Experiment 8ppt

Experiment 8: GALVANIC CELLS

C. Electromotive Force of Galvanic Cells  Methodology

Two test electrodes were coupled and connected through a salt bridge. Cells A to D were prepared and their respective potential differences measured by means of a voltmeter. Below is the schematic diagram for cells A to B:  Cell A: Zn l Zn2+ ll Pb2+ l Pb  Cell B: Zn l Zn2+ ll Fe2+ l Fe3+ l C  Cell C: Pb l Pb2+ ll Fe2+ l Fe3+ l C  Cell D: Zn l Zn2+ ll PbSO4(s), SO42- l Pb 

C. Electromotive Force of Galvanic Cells  Results

Cell A: Zn l Zn2+ ll Pb2+ l Pb Anode: Zn (s)  Zn2+ + 2e- ε°= -0.7618 V Cathode: Pb2+ + 2e-  Pb (s) ε°= -0.1265 V Cell: Zn (s) + Pb2+ Pb (s) + Zn2+ ε°theo = -0.1265 V – (-0.7618 V) = ε°theo = 0.6353 V ε°obs= 0.1092 V

C. Electromotive Force of Galvanic Cells  Results

Cell B: Zn l Zn2+ ll Fe2+ l Fe3+ l C Anode: Zn (s)  Zn2+ + 2eε°= -0.7618 V Cathode: 2(Fe3+ + e-  Fe2+) ε°= 0.771 V Cell: Zn (s) + 2 Fe3+ 2Fe2+ + Zn2+ ε°theo = 0.771 V –(-0.7618 V) ε°theo = 1.5328 V ε°obs= 9.33 x 10-4 V

C. Electromotive Force of Galvanic Cells  Results

Cell C: Pb l Pb2+ ll Fe2+ l Fe3+ l C Anode: Pb (s)  Pb2+ + 2eε°= -0.1265 V Cathode: 2(Fe3+ + e-  Fe2+) ε°= 0.771 V Cell: Pb (s) + 2Fe3+  Pb2+ + 2 Fe2+ ε°theo = 0.771 V – (-0.1265 V) ε°theo = 0.8975V ε°obs= 8.02 x 10-4 V

C. Electromotive Force of Galvanic Cells  Results Cell D: Zn l Zn2+ ll PbSO4(s), SO42- l Pb Anode: Zn (s)  Zn2+ + 2eε°= -0.7618 V Cathode: PbSO4(s) + 2e-  Pb (s) + SO42ε°=-0.3546 V Cell:PbSO4(s) +Zn(s) Pb (s) + SO42- + Zn2+ ε°theo = -0.3546 V–(-0.7618V) ε°theo = 0.4072V ε°obs= 0.385 V

C. Electromotive Force of Galvanic Cells 

Discussions   





For current to flow, an emf is required. Emf causes differences of potential to exist between points in the circuit causing charges to move. Standard electrode potential measures the relative force tending to drive the half reaction from a state in which the reactants and products are at unit activity to a state in which the reactants and products are at their equilibrium activities . The more positive the electrode potential, the greater is its tendency to be reduced; therefore it is a stronger oxidizing agent/ oxidant The electrical work done on the external load is equal to  ∆G = -nFε cell

E. Determination of Solubility Product Constant  Methodology A

Pb I Pb2+ II PbSO4(s), SO42- I Pb cell was prepared and its voltage was measured using a voltmeter.

E. Determination of Solubility Product Constant  Results

and Computations

Pb I Pb2+ II PbSO4(s), SO42- I Pb Anode Half Reaction: -( Pb2+ + 2e-  Pb(s) ) ε°= -0.126 V Cathode Half Reaction: PbSO4(s) + 2e-  Pb(s) + SO42- ε°= -0.350 V Cell Reaction: PbSO4(s) = Pb2+ + SO42εcell°= -0.350 V-( -0.126 V) = -0.224 V Ksp = [Pb2+][ SO42-] = 1.6 x 10-8 Ksp from εcell°= 2.70665207 x 10-8 Ksp from experiment = 0.9618516805

E. Determination of Solubility Product Constant  Computations

∆G° = - RT ln K ln K = (∆G°/ -RT) = (-nF εcell°/ -RT) = (-n 96485 εcell°/ -8.314 x 298.15) log K = (nεcell°/0.05916) K = 10 ^ (nεcell°/0.05916) Ksp = 10 ^ (2 x (-0.5 x10-3)/0.05916) = 0.9618516805

E. Determination of Solubility Product Constant 

Discussion Equilibrium constants can be evaluated from the standard cell potential by applying the Nernst equation. Since our system is at equilibrium ε = 0, therefore 0 = εcell° – {(RT)/nF} ln K {(RT)/nF} ln K = εcell° ln K = (nF εcell°)/RT K = exp((nF εcell°)/RT) Consider a saturated solution of a slightly soluble solute which is in contact with the undissolved solid. Cell reaction : AxBy (s) = x Ay+(aq) + yBx-(aq) Ka = a(Ay+)x a(Bx-)y , in very dilute solutions, the activity coefficient is almost equal to 1.00 and activity is taken to be equal to molar concentration Ksp = [Ay+]^x[Bx-]^y The Ksp can be solved from the cell potential of the cell reaction given above.

Determination of Activity Coefficient  Know

the half reactions and overall reaction process  Measure εcell  Compute  Obtain

ε° cell

ionic strength at different concentrations  Compute activity coefficient using Nernst law

 AHR:

Fe(s) → Fe2+ + 2e CHR: PbSO4(s) + 2e- → Pb (s) + SO42 Overall

rxn: PbSO4(s) + Fe(s) → Pb (s) + Fe2+ + SO42-

OR PbSO4(s) + Fe(s) → Pb (s) + Fe2+ (aFe2+ )+ SO42-(aSO42-)

 The

general emf equation follows as:

ε = ε° - (RT/nF) ln aFe2+ + aSO4OR ε = ε° - (RT/nF) ln aFeSO4

[FeSO4], M

Εcell (V)

0.100

15.5 x 10-3

0.080

69.8 x 10-3

0.060

20.0 x 10-3

0.040

29.2 x 10-3

0.020

-12 x 10-3

[FeSO4], M

Εcell + (RT/nF)ln[FeSO4]

0.100

-0.01406

0.080

0.03737

0.060

-0.01592

0.040

-0.01213

0.020

-0.06223

[FeSO4], M

Ionic Strength, μ

0.100

0.400

0.080

0.320

0.060

0.240

0.040

0.160

0.020

0.080

Ionic strength, μ = ½[ΣCiZi2]

ε° = ε + RT/nF ln [FeSO4] ε° = ε + RT/nF ln aFeSO4 a= (aFe2+)x(aSO42-) = (γm)Fe2+(γm)SO42if m=conc. of FeSO4=M FeSO4 m = mFe2+ = mSO42a= (γ+m )●(γ+m )= γ2m2 =(γm ) - 2 ε = ε° - 2(RT/nF) ln γm

ε + (2RT/nF)lnm= ε°- (2RT/nF)lnγ

0.06 y = 0.1814x - 0.0994

ε + (2RT/nF)lnm

0.04 0.02

Series1

0 -0.02

0

0.2

0.4

0.6

0.8

Linear (Series1)

-0.04 -0.06 -0.08 μ^(1/2)

R = 0.708

b = ε° = -0.0994

 Final

Equation:

lnγ = [(ε°- ε)/ (2RT/nF)] - lnm

[FeSO4]

εcell (Volts)

εcell + (RT/nF) ln[FeSO4]

Ionic Strength, µ

Activity Coefficient,γ

0.100 M

15.5 x 10-3

-0.01406

0.40

0.01725

0.800 M

69.8 x 10-3

0.03737

0.32

0.15852

0.600 M

20.2 x 10-3

-0.01592

0.24

0.11751

0.400 M

29.2 x 10-3

-0.01213

0.16

0.6650

0.200 M

-12 x 10-3

-0.06223

0.08