + 2e-


36/65

* What happened:

C

KCl + Cl2

Cl-(1 M), Cl2|C

KBr + Br2

Br-(1 M), Br2|C

C C C

batt batt

Bubbles Bubbles

Yell ish

2Cl- Cl2 + 2e- 2Br- Br2 + 2e

-


37/65

3 Connect the electrolyzed solutions tothe Cu half cell

Volt-

meter

CCu

KX, X2

Salt bridge

CuSO4

X2 + 2e- 2X-


38/65

3 Record voltmeter reading

Volt-meter

CCu

KX, X2

Salt bridge

CuSO4


39/65


40/65

Cell Notation Volts

Cu|Cu2+||Cu2+|Cu 0.0896

Zn|Zn2+||Cu2+|Cu 1.08

Cu|Cu2+||Fe3+|Fe 0.414

C|Cl-,Cl2||Cu2+|Cu 0.468

C|Br-,Br2||Cu2+|Cu 0.269

C|I-,I2||Cu2+|Cu 0.121


41/65

Finding for Eanode:

Take second set-up:

Zn|Zn2+||Cu2+|Cu

Given:

Cu2+ (aq) + 2e- Cu(s) , E = 0.34


42/65

Finding for Eanode:

Take second set-up:

Zn|Zn2+||Cu2+|Cu

Manipulating the equation:

E cathode - E anode = E cell0.34 - E anode = E cell

E anode = 0.34 - E cell


43/65

Zn|Zn2+

||Cu2+

|CuExperimental E cell = 1.08 V

E anode = 0.34 - 1.08

E anode

= - 0.74Zn2+ (aq) + 2e

- Zn(s) , E = 0.763

% error = [-0.763 (- 0.74)]/-0.763

= 3.01%

Finding for Eanode:

Take second set-up:


44/65

Finding for Ecathode:

Take third set-up:

Cu|Cu2+||Fe3+, Fe2+|C

Given:

Cu2+ (aq) + 2e- Cu(s) , E = 0.34


45/65


Take third set-up:

Cu|Cu2+||Fe3+, Fe2+|C

Manipulating the equation:

E cathode - E anode = E cell

E cathode = E cell + 0.34

E cathode = 0.34 + E cell


46/65


Take third set-up:

Cu|Cu2+

||Fe3+

, Fe2+

|CExperimental E cell = 0.414 V

E cathode = 0.414 + 0.34

E cathode = 0.754 V

Fe3+ + e- Fe2+ E= 0.771

% error = [0.771- 0.754)]/0.771= 2.2 %


47/65


48/65

Voltage reading: E = 0.468 V

Ecathode = Ecell + 0.34

Ecathode = 0.468 + 0.34 =0.808V


49/65

Take 4th set-up:

C|Cl-, Cl2||Cu2+|Cu

Data:

Ampere = 0.065 A

Time = 270 seconds

Substituting:

C = 0.065 x 270 = 17.55 Coulombs


50/65

Take 4th set-up:

C|Cl-, Cl2||Cu2+|Cu

From the equation,

2 mol e- = 2 mol Cl- transformed to Cl2Hence,

1.818 x 10-4 mol e- = 1.818 x 10-4 mol Cl-

2Cl- (aq) Cl2+ 2e-

17.55 C x(1 mole e-/96,500 C) = 1.818 x 10-4


51/65

Take 4th set-up:

C|Cl-, Cl2||Cu2+|Cu

0.025 mol Cl- - 1.818 x 10-4 mol Cl- = 0.0248182 mol Cl- left0.0248182 mol Cl-/0.025 L = 0.992728 M

[Cl-] = 0.992728 M

Cl2 is a gas, so the activity of Cl2 is taken as 1,

so Q = __1__[Cl-]2

2Cl- (aq) Cl2+ 2e-


52/65

E = E - 0.0592 log Q

n

Substituting E, n and Q in the equation

0.808 = E - 0.0592 log (1.014704)2

E = 0.808 + 0.0001876479346

= 0.8081876479V


53/65

Cell Notation Cathode Anode

Cu|Cu2+||Cu2+|Cu Copper Copper

Zn|Zn2+||Cu2+|Cu Copper Zinc

Cu|Cu2+||Fe3+|Fe Iron Copper

C|Cl-,Cl2||Cu2+|Cu Copper Chlorine

C|Br-,Br2||Cu2+|Cu Copper Bromine

C|I-,I2||Cu2+|Cu Copper Iodine


54/65

Cell Notation Cathode Anode

Cu(s)|Cu2+

(aq) (1M)|| Cu2+

(aq)

(1M)|Cu(s)

Cu2+(aq)(1M)|Cu(s)

Cu(s)|Cu2+

(aq) (1M)

Zn(s)|Zn2+

(aq) (1M)|| Cu2+

(aq)

(1M)|Cu(s)

Cu2+(aq)

(1M)|Cu(s)

Zn(s)|Zn2+

(aq) (1M)

Cu(s)|Cu2+

(aq) (1M) ||Fe3+

(aq) (1M),

Fe2+(aq) (1M)|C(graphite)

Fe3+(aq) (1M),

Fe2+(aq)(1M)|C(graphite)

Cu(s)|Cu2+

(aq) (1M)

C(graphite)| Cl2(g) |Cl-(aq) (1M)||Cu

2+(aq)

(1M)|Cu(s)

Cu(s)|Cu2+

(aq)

(1M)

C(graphite)| Cl2(g)

|Cl-(aq) (1M)

C(graphite)| Br2(l),Br-(aq) (1M)||Cu

2+(aq)

(1M)|Cu(s)

Cu(s)|Cu2+

(aq)

(1M)

C(graphite)| Br2(l),Br-

(aq) (1M)

C(graphite)| I2(s),I-(aq) (1M)||Cu

2+(aq)

(1M)|Cu(s)

Cu(s)|Cu2+

(aq)

(1M)

C(graphite)| I2(s),I-(aq)

(1M)


55/65

Variable Half-

Cell Notation

Calculated

Volts

Book

Value

% Error

Cu|Cu2+ 0.2504 0.34 26 %

Zn|Zn2+ - 0. 4 - 0. 63 3.0 %

Fe3+|Fe 0. 54 0. 2.2 %

C|Cl-,Cl2 *0.808 .359 40.54 %

C|Br-,Br2 *0.609 .08 43.9 %

C|I-,I2 *0.46 0.6 5 25.04 %

* assuming concentration did not change with electrolysis


56/65

More (-) Ered Higher tendency to be oxidized

Higher reducing power

When placed as anode, reaction is spontaneous

More (+) Ered

Higher tendency to be reduced Higher oxidizing power

When placed as cathode, reaction is spontaneous


57/65

Take Electrolysis for 4th set-up:

C|Cl-, Cl2

Voltage is applied

Vapplied = Eanode - Ecathode

2Cl- (aq) Cl2+ 2e-

Reduction potential = 1.359 V spontaneous

Oxidation potential = -1.359 V non-spontaneous


58/65

2Cl- (aq) Cl2+ 2e-

At equilibrium,Eanode = EcathodeV = 0Oxidation and reduction at same rate


59/65

If Eanode = Ecathode system at equilibriumIf Eanode > Ecathode Ecathode,EanodeIf Eanode < Ecathode Ecathode,Eanode

2Cl- (aq) Cl2+ 2e-

In the experiment, we want to produce more Cl2

oxidation should predominate

Eanode should increase

should be Eanode > Ecathode

Vapplied = Eanode - Ecathode


60/65

Ecathode = E - 0.0592 log [Cl-]2

n [Cl2]

2Cl- (aq) Cl2+ 2e-

Eanode

= E - 0.0592 log [Cl2

]n [Cl-]2

Cl2 = Ecathode = Ered = spontaneity


61/65

Concentration changed as the halogen

solutions were electrolyzed

Electrolysis did not produce enough

halogen, X2 that will be reduced

Slightly different concentrations (not both

exactly 1 M)

increase in voltage reading(concentration difference can also cause

potential diff)


62/65

Example: Calculate the mass ofcopper metal produced at the

cathode during the passage of 2.50

amperes of current through a solution

of copper(II) sulfate for 50.0 minutes.

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

Current X

Time

no. of

Coulombs

Useful

Conversions:

1 J = 1 CV

1 A = 1 C/s

Mass of

substanceMol. of e-

passed

Cug47.2emol2

Cug5.63

C485,96

emol1

s

C50.2

min1

s60min0.50Cug !

!


63/65

A galvanic cell is to be constructed using a Fe3+/Fe2+ halfcell and a Br2/Br

- half-cell. The Br2/Br- half-cell was

constructed by electrolyzing 50 mL of 0.5 M KBr using a 2.5A current for 3 minutes. The Fe3+/Fe2+ half cell wasconstructed by mixing 25 mL of 2 M FeSO4 and 25 mL of 2 M

FeCl3. (MW Br2: 159.8 g/mol)

Fe3+ (aq) + e- Fe2+

(aq) Eo

red = +0.771 VBr2(aq) + 2e

- 2Br-(aq) Eo

red = +1.087 V

a) Assuming that the bromide ions are converted to Br2(aq),

crcalculate [Br2] and [Br

-

] upon electrolysis.b) Write the cell notation (with the correct phases andconcentrations of ions) for the galvanic cell

c) Calculate the Ecell for the galvanic cell.


64/65

Answers:

a) [Br2] = 0.046638 M, [Br-] = 0.406722 M

b) Fe2+(aq) (1M), Fe3+

(aq)

(1M)||Br2(aq)(0.046638 M), Br-(aq)

(0.406722 M)

c) 0.2997243 V


65/65

Belcher, R. Quantitative Inorganic Analysis,1970

Christian, G.D. Analytical Chemistry, 1986

Day, Underwood, et al. Quantitative Analysis,

1967Haenisch, Pierce, et al. Quantitative Analysis,

1958

Skoog, et al., Fundamentals of Analytical

Chemistry, Eighth edition, 2004 http://academic.pgcc.edu/psc/chm103/103_m

anual.pdf

Determination of Electrode Potentials Oral Report

Documents

Transcript of Determination of Electrode Potentials Oral Report