option c nernst equation, voltaic cell and concentration cell

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Prepared by Lawrence Kok

Tutorial on Voltaic Cell, Nernst Equation and concentration cell.

http://lawrencekok.blogspot.com/

Types voltaic cell

Conversion electrical energy to chemical energy

Electrochemistry

Electrolytic cellVoltaic cell

NH4CI and ZnCI2

Redox rxn (Oxidation/reduction)Movement electronProduce electricity

Conversion chemical energy to electrical energy

Electrodes – different metal (Half cell) Electrodes – same metal (Half cell)

Daniell cell Alkaline cellDry cell Nickel cadmium cell

Primary cell (Non rechargeable)

MnO2 and KOH

Secondary cell (Rechargeable)

Current – measured Amperes or Coulombs per second 1A = 1 Coulomb charge pass through a point in 1 s = 1C/s1 Coulomb charge (elec) = 6.28 x 10 18 elec passing in 1 s

1 elec/proton carry charge of – 1.6 x 10 -19 C ( very small)6.28 x 10 18 elec carry charge of - 1 C

Electric current

Flow electric charges (elec, -ve)From High to low electric potential Potential Diff – measure with ammeter

ondelectron

ondCoulombA

sec.1.1028.6

sec111

18

Current Electric Current – moving charges in solid wire or solution

Flow of charges

---

Solid/WireSolution/Electrolyte

Electron move in random No current flow cause

No potential difference

Electrons & Protons

--

++

1A = 6.28 x 1018 e 1 s

Potential Difference across wireElectron move in one direction

Current flow

+ve ions -ve ions(cations) (anions)

Potential Diff applied/Battery

ItQ t = Time/ s

Find amt charges pass through if Current is 2.ooA, time is 15 min

ItQ

Current flow

Q = Amt Charges/ C I = Current/ A

CQ 1800601500.2

http://www.howequipmentworks.com/electricity_basics/

Electric Potential

CJVolt 11

-Measured in Volt with Voltmeter- 1 V = 1 Joule energy released when 1 Coulomb charge pass through 1 point- 1 V = 1 J/C

V = Potential DiffI = CurrentR = Resistance

Potential diff bet 2 points is 1 V ↓1 J energy released when 1 C charge passes through

Voltmeter across1Volt

1 V

+ -

1 Ω 2 Ω

Charges (-ve) flow down

ARVI

RIV

236

VVRIV212

-+-

+

VVRIV422

Total current

Potential Diff(PD) vs CurrentPD = Water Pressure

PD = 1.5V – 1.5J energy released 1C charge flow downPD – cause charge flow = CURRENT

Potential Diff(PD) vs Current

1.5V = 1.5J/CA

DElectric potential/PD/Voltage = Electric Pressure = VoltElectric Current = Charge flow = Amp

Electric Potential Energy = Work done to bring a charge to a point = JouleVoltage NOT same as energy, Voltage = energy/charge

Battery lift charges, Q to higher potentialPotential Energy bet 2 terminals in battery stored as chemical energy

2A 2A

Potential Diff/VoltagePotential Diff/Voltage

EMF vs PD

V = Potential DiffI = CurrentR = Resistance

Max potential diff bet two

electrodes of battery source. + -

1 Ω 2 Ω

ARVI

RIV

236

VVRIV212

VV

RIV422

Total current

Current flow Circuit complete

Circuit complete↓

Current flow↓

Internal resistance(battery - 1Ω)

↓Terminal PD = 8V

(Voltage drop)

Potential Diff/Voltage in Volt

Symbol for EMF = E / ℰ

No Current flow in circuit

EMF (Electromotive Force) VoltBattery = EMF = 9V

9 Volt

).(9 currentnoVEMFVIRV

EMF Internal resistance Ir

Place voltmeter across – EMF= 9V

No current flow.

ArR

EI

rRIEIrIREMFE

199

)18(9

)(

)()(

VVRIV881

VVRIV111

EMF = 8V+1V

8 Volt

1 Volt

EMF (6V) = 2V + 4V

4 Volt2 Volt

Charges passing through wire

Current flow Circuit complete

Internal resistanceCollision bet + ve ions with elec

(drift velocity elec)

- +

Eθ value DO NOT depend surface area of metal electrode.E cell = Energy per unit charge. (Joule)/CE cell- 10v = 10J energy released by 1C of charge = 100J energy released by 10C of charge Eθ – intensive property– independent of amt – Ratio energy/chargeIncreasing surface area metal will NOT increase E cell

Eθ Zn/Cu = 1.10V

Surface area - 10 cm2 Total charge- 100C leave electrodeE cell = 1.1V = 1.1 J energy for 1 C (charges leaving) 1C release 1.1 J energy100 C release 110 J energyVoltmeter measure energy for 1C – 110J/100C – 1.1VE cell no change

Current – measured in Amp or Coulomb per s1A = 1 Coulomb charge pass through a point in 1 s = 1C/s1 Coulomb charge (elec) = 6.28 x 10 18 elec passing in 1 s

1 electron/proton carry charge of – 1.6 x 10 -19 C ( very small)6.28 x 10 18 electron carry charge of - 1 C

ondelectron

ondCoulombA

sec.1.1028.6

sec111

18

Surface area increase ↑

Total Energy increase ↑

Total Charge increase ↑ Current increase ↑

BUT E cell remain SAMEE cell = (Energy/charge)t

QI

tIQ

Q up ↑ – I up ↑

100C flow110J released

VEcell

Ecell

eChEnergyEcell

10.1100110arg

Surface area - 100 cm2 Total charge 1000C leave electrodeE cell = 1.1V = 1.1 J energy for 1 C (charges leaving) 1 C release 1.1J energy1000 C release 1100 J energyVoltmeter measure energy for 1C – 1100J/1000C – 1.1VE cell no change

VEcell

Ecell

eChEnergyEcell

10.110001100arg

Eθ Zn/Cu = 1.10V

1000C flow1100J released

tQI

tQI

Surface area exposed 10 cm2

Surface area exposed 100 cm2

∆G θ = -nFE θ cell

Relationship bet ∆G and Kc

cellnFEG

Relationship betEnergetics and Equilibrium

cKRTG ln STHG Enthalpy change

Entropy change

Equilibrium constant

Gibbs free energy change

HG

Relationship bet ∆G, Kc and E cell

cellnFEG STHG cKRTG ln

cK

Relationship betEnergetics and Cell Potential

G cellE

Gibbs free energy change

Cell potential

F = Faraday constant (96 500 Cmol-1)

n = number electron

Relationship bet ∆G, Kc and Ecell

ΔGθ Kc Eθ/V Extent of rxn

> 0 < 1 < 0 No ReactionNon spontaneous

ΔGθ = 0 Kc = 1

0 EquilibriumMix

reactant/product

< 0 > 1 > 0 Reaction completeSpontaneous

ΔGθ Kc Eq mixture

ΔGθ = + 200

9 x 10-36 Reactants

ΔGθ = + 10 2 x 1-2 Mixture

ΔGθ = 0 Kc = 1 Equilibrium

ΔGθ = - 10 5 x 101 Mixture

ΔGθ = - 200 1 x 1035 Products

shift to left (reactant)

shift to right (products)

cellE

G

cK∆G

θ = -RT ln

K c

KnFRTE cell ln

ΔGθ ln K Kc Eq mixture

ΔGθ -ve < 0

Positive ( + )

Kc > 1

Product(Right)

ΔGθ +ve > 0

Negative

( - )

Kc < 1

Reactant(left)

ΔGθ = 0 0 Kc = 1

Equilibrium

E cell/Voltage – depend on nature of material

QnFRTEE ln

T = Temp in KQ = Rxn Quotient

E0 = std (1M)n = # e transfer

F = Faraday constant (96 500C mol -1 )

R = Gas constant (8.31)

cKRTQRTG lnln

KRTG

KRTQRTGo

c

ln

lnln

When ratio conc, Q = 1,

all in std conc = 1M

Non std condition

01ln1

RTQ

QnFRTEE ln

QRTGG o lnNon std condition

onFEG nFEG

QRTnFEnFE ln

Nernst equation

Work or Free energy to do work depend on quantity material and surface area

E cell depend

Nature of electrodeType of metal used Conc of ion Temp of sol

Eθ Q T

Current/I depend

Surface area of contact

Salt bridge conc Size of cation/anion

Resistance high ↑ – current low ↓E cell depend

Surface area of contact Salt bridge conc

Size of cation/anion

cellnFEG

Gibbs free energy changedo do WORK

n = number electron


Cell potential

Increasing surface area → increase charge Q and I current - Work increase

Current – depend on quantity and surface area

Zn ↔ Zn2+ + 2e Eθ = +0.76Cu2+ + 2e ↔ Cu Eθ = +0.34Zn + Cu2+ → Zn 2+ + Cu Eθ = +1.10V

Zn half cell (-ve)Oxidation

Cu half cell (+ve)Reduction

Anode Cathode

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

(aq) | Cu (s)

Anode Cathode

Half Cell Half Cell(Oxidation) (Reduction)

Salt Bridge Flow electrons

Zn/Cu Cell - 1M std condition

-e -e

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = +0.34 – (-0.76) = +1.10V

Zn 2+ + 2e ↔ Zn (anode) Eθ = -0.76VCu2+ + 2e ↔ Cu (cathode) Eθ = +0.34V

Eθcell = Eθ

(cathode) – Eθ(anode)

Zn 2+ + 2e ↔ Zn Eθ = -0.76VCu2+ + 2e ↔ Cu Eθ = +0.34V

Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19H2O + e- ↔ 1/2H2 + OH- -0.83Zn2+ + 2e- ↔ Zn - 0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni -0.26Sn2+ + 2e- ↔ Sn -0.14Pb2+ + 2e- ↔ Pb -0.13H+ + e- ↔ 1/2H2 0.00Cu2+ + e- ↔ Cu+ +0.15SO4

2- + 4H+ + 2e- ↔ H2SO3 +0.17Cu2+ + 2e- ↔ Cu + 0.341/2O2 + H2O +2e- ↔ 2OH- +0.40

+

+1.10 V

Eθ Zn/Cu = 1.10V

Cu2+

----

Zn Cu++++

cellnFEG

E cell with ∆G


n = number electron

cellnFEG

kJJG

G

212212300

10.1965002

Std electrode potential - std reduction potential

STD CONDITION

Zn/Cu half cell Cell diagram

QnFRTEE ln

Ratio conc, Q = 1, all in std conc = 1M, T =

298K

VE

E

10.1

1ln965002298314.810.1

Zn ↔ Zn2+ + 2e Eθ = +0.762Ag++2e ↔ 2Ag Eθ = +0.80Zn + Ag+ → Zn 2+ + Ag Eθ = +1.56V


Ag half cell (+ve)Reduction

Anode Cathode

Zn(s) | Zn2+(aq) || Ag+

(aq) | Ag (s)

Anode Cathode

Half Cell Half Cell(Oxidation) (Reduction)

Salt Bridge Flow electrons

-e -e

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = +0.80 – (-0.76) = +1.56V

Zn 2+ + 2e ↔ Zn (anode) Eθ = -0.76VAg + + e ↔ Ag(cathode) Eθ = +0.80V

Eθcell = Eθ

(cathode) – Eθ(anode)

Zn 2+ + 2e ↔ Zn Eθ = -0.76VAg+ + e ↔ Ag Eθ = +0.80V


2- + 4H+ + 2e- ↔ H2SO3 +0.17Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.521/2I2 + e- ↔ I- +0.54Fe3+ + e- ↔ Fe2+ +0.77Ag+ + e- ↔ Ag + 0.801/2Br2 + e- ↔ Br- +1.07

+

+1.56 V

Ag

Eθ Zn/Ag = +1.56V

Ag+

----

++++

Zn

E cell with ∆G

cellnFEG

n = number electron F = Faraday constant (96 500 Cmol-1)

cellnFEG

kJJG

G

301301000

56.1965002

Cell diagram Zn/Ag half cells

Ratio conc, Q = 1, all in std conc = 1M, T =

298K

Zn/Ag Cell - 1M std condition

QnFRTEE ln

VE

E

56.1

1ln965002298314.856.1

STD CONDITION



Zn/Cu Cell

-e -e


Zn ↔ Zn2+ + 2e Eθ = +0.76VCu2+ + 2e ↔ Cu Eθ = +0.34VZn + Cu2+ → Zn 2+ + Cu Eθ = +1.10V


2- + 4H+ + 2e- ↔ H2SO3 + H2O +0.17Cu2+ + 2e- ↔ Cu + 0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.521/2I2 + e- ↔ I- +0.54

+1.10 V

Cu2+

----

Zn Cu++++

QnFRTEE ln 1M 0.1M

Zn2+

10]1.0[]1[

][][

2

2

c

c

QMM

CuZnQ

0.1 M 1 M

Using Nernst Eqn

E0 = Std condition (1M) – 1.10VR = Gas constant (8.31)n = # e transfer (2 e)F = Faraday constant (96500C mol -1 )

VEE

E

07.103.010.1

)10ln()965002()29831.8(10.1

Non std 0.1M

E cell decrease ↓ [Cu2+] decrease ↓↓

Le Chatelier’s principleCu2+ + 2e ↔ Cu

↓[Cu2+] decrease ↓

↓Shift to left ←

↓

E cell → less ↓ → Cu2+ less able ↓ to receive e- [Cu2+] ↓ E cell < Eθ

1.07 < 1.10

Zn/Cu half cell Zn +Cu2+→Zn2++Cu

NON STD CONDITION



Zn/Cu Cell

-e -e





+1.10 V

Cu2+

----

Zn Cu++++

QnFRTEE ln 1M 10M

Zn2+

1.0]10[]1[

][][

2

2

c

c

QMM

CuZnQ

10 M 1 M

Using Nernst Eqn

E0 =Std condition (1M) – 1.10VR = Gas constant (8.31)n = # e transfer (2 e)F = Faraday constant (96500C mol -1 )

VEE

E

13.103.010.1

)1.0ln()965002()29831.8(10.1

Non std 0.1M

E cell increase ↑ [Cu2+] increase ↑↓

Le Chatelier’s principleCu2+ + 2e ↔ Cu

↓[Cu2+] increase ↑

↓Shift to right →

↓E cell → more ↑→ Cu2+ more able receive e-

[Cu2+] ↑ E cell > Eθ

1.13 > 1.10

Zn/Cu half cell Zn +Cu2+→Zn2++Cu

NON STD CONDITION



Zn/Cu Cell

-e -e





+1.10 V

Cu2+

----

Zn Cu++++

QnFRTEE ln 0.1M 1M

Zn2+

1.0]1[]1.0[

][][

2

2

c

c

QMM

CuZnQ

1 M 0.1 M

Using Nernst Eqn


VEE

E

13.103.010.1

)1.0ln()965002()29831.8(10.1

Non std 0.1M

E cell increase ↑ [Zn2+] decrease ↓↓

Le Chatelier’s principleZn2+ + 2e ↔ Zn

↓[Zn2+] decrease ↓

↓Shift to left ←

↓E cell → more ↑→ Zn more able lose elec

[Zn2+] ↓ E cell > Eθ

1.13 > 1.10

Zn/Cu half cell Zn + Cu2+→ Zn2+ + Cu

NON STD CONDITION

Cu half cell (-ve)Oxidation


-e

Cu ↔ Cu 2+ + 2e Eθ = - 0.34VCu2+ + 2e ↔ Cu Eθ = +0.34V

Cu ↔ Cu2+ + 2e Eθ = - 0.34VCu2+ + 2e ↔ Cu Eθ = +0.34VCu + Cu2+ → Cu2+ + Cu Eθ = 0V

Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19H2O + e- ↔ 1/2H2 + OH- -0.83

Zn2+ + 2e- ↔ Zn -0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni -0.26Sn2+ + 2e- ↔ Sn -0.14Pb2+ + 2e- ↔ Pb -0.13H+ + e- ↔ 1/2H2 0.00Cu2+ + e- ↔ Cu+ +0.15SO4

2- + 4H+ + 2e- ↔ H2SO3 + H2O +0.17Cu2+ + 2e- ↔ Cu + 0.341/2O2 + H2O +2e- ↔ 2OH- +0.40

Cu2+

Zn Cu++++

QnFRTEE ln

0.1M

01.0]1.0[]001.0[

][][

2

2

c

cathode

anodec

QCuCu

Q

0.1 M 0.001 M

Using Nernst Eqn


VEE

E

0285.00285.00

)01.0ln()965002()29831.8(0

Cu2+/Cu half cell Cu + Cu2+ → Cu2+ + Cu -e

Cu2+

0.001M

Cu (s) Cu2+(aq) (0.001M) Cu2+

(aq) (0.1M)Cu(s)

----

Concentration cellElectrode same - diff conc

Oxi cell – anode – lower concRed cell – cathode – higher conc

cathode anode

Cu

Conc cell made of Zn/Zn2+

Conc Zn2+ - 0.11M and 0.22M. Find voltage.Zn (s) Zn2+

(aq) (0.11M) Zn2+(aq) (0.22M)Zn(s)

Zn + Zn2+ → Zn2+ + Zn

cathode anode

0.22M 0.11 M

5.0]22.0[]11.0[

][][

2

2

c

cathode

anodec

QZnZn

Q QnFRTEE ln

VE

E

0089.0

)5.0ln()965002()29831.8(0

Fe half cell (-ve)Oxidation

Fe half cell (+ve)Reduction

-e

Fe ↔ Fe 2+ + 2e Eθ = + 0.45VFe2+ + 2e ↔ Fe Eθ = - 0.45V

Fe ↔ Fe2+ + 2e Eθ = + 0.45VFe2+ + 2e ↔ Fe Eθ = - 0.45 VFe + Fe2+ → Fe2+ +Fe Eθ = 0V

Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- ↔ Li -3.04K+ + e- ↔ K -2.93Ca2+ + 2e- ↔ Ca -2.87Na+ + e- ↔ Na -2.71Mg 2+ + 2e- ↔ Mg -2.37Al3+ + 3e- ↔ AI -1.66Mn2+ + 2e- ↔ Mn -1.19H2O + e- ↔ 1/2H2 + OH- -0.83

Zn2+ + 2e- ↔ Zn -0.76Fe2+ + 2e- ↔ Fe -0.45Ni2+ + 2e- ↔ Ni -0.26Sn2+ + 2e- ↔ Sn -0.14Pb2+ + 2e- ↔ Pb -0.13H+ + e- ↔ 1/2H2 0.00Cu2+ + e- ↔ Cu+ +0.15SO4

2- + 4H+ + 2e- ↔ H2SO3 + H2O +0.17

Fe2+

Zn Fe++++

QnFRTEE ln

0.1M

1.0]1.0[]01.0[

][][

2

2

c

cathode

anodec

QFeFe

Q

0.1 M 0.01 M

Using Nernst Eqn


VEE

E

029.0029.00

)1.0ln()965002()29831.8(0

Fe2+/Fe half cell Fe + Fe2+ → Fe2++ Fe -e

Fe2+

0.01M

Fe(s)Fe2+(aq) (0.01M) Fe2+

(aq) (0.1M)Fe(s)

----

Concentration cellElectrode same - in diff concOxi cell – anode – lower conc

Red cell – cathode – higher conc

cathode anode

Fe

Find cell potentialMn (s) Mn2+

(aq) (0.1M) Pb2+(aq) (0.0001M)Pb(s)

Mn + Pb2+ → Mn2+ + Pb

0.0001M 0.1 M

cathode anode 001.0]0001.0[]1.0[

][][

2

2

c

cathode

anodec

QPbMn

Q QnFRTEE ln

VE

E

96.0

)001.0ln()965002()29831.8(05.1

Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/http://spmchemistry.onlinetuition.com.my/2013/10/electrolytic-cell.htmlhttp://www.chemguide.co.uk/physical/redoxeqia/introduction.htmlhttp://educationia.tk/reduction-potential-tablehttp://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s23-electrochemistry.html

Prepared by Lawrence Kok

Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com

option c nernst equation, voltaic cell and concentration cell

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