IB Chemistry on Standard Reduction Potential, Standard Hydrogen Electrode and Electrochemical Series

http://lawrencekok.blogspot.com

Prepared by Lawrence Kok

Tutorial on Standard Electrode Potential, Standard Reduction Potential and Electrochemical Series.

http://lawrencekok.blogspot.com/

Potential Diff bet Zn/Zn2+

Electrode potential Zn/Zn2+ = -ve

-

Electrode Potential

Redox Equilibrium

Zn2+

Zn → Zn 2+ + 2e (Oxidation)Zn 2+ + 2e → Zn (Reduction)Zn 2+ + 2e ↔ Zn (At equilibrium)

Metal Zn placed in its sol Zn2+ ion

Equilibrium bet Zn/Zn2+

Zn metal reactive lose e form Zn2+

Equilibrium shift to right ←

Potential Diff form bet Zn/Zn2+

Potential Diff Electrode potential = -ve

Zn2+

Zn2+

Zn

Zn2+

Zn

Zn2+

Zn2+ Zn2+

Zn 2+ + 2e ↔ ZnEqui shift to ←

-

--

Zn

---

-

++

+

++ +

+ +

+

Voltage of Zn/Zn2+ can’t be measured. Abs electrode potential can’t measured. Only Diff in electrode potential can be measured.

Cannot measure Abs Potential

Metal Cu placed in its sol Cu2+ ion

Equilibrium bet Cu/Cu2+

Cu2+ ion gain -2e form Cu

Equilibrium shift to left ←

Potential Diff form bet Cu/Cu2+

Potential Diff Electrode potential = +ve

Cu

Cu2+

Cu2+

Cu2+

Cu2+

Cu → Cu2+ + 2e (Oxidation) Cu2+ + 2e → Cu (Reduction)Cu2+ + 2e ↔ Cu(At equilibrium)

Cu

-e

-e-eCu2+

Cu2+

Cu2+

Cu2+ + 2e ↔ CuEqui shift to →

Zn Half Cell

+

++

Cu

++

+

----

--- ------

Potential Diff bet Cu/Cu2+

Electrode potential Cu/Cu2+ = +ve

Cannot measure Abs Potential

Voltage of Cu/Cu 2+ can’t be measured. Abs electrode potential can’t measured. Only Diff in electrode potential can be measured.

PDF version

Online version

Click here chem database (std electrode potential)


Click here interactive ECS Click here pdf version ECS

Cu Half Cell

http://www.fptl.ru/biblioteka/spravo4niki/handbook-of-Chemistry-and-Physics.pdf

http://www.hbcpnetbase.com/

http://www2.hkedcity.net/sch_files/a/lsc/lsc-chem/public_html/nss/redox/ecs.html

http://downloads.gphysics.net/UACh/Medicina/TablasIones.pdf


http://www.hbcpnetbase.com/


http://downloads.gphysics.net/UACh/Medicina/TablasIones.pdf

Potential Diff Cu/Cu2+


Potential Diff Zn/Zn2+


Zn2+


Zn2+

Zn2+

Zn

Zn2+

Zn

Zn2+

Zn2+ Zn2+


--

-

Zn

---

-+

+++

+ +

++

+Can’t measure Abs Potential

Cu

Cu2+

Cu2+

Cu2+

Cu2+


Cu

-e

-e-e

Cu2+

Cu2+

Cu2+


Zn Half Cell

++

+

Cu

+++

-

Cu Half Cell

Zn/Cu Voltaic Cell

External circuit – flow of electronsComplete circuit

------

------- -

Connect 2 Half Cell with wire/ salt bridge

Zn half cell (-ve)Oxidation

Cu half cell (+ve)Reduction

Salt Bridge – flow of ionsComplete the circuit

Cu2+ + 2e → Cu Zn → Zn 2+ + 2e

Zn + Cu2+ → Zn2+ + Cu

Anode Cathode

Maintain electrical neutrality

Salt bridge – saturated KNO3

Zn2+ increase ↑

NO3- flow in to balance excess Zn2+

Cu2+ decrease ↓, excess –ve ion ↑

K+ flow in to balance loss of Cu2+

Zn Cu

----

Zn2+Zn2+

Zn2+

Excess of Zn2+ ion

++

++

---

-

------

--

Excess of –ve ion

+++

+++

+

Without Salt Bridge

-++

++

With Salt Bridge

(electron unable to flow due to ESF)

NO3-

NO3-

NO3-

NO3-

+

+

+ K+

K+

K+

-

--

K+ flow in to balance excess of – ion

NO3- flow in to balance

excess of + ion

2 Half Cell to make a Voltaic Cell

-e -e

----

++++





Zn2+


Zn2+

Zn2+

Zn

Zn2+

Zn

Zn2+

Zn2+ Zn2+


--

-

Zn

---

-+

+++

+ +

++


Cu

Cu2+

Cu2+

Cu2+

Cu2+


Cu

-e

-e-e

Cu2+

Cu2+

Cu2+


++

+

Cu

+++

-


------

------- -




Voltmeter – High resistance (No current flow) Salt Bridge – flow of ions

Complete the circuit

Cu2+ + 2e → Cu Zn → Zn 2+ + 2e

1.10VoltPotential diff can be measured.

Voltmeter across – EMF

1.10 Volt

Zn + Cu2+ → Zn2+ + Cu

Anode Cathode

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

(aq)| Cu (s)

Cell diagram

Anode Cathode

Half Cell Half Cell(Oxidation) (Reduction)

Phase boundarySalt Bridge Flow electrons



Zn2+ increase ↑


Cu2+ decrease ↓

K+ flow in to balance loss of Cu2+

Zn/Cu Voltaic Cell 2 Half Cell to make a Voltaic CellZn Half Cell Cu Half Cell

-e -e

----

++++

Potential Diff Ag/Ag2+

Electrode potential Ag/Ag2+ = +ve



Zn2+


Zn2+

Zn2+

Zn

Zn2+

Zn

Zn2+

Zn2+ Zn2+


--

-

Zn

---

-+

+++

+ +

++


Ag

Ag+

Ag+

Ag+

Ag+

Ag → Ag+ + e (Oxidation) Ag+ + e → Ag (Reduction)Ag+ + e ↔ Ag(At equilibrium)

Ag

-e

-e-e

Ag+

Ag+

Ag+

Ag+ + e ↔ AgEqui shift to →

++

+

Ag

+++

-


------

------- -



Ag half cell (+ve)Reduction



Ag+ + e → Ag Zn → Zn 2+ + 2e



1.56 Volt

Zn + 2Ag+ → Zn2+ + 2Ag

Anode Cathode

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

(aq)| Ag (s)

Cell diagram

Anode Cathode





Zn2+ increase ↑


Ag+ decrease ↓

K+ flow in to balance loss of Ag+

Zn/Ag Voltaic Cell 2 Half Cell to make a Voltaic CellZn Half Cell Ag Half Cell

Ag

Ag+

-e -e

----

++++

Potential Diff Ag/Ag2+

Electrode potential Ag/Ag2+ = +ve


Electrode potential Cu/Cu2+ = -ve

Cu2+

Cu → Cu 2+ + 2e (Oxidation)Cu 2+ + 2e → Cu (Reduction)Cu 2+ + 2e ↔ Cu (At equilibrium)

Cu2+

Cu2+

Cu

Cu2+

Cu

Cu2+

Cu2+ Cu2+

Cu 2+ + 2e ↔ CuEqui shift to ←

--

-

Cu

---

-+

+++

+ +

++


Ag

Ag+

Ag+

Ag+

Ag+

Ag → Ag+ + e (Oxidation) Ag+ + e → Ag (Reduction)Ag+ + e ↔ Ag(At equilibrium)

Ag

-e

-e-e

Ag+

Ag+

Ag+

Ag+ + e ↔ AgEqui shift to →

++

+

Ag

+++

-


------

------- -


Cu half cell (-ve)Oxidation




Ag+ + e Ag →Cu → Cu 2+ + 2e



0.46 Volt

Cu + 2Ag+ → Cu2+ + 2Ag

Anode Cathode

Cu(s) | Cu2+(aq) || Ag+

(aq)| Ag (s)

Cell diagram

Anode Cathode





Cu2+ increase ↑

NO3- flow in to balance excess Cu2+

Ag+ decrease ↓

K+ flow in to balance loss of Ag+

Cu/Ag Voltaic Cell 2 Half Cell to make a Voltaic CellCu Half Cell Ag Half Cell

Ag

Ag+

Cu

Cu2+

-e -e

----

++++

Standard Electrode Potential

Standard Hydrogen Electrode (SHE)

Platinum coat with Platinum oxide/black – increase surface area for adsorption H2 - catalyze equilibrium bet H2 /H+

- H2 ↔ 2H+ + 2e-

Eθ

Standard Reference electrodeAll Cell Potential are measured against

• Conc ( 1M)• Pressure (1 atm)• Temp (298K)• Platinum- inert electrode (sys without metal)

Standard condition

H2 at 1 atm

Platinum

H2 gas

Pt wire

Platinum

2H+ + 2e ↔ H2

Eθ = 0V

Types of Half Cells

Metal/ Ion (M/M+)

Gas/ Ion (M/M-)

Ion/ Ion (Fe3+/Fe2+)

• Pure Zn metal• Conc (1M Zn2+)• Pressure (1 atm)• Temp (298K)

Condition Std Zn/Zn2+

Condition Std CI2/CI-

• CI2 gas• Platinum electrode• Conc (1M CI-)• Pressure (1 atm)• Temp (298K)

• Platinum electrode• Conc (1M Fe3+/Fe2+)• Pressure (1 atm)• Temp (298K)

Condition Std Fe3+/ Fe2+

Zn2+

Zn

Fe3+/Fe2+

CI-

Condition for Standard CANT

MEASURE

ABS

POTENTIAL

1

2

3

How to measure electrode potential ?

Pt

1M H+

MeasureDifference?

http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s23-electrochemistry.html


Std Hydrogen Electrode (SHE)

Eθ = 0V

Types of Half Cells

Metal/ Ion (M/M+)

Gas/ Ion (M/M+)

Ion/ Ion (Fe3+/Fe2+)

• Pure Zn metal• Conc (1M Zn2+)• Pressure (1 atm)• Temp (298K)

Condition Std Zn/Zn2+

Condition Std CI2/CI-

• CI2 gas• Platinum electrode• Conc (1M CI-)• Pressure (1 atm)• Temp (298K)

• Platinum electrode• Conc (1M Fe3+/Fe2+)• Pressure (1 atm)• Temp (298K)

Condition Std Fe3+/ Fe2+

Zn2+

Zn

Fe3+/Fe2+

1

2

3

Connect to SHE

Connect to SHE

Connect to SHE

Eθ = 0V

Eθ = 0V

Eθ = -0.76VStandard electrode potential Zn/Zn2+ = -0.76V Eθ

cell = -0.76V

Eθ = +0.77V

Eθ = +1.35V

Standard electrode potential Fe3+/Fe2+ = +0.77V Eθ

cell = +0.77V

Standard electrode potential CI2 /CI- = +1.35V Eθ

cell = +1.35V

Eθ= -0.76V

Eθ= +0.77V

Eθ= +1.35V

2 Half Cell with SHE as reference electrode

CI-

Pt

+

+

+

Pt


Std Electrode Potential diff systems

Eθ = 0V

Eθ = 0V

Eθ = 0V

Eθ = -0.76V

Standard electrode potential Zn/Zn2+ = -0.76V Eθ

cell = -0.76V

Eθ = +0.77V

Eθ = +1.35V

Standard electrode potential Fe3+/Fe2+ = +0.77V Eθ

cell = +0.77V

Standard electrode potential CI2 /CI- = +1.35V Eθ

cell = +1.35V

Eθ= -0.76V

Eθ= +0.77V

Eθ= +1.35V

STANDARD Reduction potential – Hydrogen as std

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- 1/2H↔ 2 -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ -0.26Sn2+ + 2e- Sn ↔ -0.14Pb2+ + 2e- Pb ↔ -0.13H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4

2- + 4H+ + 2e- H↔ 2SO3 + H2O +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- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.35MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

-ve reduction potential

+ve reduction potential

Click here std analogy video

Click here std analogy

PDF version


Compared to H2 as std

Eθ cell/Cell Potential = EMF in voltEMF prod when half cell connect to SHE at std conditionStd electrode potential written as std reduction potential

https://www.youtube.com/watch?v=0dLBwyTh9Dg

http://www.chemguide.co.uk/physical/redoxeqia/introduction.html


https://www.youtube.com/watch?v=0dLBwyTh9Dg

http://www.chemguide.co.uk/physical/redoxeqia/introduction.html



H2 half cell (+ve)Reduction

Anode Cathode

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

(aq) , H2(g) | Pt (s)

Cell diagram

Anode Cathode


Salt Bridge Flow electrons

Eθcell = Eθ

(cathode) – Eθ (anode)

Eθcell = 0.00 – ( Eθ Zn )

0.76 = 0.00 - Eθ Zn Eθ Zn = -0.76V

Zn2+ + 2e Zn E↔ θ = ?2H+ + 2e ↔ H2 Eθ = 0.00V

Std electrode potential as std reduction potential

Find Eθcell (use formula)

Eθcell = Eθ

(cathode) – Eθ(anode)

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- 1/2H↔ 2 + OH- -0.83

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

2- + 4H+ + 2e- H↔ 2SO3 + H2O +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- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

-0.76V

+ve/high electrode potential is cathode (+)-ve/ low electrode potential is anode (-)Electrons flow from anode (- ) to cathode (+ )

Eθ Zn/H2 = 0.76V

Zn/H2

Zn

Zn2+H+

Pt

H2

--- +

-e

Zn/H2 Cell Determine Eθ cell Zn/Zn2+

Eθ value DO NOT depend on stoichiometric coefficient (Independent of stoichiometric eqn)

H2 half cell (-ve)Oxidation

Fe3+/2+ half cell (+ve)Reduction

Anode Cathode

Pt(s) | H2, H+(aq) || Fe3+

Fe2+ | Pt (s)

Cell diagram

Anode Cathode





Eθcell = Eθ



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

2- + 4H+ + 2e- H↔ 2SO3 + H2O +0.17Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.521/2I2 + e- ↔ I- +0.54Fe3+ + e- ↔ Fe2+ ?????Ag+ + e- ↔ Ag +0.801/2Br2 + e- ↔ Br- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+0.77V


Eθ H2 /Fe3+ = 0.77V

Pt

Fe3+

H+

PtH2

+++--

-e

H2 /Fe3+,Fe2+ Cell

H2 /Fe3+,Fe2+

2H+ + 2e ↔ H2 Eθ = 0.00VFe3+ + e Fe↔ 2+ Eθ = ????

Eθcell = Eθ


Eθcell = Eθ Fe3+ – (-0.00)

0.77 = Eθ Fe3+

Determine Eθ cell Fe 3+/Fe2+



CI2 half cell (+ve)Reduction

Anode

Pt(s) | H2, H+(aq) || CI2

,CI-

| Pt (s)

Cell diagram

Anode Cathode





Eθcell = Eθ





2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- ?????MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+1.35V


Eθ H2 /CI2

= 1.35V

H+

PtH2 --

-e

H2 /CI2 Cell

2H+ + 2e ↔ H2 Eθ = 0.00VCI + e CI↔ - Eθ = ?????

Eθcell = Eθ


Eθcell = Eθ CI2 – (-0.00)

1.35 = Eθ CI2

H2 /CI2 Cell

+Pt

CI - CI2

Determine Eθ cell H2 /CI2




Anode Cathode

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

(aq) | Cu (s)

Cell diagram

Anode Cathode



Zn/Cu Voltaic Cell

-e -e

Zn/Cu half cells

Eθcell = Eθ


Eθcell = +0.34 – (-0.76) = +1.10V

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


Find Eθcell (use reduction potential)Find Eθ

cell (use formula)

Zn + Cu2+ Zn→ 2+ + Cu Eθ = ?????

Eθcell = Eθ


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

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


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

2- + 4H+ + 2e- H↔ 2SO3 + H2O +0.17

Cu2+ + 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- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.35MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+1.10 V

Eθ Zn/Cu = 1.10V

Cu2+


----

Zn Cu

++++




Anode Cathode

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

(aq) | Ag (s)

Cell diagram

Anode Cathode



Zn/Ag Voltaic Cell

-e -e

Zn/Ag half cells

Eθcell = Eθ


Eθcell = +0.80 – (-0.76) = +1.56V

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



cell (use formula)

Zn + Ag+ Zn→ 2+ + Ag Eθ = ?????

Eθcell = Eθ


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

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



2- + 4H+ + 2e- H↔ 2SO3 + H2O +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- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+1.56 V

Ag

Eθ Zn/Ag = 1.56V

Ag+


----

++++

Zn




Anode Cathode

Cu(s) | Cu2+(aq) || Ag+

(aq) | Ag (s)

Cell diagram

Anode Cathode



Cu/Ag Voltaic Cell

-e -e

Cu/Ag half cells

Eθcell = Eθ


Eθcell = +0.80 – (+0.34) = +0.46V

Cu 2+ + 2e Cu (anode) E↔ θ = +0.34VAg + + e Ag(cathode) E↔ θ = +0.80V



cell (use formula)

Cu + 2Ag+ Cu→ 2+ + 2Ag Eθ = ?????

Eθcell = Eθ


Cu 2+ + 2e Cu E↔ θ = +0.34VAg+ + e Ag E↔ θ = +0.80V

Cu Cu↔ 2+ + 2e Eθ = -0.342Ag+ + 2e 2Ag E↔ θ = +0.80Cu + 2Ag+ Cu → 2+ + 2Ag Eθ = +0.46V

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- 1/2H↔ 2 + OH- -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ -0.26Sn2+ + 2e- Sn ↔ -0.14Pb2+ + 2e- Pb ↔ -0.13H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4


2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+0.46V

AgCu

Cu2+

Half cell- high electrode potential is cathode (+)Half cell - low electrode potential is anode (-)Electrons flow from anode (- ) to cathode (+ )

Eθ Cu/Ag = 0.46V

Ag+

----

++++


Mn half cell (-ve)Oxidation

Ni half cell (+ve)Reduction

Anode Cathode

Mn(s) | Mn2+(aq) || Ni2+

(aq) | Ni (s)

Cell diagram

Anode Cathode



Mn/Ni Voltaic Cell

-e -e

Mn/Ni half cells

Eθcell = Eθ


Eθcell = -0.26 – (-1.19) = +0.93V

Mn 2+ + 2e Mn (anode) E↔ θ = -1.19VNi2+ + 2e Ni (cathode) E↔ θ = -0.26V



cell (use formula)

Mn + Ni2+ Mn→ 2+ + Ni Eθ = ?????

Eθcell = Eθ


Mn 2+ + 2e Mn E↔ θ = -1.19VNi2+ + 2e Ni E↔ θ = -0.26V

Mn Mn↔ 2+ + 2e Eθ = +1.19Ni2+ + 2e Ni E↔ θ = -0.26Mn + Ni2+ Mn→ 2+ + Ni Eθ = +0.93V

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- 1/2H↔ 2 + OH- -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ - 0.26Sn2+ + 2e- Sn ↔ -0.14Pb2+ + 2e- Pb ↔ -0.13H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4


2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+0.93 V

Eθ Mn/Ni = 0.93V

Ni2+


----

NiMn

++++

Mn2+


Fe half cell (-ve)Oxidation

MnO4- half cell (+ve)Reduction

Anode Cathode

Fe(s) | Fe2+(aq) || MnO4

- ,H+, Mn2+ | Pt (s)

Cell diagram

Anode Cathode



Fe/MnO4- Voltaic Cell

-e -e

Fe/MnO4- half cells

Eθcell = Eθ


Eθcell = +1.51 – (-0.45) = +1.96V

Fe2+ + 2e Fe E↔ θ = -0.45VMnO4

- + 5e ↔ Mn2+ + 4H2O Eθ = +1.51V



cell (use formula)

5Fe + 2MnO4- + 16H+ 5Fe→ 2+ +2Mn2+ + 8H2O Eθ = ?

Eθcell = Eθ


Fe 2+ + 2e Fe E↔ θ = -0.45VMnO4

- + 5e ↔ Mn2+ + 4H2O Eθ = +1.51V

Fe Fe↔ 2+ + 2e Eθ = +0.45MnO4

- +5e Mn↔ 2+ + 4H2O Eθ = +1.51Fe + MnO4

- Mn→ 2+ + Fe2+ Eθ = +1.96V

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- 1/2H↔ 2 + OH- -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ -0.26Sn2+ + 2e- Sn ↔ -0.14Pb2+ + 2e- Pb ↔ -0.13H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4


2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36

MnO4- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

1/2F2 + e- ↔ F +2.87

+

+1.96V

PtFe

Fe2+

Eθ Fe/MnO4- = 1.96V

MnO4-

Mn2+

Using platinum electrode


----

++++




Anode Cathode

Zn(s) | Zn2+(aq) || Fe3+ , Fe2+

(aq) | Pt (s)

Cell diagram

Anode Cathode



Zn/Fe3+,Fe2+ Cell

-e -e

Eθcell = Eθ


Eθcell = +0.77 – (-0.76) = +1.53V

Zn2+ + 2e Zn E↔ θ = -0.76VFe3+ + e ↔ Fe2+ Eθ = +0.77V



cell (use formula)

Zn + 2Fe3+ Zn→ 2+ +2Fe2+ Eθ = ?

Eθcell = Eθ


Zn 2+ + 2e Zn E↔ θ = -0.76VFe3+ + e ↔ Fe2+ Eθ = +0.77V

Zn Zn↔ 2+ + 2e Eθ = +0.762Fe3 +2e 2Fe↔ 2+ Eθ = +0.77Zn + 2Fe3+ Zn→ 2+ + 2Fe2+ Eθ = +1.53V




2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36

MnO4- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

1/2F2 + e- ↔ F +2.87

+

+1.53V

PtZn

Zn2+


Eθ Zn/Fe3+ = 1.53V

Fe3+-

Fe2+


Zn/Fe3+,Fe2+

----

++++



I2 half cell (+ve)Reduction

Anode Cathode

Zn(s) | Zn2+(aq) || I2 , I-

(aq) | Pt (s)

Cell diagram

Anode Cathode



Zn/I2 , I- Cell

-e -e

Eθcell = Eθ


Eθcell = +0.54 – (-0.76) = +1.30V

Zn2+ + 2e Zn E↔ θ = -0.76VI2

+ 2e ↔ 2I- Eθ = +0.54V



cell (use formula)

Zn + I2 Zn→ 2+ +2I- Eθ = ?

Eθcell = Eθ


Zn Zn↔ 2+ + 2e Eθ = +0.76I2

+ 2e 2I↔ - Eθ = +0.54Zn + I2

Zn→ 2+ + 2I- Eθ = +1.30V



2- + 4H+ + 2e- H↔ 2SO3 + H2O +0.17Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.521/2I2 + e- ↔ I- +0.54

Fe3+ + e- ↔ Fe2+ + 0.77Ag+ + e- ↔ Ag +0.801/2Br2 + e- ↔ Br- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+1.30V

PtZn

Zn2+


Eθ Zn/I2 = 1.30V

I--

I2


----

++++

Zn/I2 , I-

Zn2+ + 2e Zn E↔ θ = -0.76VI2

+ 2e ↔ 2I- Eθ = +0.54V



H2 half cell (+ve)Reduction

Anode Cathode

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

(aq) , H2(g) | Pt (s)

Cell diagram

Anode Cathode



Eθcell = Eθ


Eθcell = 0.00 – (-0.76) = +0.76V

Zn2+ + 2e Zn E↔ θ = -0.76V2H+ + 2e ↔ H2 Eθ = 0.00V



cell (use formula)

Zn + 2H+ Zn→ 2+ + H2 Eθ = ?

Eθcell = Eθ


Zn 2+ + 2e Zn E↔ θ = -0.76V2H+ + 2e ↔ H2 Eθ = 0.00V

Zn Zn↔ 2+ + 2e Eθ = +0.762H+ +2e H↔ 2

Eθ = 0.00Zn + 2H+ Zn→ 2+ + H2 Eθ = +0.76V




2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+0.76V


Eθ Zn/H2 = 0.76V

Using platinum electrode/H2

Zn/H2

Zn

Zn2+H+

Pt

H2

--- +

-e

Zn/H2 Cell




Anode Cathode

Pt(s) | H2, H+(aq) || Ag+

(aq) | Ag (s)

Cell diagram

Anode Cathode



H2/Ag Cell

Eθcell = Eθ


Eθcell = +0.80 – (-0.00) = +0.80V

2H+ + 2e ↔ H2 Eθ = 0.00VAg+ + e Ag E↔ θ = +0.80V



cell (use formula)

H2 + 2Ag+ 2H→ + + 2Ag Eθ = ?

Eθcell = Eθ


H2 2H↔ + + 2e Eθ = +0.002Ag+ +2e 2Ag E↔ θ = +0.80H2 + 2Ag+ 2H→ + + 2Ag Eθ = +0.80V




2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+0.80V


Eθ H2 /Ag = 0.80V


H2/Ag

Ag

Ag+

H+

PtH2

2H+ + 2e ↔ H2 Eθ = 0.00VAg+ + e Ag E↔ θ = +0.80V

+++--

-e




Anode Cathode

Pt(s) | H2, H+(aq) || Fe3+

Fe2+ | Pt (s)

Cell diagram

Anode Cathode





cell (use formula)

H2 + 2Fe3+ 2H→ + + 2Fe 2+ Eθ = ?

Eθcell = Eθ


H2 2H↔ + + 2e Eθ = +0.002Fe3+ +2e 2Fe↔ 2+ Eθ = +0.77H2 + 2Fe3+ 2H→ + + 2Fe2+ Eθ = +0.77V




2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+0.77V


Eθ H2 /Fe3+ = 0.77V


Pt

Fe3+

H+

PtH2

+++--

-e

H2 /Fe3+,Fe2+ Cell

H2 /Fe3+,Fe2+

2H+ + 2e ↔ H2 Eθ = 0.00VFe3+ + e Fe↔ 2+ Eθ = +0.77V

2H+ + 2e ↔ H2 Eθ = 0.00VFe3+ + e Fe↔ 2+ Eθ = +0.77V

Eθcell = Eθ


Eθcell = +0.77– (-0.00) = +0.77V



CI2 half cell (+ve)Reduction

Anode Cathode

Pt(s) | H2, H+(aq) || CI2

,CI-

| Pt (s)

Cell diagram

Anode Cathode





cell (use formula)

CI2 + H2 2CI→ - + 2H+ Eθ = ?

Eθcell = Eθ


H2 2H↔ + + 2e Eθ = +0.00CI2

+2e 2CI↔ - Eθ = +1.35H2 + CI2

2H→ + + 2CI- Eθ = +1.35V




2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.33

1/2CI2 + e- ↔ CI- +1.35MnO4

- + 8H+ + 5e- ↔ Mn2+ +1.511/2F2 + e- ↔ F +2.87

+

+1.35V


Eθ H2 /CI2

= 1.35V



H+

PtH2 --

-e

H2 /CI2 Cell

2H+ + 2e ↔ H2 Eθ = 0.00VCI + e CI↔ - Eθ = +1.35V

Eθcell = Eθ


Eθcell = +1.35 – (-0.00) = +1.35V

H2 /CI2 Cell

2H+ + 2e ↔ H2 Eθ = 0.00VCI + e CI↔ - Eθ = +1.35V

+Pt

CI - CI2


STANDARD Reduction potential – H2 as std

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- H↔ 2+OH- -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ -0.26Sn2+ + 2e- Sn ↔ -0.14Pb2+ + 2e- Pb ↔ -0.13H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4

2- + 4H+ + 2e- H↔ 2SO3 +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- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ +7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F- +2.87

-ve reduction potential

+ve reduction potential

Compared to H2 as std

Eθ cell/Cell Potential = EMF in voltEMF when half cell connect to SHE std conditionStd potential written as std reduction potential

TOP right• High ↑ tendency lose e• Li Li → + + e• Eθ Li = +3.04V• STRONG reducing Agent•Oxi favourable (Eθ =+ve)


STRONG Reducing Agent

WEAK Reducing Agent

BOTTOM right• Low ↓ tendency lose e• F - 1/2F→ 2 + e• Eθ F2 = - 2.87V• WEAK reducing Agent•Oxi NOT favourable (Eθ =-ve)

BOTTOM right• Low ↓ tendency lose e• F - 1/2F→ 2 + e

• Eθ F2 = - 2.87V• WEAK reducing Agent•Oxi NOT favourable (Eθ =-ve)

WEAK Oxidizing Agent

STRONGOxidizing Agent

TOP left• Low ↓ tendency gain e• Li+ + e Li→• Eθ Li= - 3.04V• WEAK oxidizing Agent• Red NOT favourable (Eθ =-ve)

TOP left• Low ↓ tendency gain e• Li+ + e Li→• Eθ Li= - 3.04V• WEAK oxidizing Agent• Red NOT favourable (Eθ =-ve)

BOTTOM left• High ↑ tendency gain e• F2 + 2e 2F→ -

• Eθ F2= +2.87V• STRONG oxidizing Agent•Red favourable (Eθ =+ve)



О О

О О


STANDARD Reduction potential – H2 as std


2- + 4H+ + 2e- H↔ 2SO3 +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- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ +7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F - +2.87

Eθ cell/Cell Potential = EMF in voltEMF when half cell connect to SHE std conditionStd potential written as std reduction potential









Li Li → + + e Eθ Li = +3.04VLi Li → + + e Eθ Li = +3.04V

1/2F2 + e F→ - Eθ F2 = + 2.87V1/2F2 + e F→ - Eθ F2 = + 2.87V

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2-+14H+ +6e- ↔ 2Cr3+ +7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F - +2.87

-ve potential

+ve potential

Uses of Standard Electrode Potential (SEP) Data1

TOP left• Low ↓ tendency gain e• Li+ + e Li→• Eθ Li= - 3.04V• Red NOT favourable (Eθ =-ve)

TOP left• Low ↓ tendency gain e• Li+ + e Li→• Eθ Li= - 3.04V• Red NOT favourable (Eθ =-ve)



О TOP right• High ↑ tendency lose e• Li Li → + + e• Eθ Li = +3.04V•Oxi favourable (Eθ =+ve)

TOP right• High ↑ tendency lose e• Li Li → + + e• Eθ Li = +3.04V•Oxi favourable (Eθ =+ve)

STRONG Reducing AgentО

WEAK Reducing Agent


• Eθ F2= +2.87V•Red favourable (Eθ =+ve)



О

BOTTOM right• Low ↓ tendency lose e• F - 1/2F→ 2 + e• Eθ F2 = - 2.87V•Oxi NOT favour (Eθ =-ve)

BOTTOM right• Low ↓ tendency lose e• F - 1/2F→ 2 + e• Eθ F2 = - 2.87V•Oxi NOT favour (Eθ =-ve)

О

Relative strength of Oxidizing/Reducing Agent

Eθ = +ve SEP ↓

Strong oxidizing↓

Weak reducing agent↓

F2 strongest oxidizing agent

↓F- ion weakest reducing agent

Eθ = -ve SEP ↓

Weak oxidizing↓

Strong reducing agent↓

Li+ ion weakest oxidizing agent

↓Li metal strongest

reducing agent

Reaction to happen↓

1 Oxidizing + 1 Reducing agent (Strong) (Strong)

from both side

Reaction NEVER happen↓

TWO Oxidizing agent from same sides

orTWO Reducing agent from same sides

Oxidized sp ↔ Reduced sp Eθ/VLi+ + e- Li↔ -3.04

Ca2+ + 2e- Ca ↔ -2.87Na+ + e- Na↔ -2.71

Al3+ + 3e- AI ↔ -1.66Mn2+ + 2e- Mn ↔ -1.19H2O + e- H↔ 2+OH- -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ -0.26Sn2+ + 2e- Sn ↔ -0.14Pb2+ + 2e- Pb ↔ -0.13H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4

2- + 4H+ + 2e- H↔ 2SO3 + H2O +0.17Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.521/2I2 + e- ↔ I- +0.54Fe3+ + e- Fe↔ 2+ +0.77Ag+ + e- ↔ Ag +0.80

1/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ +7H2O +1.331/2CI2 + e- ↔ CI- +1.36

1/2F2 + e- ↔ F - +2.87


TOP left• Low ↓ tendency gain e• Na+ + e Na→• Eθ Na = - 2.71V• Red NOT favourable (Eθ =-ve)

TOP left• Low ↓ tendency gain e• Na+ + e Na→• Eθ Na = - 2.71V• Red NOT favourable (Eθ =-ve)






WEAK Reducing Agent





BOTTOM right• Low ↓ tendency lose e• Ag Ag→ + + e• Eθ Ag = - 0.80V•Oxi NOT favour (Eθ =-ve)

BOTTOM right• Low ↓ tendency lose e• Ag Ag→ + + e• Eθ Ag = - 0.80V•Oxi NOT favour (Eθ =-ve)

О


ОО

Rxn feasible

Rxn not feasible

Rxn not feasible

Rxn feasibleО



from both side

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 ↔ -1.19

Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ -0.26Sn2+ + 2e- Sn ↔ -0.14

H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4

2- + 4H+ + 2e- H↔ 2SO3 + H2O +0.17Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40

1/2I2 + e- ↔ I- +0.54Fe3+ + e- Fe↔ 2+ +0.77Ag+ + e- ↔ Ag +0.80Pb2+ + 2e- Pb ↔ -0.131/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ +1.331/2CI2 + e- ↔ CI- +1.36

1/2F2 + e- ↔ F - +2.87





О

Zsign

Zn Zn↔ 2+ + 2e Eθ = +0.76VSn2+ + 2e Sn E↔ θ = -0.14VZn + Sn2+ Zn→ 2+ + Sn Eθ = +0.62V

Rxn bet Zn + Sn2+

Will it happen ?

Eθ = +0.62V+ve (spontaneous)

О

О

О Zsign



from both side

Rxn bet CI2 + I-

Will it happen ?

2I- I↔ 2 + 2e Eθ = -0.54VCI2 + 2e 2CI↔ - Eθ = +1.36VCI2 + 2I- 2CI→ - + I2

Eθ = +0.82V


Zn CI2

Both gaining electron NON spontaneous

Oxidized sp ↔ Reduced sp Eθ/V

K+ + e- K↔ -2.93Ca2+ + 2e- Ca ↔ -2.87Na+ + e- Na↔ -2.71Mg 2+ + 2e- Mg ↔ -2.37

Mn2+ + 2e- Mn ↔ -1.19


H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4

2- + 4H+ + 2e- H↔ 2SO3 + H2O +0.17Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40

I2 + e- ↔ I- +0.54Fe3+ + e- Fe↔ 2+ +0.77Ag+ + e- ↔ Ag +0.80Pb2+ + 2e- Pb ↔ -0.131/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ +7H2O +1.33 CI2 + e- ↔ CI- +1.36

1/2F2 + e- ↔ F - +2.87





WEAK Reducing Agent


О

О Rxn bet CI2 + I2

Will it happen ?

О

Rxn NEVER happen ↓TWO Oxidizing agent from same sides

Rxn NEVER happen ↓TWO Reducing agent from same sides

Rxn bet Zn + Sn Will it happen ?

Both losing electron NON spontaneous

О

Rxn NEVER happen↓

1 Oxidizing + 1 Reducing agent (WEAK) (WEAK)

from both side

Rxn bet Mg + K +

Will it happen ?

О

О

Eθ = -ve-ve (Non spontaneous)



Anode Cathode

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

(aq) | Cu (s)

Anode Cathode



-e -e

Zn/Cu half cells

Eθcell = Eθ


Eθcell = +0.34 – (-0.76) = +1.10V

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



cell (use formula)

Zn + Cu2+ Zn→ 2+ + Cu Eθ = ?????

Eθcell = Eθ


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

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



2- + 4H+ + 2e- H↔ 2SO3 + H2O +0.17

Cu2+ + 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- +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.35MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+1.10 V

Eθ Zn/Cu = 1.10V

Cu2+

----

Zn Cu

++++


Find Eθ using std electrode potential data for Zn/Cu half cell



Anode Cathode

Cu(s) | Cu2+(aq) || Ag+

(aq) | Ag (s)

Anode Cathode



-e -e

Eθcell = Eθ


Eθcell = +0.80 – (+0.34) = +0.46V

Cu 2+ + 2e Cu (anode) E↔ θ = +0.34VAg + + e Ag(cathode) E↔ θ = +0.80V



cell (use formula)

Cu + 2Ag+ Cu→ 2+ + 2Ag Eθ = ?????

Eθcell = Eθ


Cu 2+ + 2e Cu E↔ θ = +0.34VAg+ + e Ag E↔ θ = +0.80V

Cu Cu↔ 2+ + 2e Eθ = -0.342Ag+ + 2e 2Ag E↔ θ = +0.80Cu + 2Ag+ Cu→ 2+ + 2Ag Eθ = +0.46V

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- 1/2H↔ 2 -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ -0.26Sn2+ + 2e- Sn ↔ -0.14Pb2+ + 2e- Pb ↔ -0.13H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4


2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+0.46V

AgCu

Cu2+

Eθ Cu/Ag = 0.46V

Ag+

----

++++


Find Eθ using std electrode potential data for Cu/Ag half cell

Mn half cell (-ve)Oxidation

Ni half cell (+ve)Reduction

Anode Cathode

Mn(s) | Mn2+(aq) || Ni2+

(aq) | Ni (s)

Anode Cathode



-e -e

Eθcell = Eθ


Eθcell = -0.26 – (-1.19) = +0.93V

Mn 2+ + 2e Mn (anode) E↔ θ = -1.19VNi2+ + 2e Ni (cathode) E↔ θ = -0.26V



cell (use formula)

Mn + Ni2+ Mn→ 2+ + Ni Eθ = ?????

Eθcell = Eθ


Mn 2+ + 2e Mn E↔ θ = -1.19VNi2+ + 2e Ni E↔ θ = -0.26V

Mn Mn↔ 2+ + 2e Eθ = +1.19Ni2+ + 2e Ni E↔ θ = -0.26Mn + Ni2+ Mn→ 2+ + Ni Eθ = +0.93V

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- 1/2H↔ 2 + OH- -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ - 0.26Sn2+ + 2e- Sn ↔ -0.14Pb2+ + 2e- Pb ↔ -0.13H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4


2-+14H+ +6e- ↔ 2Cr3+ + 7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.511/2F2 + e- ↔ F +2.87

+

+0.93 V

Eθ Mn/Ni = 0.93V

Ni2+

----

NiMn

++++

Mn2+

2 Uses of Standard Electrode Potential (SEP) Data

Find Eθ using std electrode potential data for Mn/Ni half cell

Eθ = -0.20V-ve (NON spontaneous)



from both side



H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15

Cu2+ + 2e- ↔ Cu +0.341/2O2 + H2O +2e- ↔ 2OH- +0.40Cu+ + e- ↔ Cu +0.52I2 + 2e- ↔ I- +0.54

Ag+ + e- ↔ Ag +0.801/2Br2 + e- Br↔ - +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ +7H2O +1.331/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- Mn↔ 2+ + 4H2O +1.511/2F2 + e- ↔ F - +2.87





WEAK Reducing Agent

О

ZZn Zn↔ 2+ + 2e Eθ = +0.76Sn2+ + 2e Sn E↔ θ = -0.14Zn + Sn2+ Zn→ 2+ + Sn Eθ = +0.62V

Rxn bet Zn + Sn2+

Will it happen ?




from both side

Rxn bet Cu2+ +I-

Will it happen ?

О

Rxn feasible

О

О

2I- I↔ 2 + 2e Eθ = -0.54Cu2+ + 2e Cu E↔ θ = +0.342I- + Cu2+ Cu→ + I2

Eθ = -0.20V


Rxn not feasible

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

(aq) | Sn (s)

(Oxidation) (Reduction)

Anode Cathode


Eθcell = Eθ


Eθcell = -0.14 – (-0.76) = +0.62V


Pt(s) | I-, I2 || Cu2+(aq) | Cu (s)

Anode Cathode



Eθcell = Eθ


Eθcell = +0.34 – (+0.54) = -0.20V

Determine spontaneity rxn. Will it HAPPEN ?




from both side



2- + 4H+ + 2e- H↔ 2SO3 +0.17Cu2+ + 2e- ↔ Cu +0.34

I2 + 2e- ↔ I- +0.54Fe3+ + e- Fe↔ 2+ +0.77Ag+ + e- ↔ Ag +0.801/2Br2 + e- Br↔ - +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ +7H2O +1.331/2CI2 + e- ↔ CI- +1.36

1/2F2 + e- ↔ F - +2.87





WEAK Reducing Agent

О

Z

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

Rxn bet Zn + Cu2+

Will it happen ?




from both side

Rxn bet I2 +CI-

Will it happen ?

О

Rxn feasibleО

О

2CI- CI↔ 2 + 2e Eθ = -1.36I2

+ 2e 2I↔ - Eθ = +0.54I2 + 2CI- 2I-→ + CI2

Eθ = -0.82V


Rxn not feasible

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

(aq) | Cu (s)


Anode Cathode


Eθcell = Eθ


Eθcell = 0.34 – (-0.76) = +1.10V


Pt(s) | CI-, CI2 || I2 I- | Pt (s)

Anode Cathode



Eθcell = Eθ


Eθcell = +0.54 – (+1.36) = -0.82V



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- H↔ 2 + OH- -0.83Zn2+ + 2e- Zn ↔ -0.76Fe2+ + 2e- Fe ↔ -0.45Ni2+ + 2e- Ni ↔ -0.26Sn2+ + 2e- Sn ↔ -0.14

H+ + e- 1/2H↔ 2 0.00Cu2+ + e- Cu↔ + +0.15SO4

2- + 4H+ + 2e- H↔ 2S +0.17Cu2+ + 2e- ↔ Cu +0.34

Cu+ + e- Cu ↔ +0.52

Fe3+ + e- Fe↔ 2+ +0.77Ag+ + e- ↔ Ag +0.801/2Br2 + e- Br↔ - +1.071/2O2 + 2H+ +2e- ↔ H2O +1.23Cr2O7

2-+14H+ +6e- ↔ 2Cr3+ +1.331/2CI2 + e- ↔ CI- +1.36

1/2F2 + e- ↔ F - +2.87





WEAK Reducing Agent

Cu Cu↔ 2+ + 2e Eθ = -0.342H+ + 2e H↔ 2 Eθ = +0.00Cu + 2H+ Cu→ 2+ +H2

Eθ = -0.34V

Rxn bet Cu + H+

Will it happen ?


Reaction NEVER happen↓1 Oxidizing + 1 Reducing agent (WEAK) (WEAK) from both side

Rxn bet Fe3+ +CI-

Will it happen ?

О

О

О

2CI- CI↔ 2 + 2e Eθ = -1.362Fe3+ + 2e 2Fe↔ 2+ Eθ = +0.772Fe3+ + 2CI- 2Fe→ 2++CI2

Eθ = -0.59V


Rxn not feasible

Cu(s) | Cu2+(aq) || H+

H2 | Pt (s)


Anode Cathode


Eθcell = Eθ


Eθcell = 0.00 – (+0.34) = -0.34V

Eθ = -0.34V-ve (spontaneous)

Pt(s) | CI-, CI2 || Fe3+ ,Fe2+ |Pt (s)

Anode Cathode



Eθcell = Eθ


Eθcell = +0.77 – (+1.36) = -0.59V

О

Rxn not feasible

Reaction NEVER happen↓1 Oxidizing + 1 Reducing agent (WEAK) (WEAK) from both side


Eθ value DO NOT depend surface area of metal electrode.EMF = Energy per unit charge. (Joule)/CEMF 10v = 10J energy released by 1C of charge flowing = 100J energy released by 10C of charge flowing Eθ – intensive property– independent of amt – ratio energy/charge

Increasing surface area metal will NOT increase EMF

Eθ Zn/Cu = 1.10V

Surface area exposed 10 cm2 Total charges 100C leave electrodeEMF = 1.10V = 1.1 J energy for 1 C (charges leaving)1C release 1.1J energy100 C release 110 J energyVoltmeter measure energy for 1C – 110J/100C – 1.1VEMF no change

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

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

ond

electron

ond

CoulombA

sec.1

.1028.6

sec1

11

18×==

Surface area increase ↑

Total Energy increase ↑

Total Charge increase ↑ Current increase ↑

BUT EMF remain SAMEEMF = (Energy/charge)t

QI

tIQ

=

×=

Q up ↑ – I up ↑

100C flow110J released

VEMF

EMF

eCh

EnergyEMF

10.1100

110

arg

=

=

=

Surface area exposed 10 cm2

Surface area exposed 100cm2

Surface area exposed 100 cm2 Total charges 1000C leave electrodeEMF = 1.10V = 1.1 J energy for 1 C (charges leaving)1C release 1.1J energy1000 C release 1100 J energyVoltmeter measure energy for 1C – 1100J/1000C – 1.1VEMF no change

VEMF

EMF

eCh

EnergyEMF

10.11000

1100

arg

=

=

=

Eθ Zn/Cu = 1.10V

1000C flow

1100J released

t

QI =

t

QI =

Iron rust in presence of water + oxygen Iron galvanized/coated with zinc.

Oxidized sp ↔ Reduced sp Eθ/VZn2+ + 2e- Zn↔ - 0.76Fe2+ + 2e- ↔ Fe -0.44O2 + 2H2O + 4e ↔ 4OH- +0.40

Iron rustingRusting Process happen

Eθ/VFe2+ + 2e- ↔ Fe -0.44O2 + 2H2O + 4e ↔ 4OH- +0.40O2 + 4H+ + 4e ↔ 2H2O + 1.23

H2O + O2 less reactive (cathode region) – reduction – gain e

Fe more reactive (anode region) – oxidation - lose e

OxidationReduction

Fe2+ + 2e- Fe -0.44↔O2 +2H2O+4e ↔ 4OH- +0.40

Fe Fe↔ 2+ + 2e Eθ = +0.44O2+2H2O+4e ↔ 4OH- Eθ = +0.402Fe +O2

+2H2O→2Fe2++4OH- Eθ = +0.84V

Eθ = +0.84V +ve (spontaneous)

ОО

Dissolve O2 in water

Dissolve O2 in acid

How galvanizing reduces rusting

Iron Galvanized with Zn

Iron/Steel Galvanized with tin

Zn more reactive – lose e instead of Fe

Zn as Sacrificial metal/ Cathodic Protection

Electron flow to O2/H2O region

Prevent Fe rusting/lose e

O2 gain e

Fe

O2 + 2H2O + 4e ↔ 4OH-

flow e-

Zn oxidation/lose e

Zn2+ + 2e- ↔ Zn -0.76O2 +2H2O+4e ↔ 4OH- +0.40

Zn lose e- (Stronger RA)

Zn ↔ Zn2+ + 2e Eθ = +0.76O2+2H2O+4e ↔ 4OH- Eθ = +0.402Zn +O2

+2H2O→2Zn2++4OH- Eθ = +1.16V

Eθ = +1.16 +ve (spontaneous)

water

ОО

Anodic region

Cathodic region

Zn Zn

FeFeFe


Iron rust in presence of water + oxygenIron can coated with tin widely used in canningTin corrodes less readily than iron (protect iron)

Oxidized sp ↔ Reduced sp Eθ/VFe2+ + 2e- ↔ Fe -0.44Sn2+ + 2e- Sn -0.14↔O2 + 2H2O + 4e ↔ 4OH- +0.40

Iron rusting

If tin coat broken, iron rust faster as it will displace tin ions from its solutionWill iron rust spontaneously, if Sn2+ (tin ions) are formed.

Rusting Process happen

Eθ/VFe2+ + e- ↔ Fe -0.44O2 + 2H2O + 4e ↔ 4OH- +0.40O2 + 4H+ + 4e ↔ 2H2O + 1.23

H2O + O2 less reactive (cathode region) – reduction – gain e

Fe more reactive (anode region) – oxidation - lose e

OxidationReduction

Fe2+ + 2e- Fe -0.44↔O2 +2H2O+4e ↔ 4OH- +0.40


+2H2O 4Fe→ 2++4OH- Eθ = +0.84V


ОО

Dissolve O2 in water

How coating reduces rustingIron/Steel coated with tin/Sn

BUT if it is exposed - Fe will rust Fe more reactive Sn

Tin/Sn protect Fe metal

Electron flow Fe to O2/H2O region

water

Fe oxidation/lose e

flow e-

O2 + 2H2O + 4e ↔ 4OH-

Iron metal

waterO2 gain e

SnSn2+

Oxidized sp ↔ Reduced sp Eθ/VFe2+ + 2e- ↔ Fe -0.44Sn2+ + 2e- Sn -0.14↔O2 + 2H2O + 4e ↔ 4OH- +0.40


+2H2O 4Fe→ 2++4OH- Eθ = +0.84V

Fe Fe↔ 2+ + 2e Eθ = +0.44Sn2+ + 2e ↔ Sn Eθ = -0.14Fe + Sn2+ → Fe2++ Sn Eθ = +0.30V


О

О

О

О

Sn SnSn

FeFeFe Fe

State which is able to convert Fe2+ to Fe3+

Oxidized sp ↔ Reduced sp Eθ/VAI3+ + 3e- AI -1.66↔I2 + 2e- ↔ 2I- +0.54Fe3+ + e- ↔ Fe2+ +0.77H2O2 + 2H+ + 2e ↔ 2H2O +1.07Co3+ + e ↔ Co2+ +1.51

2Fe2+ 2Fe↔ 3+ + 2e Eθ = -0.77H2O2 + 2H+ + 2e 2H↔ 2O Eθ =+1.072Fe2+ + H2O2

+ 2H+ 2Fe→ 3+ + 2H2O Eθ = +0.30V


Fe2+ Fe↔ 3+ + e Eθ = -0.77Co3+ + e ↔ Co 2+ Eθ =+1.51Fe2+ + Co3+ Fe→ 3+ + Co2+ Eθ = +0.74V


Eθ cell = EMF in V (std condition)Eθ = Show ease/tendency of species to accept/lose electronEθ = +ve std electrode potential = stronger oxidizing agent – weaker reducing agent – accept eEθ = - ve std electrode potential = stronger reducing agent - weaker oxidizing agent – lose eEMF when half cell connect to SHE std conditionStd potential written as std reduction potentialEθ value DO NOT depend on stoichiometric coefficient. EMF = Energy per unit charge. (Joule)/CEMF 10v = 10J energy released by 1C of charge flowing = 100J energy released by 10C of charge flowing Eθ , Std electrode potential – intensive property – not dependent on amt – ratio energy/chargeEθ = +ve suggest rxn feasible, does not tell rate, feasible but may be slow, give no indication rateEθ = +ve = Energetically feasible but kinetically non feasible

E = ↑ +ve ↑ (OA)

Oxidized sp ↔ Reduced sp Eθ/VFe3+ + e- ↔ Fe2+ +0.77H2O2 +2H++2e ↔ 2H2O +1.07

Oxidized sp ↔ Reduced sp Eθ/VFe3+ + e- ↔ Fe2+ +0.77Co3+ + e ↔ Co2+ +1.51

Stronger OA

Strongest OA

Redox Question

Aluminium air battery

Excellent Zn/Cu gravity cell for IA

Zinc air battery

Videos on battery making

Arrange the species in order of increasing oxidizing/reducing strength

Oxidized sp ↔ Reduced sp Eθ/VZn2+ + 2e- Zn -0.76↔Br2 + 2e- ↔ 2Br- +1.07I2 + 2e- ↔ 2I- +0.54Fe3+ + e- ↔ Fe2+ +0.77MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

Oxidizing agent (OA)MnO4

_ > Br2 > Fe3+ > I2 > Zn2+

Reducing agent (RA)Zn > I- > Fe 2+ > Br- > Mn2+

Arrange in order of increasing reducing strength. (Strongest reducing agent)

Redox Questions

1 2

E = most +ve ↑ strongest OA

E = most -ve ↑ strongest RA

Oxidized sp ↔ Reduced sp Eθ/VZn2+ + 2e- Zn -0.76↔I2 + 2e- ↔ 2I- +0.54Fe3+ + e- ↔ Fe2+ +0.77Br2 + 2e- ↔ 2Br- +1.07MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

arrange increasing ↑ E value

E = ↑ +ve ↑ (OA)

Eθ/VX 3+ + 3e- X -1.56↔Y 2+ + 2e- Y -2.70↔Z 2+ + 2e- Z +0.90↔

E = ↑ -ve ↑(RA)

E = most -ve ↑strongest RA

Reducing agentY > X > Z

arrange increasing ↑ E value

Eθ/VY 2+ + 2e- Y -2.70↔X 3+ + 3e- X -1.56↔Z 2+ + 2e- Z +0.90↔

E = ↑ -ve ↑ (RA)

4433

Oxidized sp ↔ Reduced sp Eθ/VTi2+ + 2e- Ti -1.63↔2H+ + 2e- H↔ 2 0.00

Rxn bet Ti + H+

Will it happen ?

Ti Ti↔ 2+ + 2e Eθ = +1.632H+ + 2e H↔ 2 Eθ = 0.00Ti + 2H+ Ti→ 2+ + H2

Eθ = +1.63V Eθ = +1.63V+ve (spontaneous)

What happen when gold added to acid

Oxidized sp ↔ Reduced sp Eθ/V2H+ + 2e- H↔ 2 0.00Au3+ + 3e ↔ Au +1.58

Rxn bet Au + H+

Will it happen ?

What happen when titanium added to acid

2Au 2↔ Au3+ + 6e Eθ = -1.586H+ + 6e 3H↔ 2 Eθ = 0.002Au + 6H+ 2Au→ 3+ + 3H2

Eθ = -1.58V Eθ = -1.58V-ve ( NON spontaneous)

acid acid

Redox Question

6Predict if manganate will oxidize chloride ion?MnO2 + 4H+ + 2CI- Mn→ 2+ + 2H2O + CI2 Eθ = ?

55

MnO2 +4H+ + 2e- Mn↔ 2+ + 2H2O +1.23

1/2CI2 + e- ↔ CI- +1.36

2CI- CI↔ 2 + 2e Eθ = -1.36MnO2 + 4H+ + 2e Mn↔ 2+ + 2H2O Eθ = +1.23MnO2 + 4H++2CI- Mn→ 2++2H2O+CI2 Eθ= -0.13V

Eθ = -0.13V -ve (NON spontaneous)

Oxidized sp ↔ Reduced sp Eθ/VCr2O7

2-+ 14H+ + 6e- ↔ 2Cr3+ + 7H2O +1.33MnO2

+4H+ + 2e- Mn↔ 2+ + 2H2O +1.231/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

Predict if MnO4- able to oxidize aq CI- to CI2

2MnO4 + 16H+ + 10CI- 2Mn→ 2+ + 8H2O + 5CI2

E = ↑ +ve ↑ (OA)

ОО Oxidized sp ↔ Reduced sp Eθ/VCr2O7

2-+ 14H+ + 6e- ↔ 2Cr3+ + 7H2O +1.33MnO2

+4H+ + 2e- Mn↔ 2+ + 2H2O +1.231/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51

О О2CI- CI↔ 2 + 2e Eθ = -1.36MnO4 - + 8H+ + 5e Mn↔ 2+ + 4H2O Eθ = +1.512MnO4 + 16H++10CI- 2Mn→ 2++8H2O+5CI2 Eθ= +0.15V

1/2CI2 + e- ↔ CI- +1.36MnO4

- + 8H+ + 5e- ↔ Mn2+ + 4H2O +1.51


Predict if iron react with HCI a) absence air

Which is stronger OA ?

Fe Fe↔ 2+ + 2e Eθ = +0.442H+ + 2e H↔ 2 Eθ = 0.00VFe + 2H+ Fe→ 2+ + H2

Eθ = +0.44V


Oxidized sp ↔ Reduced sp Eθ/VFe2+ + 2e- Fe -0.44↔2H+ + 2e- H↔ 2 0.00O2 +2H2O+4e ↔ 4OH- +0.40


+2H2O→2Fe2++4OH- Eθ = +0.84V

Predict if iron react with HCI b) presence of air

Fe2+ + 2e- Fe -0.44↔2H+ + 2e- H↔ 2 0.00

ОО Fe2+ + 2e- Fe -0.44↔O2 +2H2O+4e ↔ 4OH- +0.40

ОО

Oxidized sp ↔ Reduced sp Eθ/VFe2+ + 2e- Fe -0.44↔2H+ + 2e- H↔ 2 0.00O2 +2H2O+4e ↔ 4OH- +0.40


Iron rusting

E = ↑ +ve ↑ (OA)

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.htmlhttp://wps.prenhall.com/wps/media/objects/4680/4792445/ch18_10.htm

Prepared by Lawrence Kok

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

Date post:	20-Jul-2015
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IB Chemistry on Standard Reduction Potential, Standard Hydrogen Electrode and Electrochemical Series

Education