STRONG Reducing Agent

WEAK Reducing Agent

WEAK Oxidising Agent

STRONG Oxidising Agent

• Species on TOP left • Low ↓ tendency to gain e • Li+ + e → Li • Eθ Li= - 3.04V • WEAK oxidising Agent • Reduction NOT favourable

• Species on TOP right • High ↑tendency to lose e • Li → Li + + e • Eθ Li = +3.04V • STRONG reducing Agent •Oxidation favourable

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

• Eθ F2= +2.87V • STRONG oxidising Agent •Reduction favourable

• Species on BOTTOM right • Low ↓tendency to lose e • F - →1/2F2 + e • Eθ F2 = -2.87V • WEAK reducing Agent •Oxidation NOT favourable

Electrochemical Series • Shows the ease/tendency of species to accept/lose electrons

• Written as standard reduction potential

Voltaic Cell for Zn/Cu electrode

Copper reduction Half Cell Zinc oxidation Half Cell

Eθcell = Eθ

(+ve) - Eθ (-ve)

Eθcell = Eθ

(cathode) (red) - Eθ (anode)(oxi)

Eθcell = Eθ

(Cu) - Eθ (Zn)

= 0.34 – (-0.76)

= +1.10V

Zn2+ + 2e → Zn Eθcell = -0.76V (Reduction potential)

Cu2+ + 2e → Cu Eθcell = +0.34V (Reduction potential)

Zn → Zn 2+ + 2e Eθ(Zn) (Oxidation potential)

Cu 2+ + 2e → Cu Eθ (Cu) (Reduction potential)

Zn + Cu 2+ → Cu + Zn 2+ Eθcell + 1.10V

Cell Diagram for Zn/Cu half cells

Electrode/electrolyte ║ Electrolyte/electrode

(-ve)/oxidation ║ (+ve)/reduction

Zn/Zn2+ ║ Cu 2+/Cu

Determine Eθcell using std electrode potential and formula Eθ

cell = Eθ (+ve) - E

θ (-ve)

Voltaic Cell for Mg/Fe electrode

Iron reduction Half Cell

Eθcell = Eθ

(+ve) - Eθ (-ve)

Eθcell = Eθ

(cathode) (red) - Eθ (anode) (oxi)

Eθcell = Eθ

(Fe) - Eθ (Mg)

= 0.44 – (-2.36)

= +1.92V

Mg2+ + 2e → Mg Eθcell = -2.36V (Reduction potential)

Fe2+ + 2e → Fe Eθcell = +0.44V (Reduction potential)

Mg → Mg 2+ + 2e Eθ(Mg) (Oxidation potential)

Fe 2+ + 2e → Fe Eθ (Fe) (Reduction potential)

Mg + Fe2+ → Fe + Mg2+ Eθcell = + 1.92V

Cell Diagram for Mg/Fe half cells

Electrode/electrolyte ║ Electrolyte/electrode

(-ve)/oxidation ║ (+ve)/reduction

Mg/Mg2+ ║ Fe2+/Fe

Determine Eθcell using std electrode potential and formula Eθ

cell = Eθ (+ve) - E

θ (-ve)

Magnesium oxidation Half Cell

STRONG Reducing Agent

WEAK Reducing Agent

WEAK Oxidising Agent

STRONG Oxidising Agent

• Species on TOP left • Low ↓ tendency to gain e • Li+ + e → Li • Eθ Li= - 3.04V • WEAK oxidising Agent • Reduction NOT favourable

• Species on TOP right • High ↑tendency to lose e • Li → Li + + e • Eθ Li = +3.04V • STRONG reducing Agent •Oxidation favourable

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

• Eθ F2= +2.87V • STRONG oxidising Agent •Reduction favourable

• Species on BOTTOM right • Low ↓tendency to lose e • F - →1/2F2 + e • Eθ F2 = -2.87V • WEAK reducing Agent •Oxidation NOT favourable

Electrochemical Series • Shows the ease/tendency of species to accept/lose electrons

• Written as standard reduction potential

Steps to follow 1. Species on top right / Eθ = (-ve) are Strong Reducing Agent (RA)

2. Species on bottom left/ Eθ = (+ve) are Strong Oxidising Agent (OA)

3. For a redox reaction to happen, choose

• One (OA) from left

• One (RA) from right

4. Arrange the redox couple in order of their Eθ values

• Eθ = (-ve) on top

• Eθ = (+ve) at bottom

5. Follow the anticlockwise rule or Z rule

Reaction between Zn + Cu2+ is possible (spontaneous)

Zn is strong RA on the right

• Undergo oxidation Zn → Zn2+ + 2e

Cu 2+ is strong OA on the left

• Undergo reduction Cu2+ +2e → Cu

• Z rule is followed and EθCell is (+ve)

•Zn + Cu2+ → Zn2+ +Cu

Using Standard Electrode Potential to compare the strength of Oxidising/Reducing agents

Is reaction between Zn + Cu2+ possible (spontaneous) ?

Eθ Cell calculation

Zn → Zn 2+ + 2e Eθ(Zn) = +0.76V (Oxidation potential)

Cu 2+ + 2e → Cu Eθ (Cu) = +.34V (Reduction potential)

Zn + Cu 2+ → Cu + Zn 2+ Eθcell = + 1.10V

SPONTANEOUS

Using Standard Electrode Potential to predict if reaction between Zn + Cu2+ is possible

Reaction between Zn + Cu2+ is spontaneous

Zn is strong RA on the right

• Undergo oxidation Zn → Zn2+ + 2e

Cu 2+ is strong OA on the left

• Undergo reduction Cu2+ +2e → Cu

• Z rule is followed and EθCell is (+ve)

• Zn + Cu2+ → Zn2+ + Cu

Is reaction between Zn + Cu2+ possible (spontaneous) ?

Zn → Zn 2+ + 2e Eθ(Zn) = +0.76V (Oxidation potential)

Cu 2+ + 2e → Cu Eθ (Cu) = +.34V (Reduction potential)

Zn + Cu 2+ → Cu + Zn 2+ Eθcell = + 1.10V

EθCell calculation

Using Standard Electrode Potential to predict if reaction between Mg + Cu2+ is possible

Reaction between Mg + Cu2+ is spontaneous

Mg is strong RA on the right

• Undergo oxidation Mg → Mg2+ + 2e

Cu2+ is strong OA on the left

• Undergo reduction Cu2+ +2e → Cu

• Z rule is followed and EθCell is (+ve)

• Mg + Cu2+ → Mg2+ + Cu

Eθ Cell calculation

Mg → Mg2+ + 2e Eθ(Mg) = +2.37V (Oxidation potential)

Cu 2+ + 2e → Cu Eθ (Cu) = +0.34V (Reduction potential)

Mg + Cu2+ → Cu + Mg2+ Eθcell = + 2.71V

Is reaction between Mg + Cu2+ possible (spontaneous) ?

SPONTANEOUS

SPONTANEOUS

Using Standard Electrode Potential to predict if reaction between Ag + Zn2+ is possible

Reaction between Ag + Zn2+ is NOT spontaneous

Ag is weak RA on the right

• Undergo oxidation Ag → Ag+ + e

Zn2+ is weak OA on the left

• Undergo reduction Zn2+ + 2e → Zn

• Z rule is NOT X followed and Eθ Cell is (-ve)

• Ag + Zn2+ → Zn2+ +Cu Х

Is reaction between Ag + Zn2+ possible (spontaneous) ?

Ag → Ag+ + e Eθ(Ag) = -0.80V (Oxidation potential)

Zn2+ + 2e → Zn Eθ (Zn) = -0.76V (Reduction potential)

Ag + Zn2+ → Zn + Ag+ Eθcell = -1.56V

Eθ Cell calculation

Using Standard Electrode Potential to predict if reaction between Fe3+ + CI - is possible

Reaction between Fe3+ + CI - is NOT spontaneous

CI - is weak RA on the right

• Undergo oxidation 2CI - → CI2 + 2e

Fe3+ is weak OA on the left

• Undergo reduction Fe3+ +e → Fe2+

• Z rule is NOT X followed and Eθ Cell is (-ve)

• Fe3+ + 2CI - → Fe2+ +CI2 Х

Eθ Cell calculation

2CI - → CI2 + 2e Eθ(CI) = -1.36V (Oxidation potential)

Fe3+ + e → Fe2+ Eθ (Fe) = +0.77V (Reduction potential)

2CI - + Fe3+ → CI2 + Fe2+ Eθcell = -0.59V

Is reaction between Fe3+ + CI - possible (spontaneous) ?

NON

SPONTANEOUS

NON

SPONTANEOUS

Using Standard Electrode Potential to predict if reaction between AI + H+ is possible

Reaction between AI + H+ (HCI) is spontaneous

AI is strong RA on the right

• Undergo oxidation AI → AI3+ + 3e

H+ is strong OA on the left

• Undergo reduction 2H+ + 2e → H2

• Z rule is followed and EθCell is (+ve)

• AI + 2H+ → Al3+ + H2

Is reaction between AI + H+ possible (spontaneous) ?

AI → AI3+ + 3e Eθ(AI) = +1.66V (Oxidation potential)

2H+ + 2e → H2 Eθ (H2) = 0.00V (Reduction potential)

AI + 2H+ → AI3+ + H2 Eθ

cell = +1.66V

Eθ Cell calculation

Using Standard Electrode Potential to predict if reaction between Cu + H + is possible

Reaction between Cu + H + (HCI) is NOT spontaneous

Cu is weak RA on the right

• Undergo oxidation Cu → Cu2+ + 2e

H+ is weak OA on the left

• Undergo reduction 2H+ + 2e → H2

• Z rule is NOT X followed and EθCell is (-ve)

• Cu + 2H + → Cu2+ +H2 Х

Eθ Cell calculation

Cu → Cu2+ + 2e Eθ(Cu) = -0.34V (Oxidation potential)

2H+ + 2e → H2 Eθ (H2) = +0.00V (Reduction potential)

Cu + 2H+ → H2 + Cu2+ Eθcell = -0.34V

Is reaction between Cu + H+ possible (spontaneous) ?

NON

SPONTANEOUS

SPONTANEOUS

Using Standard Electrode Potential to predict reaction between halogens and halides

Will Chlorine oxidise bromide to bromine ? (Cl2 + Br - → CI - + Br2)

Will Bromine oxidise iodide to iodine ? (Br2 + I - → Br - + I2 )

Will Iodine oxidise chloride to chlorine ? (I2 + CI - → I - + CI2)

Steps to follow

1. Arrange standard electrode potential in order shown below (Highest Eθ = +ve at bottom)

2. Pick one (OA) and one (RA) from both sides and follow Z rule

Reaction between Cl2 + 2I - → CI - + I2 is spontaneous

I - is strong RA on the right

• Undergo oxidation 2I - → I2 + 2e

CI2 is strong OA on the left

• Undergo reduction CI2 + 2e → 2CI -

• Z rule is followed and EθCell is (+ve)

• Cl2 + 2I - → CI - + I2

Is reaction between CI2 + I - possible (spontaneous) ?

2I - → I2 + 2e Eθ(I2) = -0.54V (Oxidation potential)

CI2 + 2e → 2CI - Eθ (CI2) = +1.33V (Reduction potential)

Cl2 + 2I - → CI -+ I2 Eθcell = +0.79V

Eθ Cell calculation

Highest Eθ (+ve) at the bottom

SPONTANEOUS

Using Standard Electrode Potential to predict if reaction between CI2 + I - is possible

Reaction between Cl2 + 2I - → CI - + I2 is spontaneous

I - is strong RA on the right

• Undergo oxidation 2I - → I2 + 2e

CI2 is strong OA on the left

• Undergo reduction CI2 + 2e → 2CI -

• Z rule is followed and EθCell is (+ve)

• Cl2 + 2I - → CI - + I2

Is reaction between CI2 + I - possible (spontaneous) ?

2I - → I2 + 2e Eθ(I2) = -0.54V (Oxidation potential)

CI2 + 2e → 2CI - Eθ (CI2) = +1.33V (Reduction potential)

Cl2 + 2I - → CI -+ I2 Eθcell = +0.79V

Using Standard Electrode Potential to predict if reaction between Br2 + I - is possible

Reaction between Br2 + 2I - → 2Br - + I2 is spontaneous

I - is strong RA on the right

• Undergo oxidation 2I - → I2 + 2e

Br2 is strong OA on the left

• Undergo reduction Br2 + 2e → 2Br -

• Z rule is followed and Eθ Cell s (+ve)

• Br2 + 2I - → 2Br - + I2

2I - → I2 + 2e Eθ(I2) = -0.54V (Oxidation potential)

Br2 + 2e → 2Br - Eθ(Br2) = +1.07V (Reduction potential)

Br2 + 2I - → 2Br - + I2 Eθcell = +0.53V

Is reaction between Br2 + I - possible (spontaneous) ?

Eθ Cell calculation

Eθ Cell calculation

SPONTANEOUS

SPONTANEOUS

Using Standard Electrode Potential to predict if reaction between I2 + CI - possible

Reaction between I2 + 2CI - → 2I - + CI2 NOT spontaneous

CI - is weak RA on the right

• Undergo oxidation 2CI - →CI2 + 2e

I2 is weak OA on the left

• Undergo reduction I2 + 2e → 2I -

• Z rule is NOT X followed and Eθ Cell is (-ve)

• I2 + 2CI - → 2I - + CI2

Is reaction between I2 + CI - possible (spontaneous) ?

2CI - → CI2 + 2e Eθ(CI2) = -1.33V (Oxidation potential)

I2 + 2e → 2I - Eθ (I2) = +0.54V (Reduction potential)

I2 + 2CI - → 2I - + CI2 Eθcell = -0.79V

Using Standard Electrode Potential to predict if reaction between I2 + CI2 possible

Reaction between I2 + CI2 is NOT spontaneous

• CANNOT have TWO species on the same side

• Z rule is NOT X followed and Eθ Cell is (-ve)

Is reaction between I2 + CI2 possible (spontaneous) ?

Eθ Cell calculation

NON

SPONTANEOUS

NON

SPONTANEOUS

Calculate the Eθ or emf for the following cells

Reaction between Mg + Sn2+ is spontaneous

Mg is strong RA on the right

• Undergo oxidation Mg → Mg2+ + 2e

Sn2+ is strong OA on the left

• Undergo reduction Sn2+ + 2e → Sn

• Z rule is followed and Eθ Cell is (+ve)

• Mg + Sn2+ → Mg2+ + Sn

Reaction between Mg + Sn2+

Fe 2+ → Fe3+ + e Eθ(Fe) = -0.777V (Oxidation potential)

MnO4 - + 8H+ + 5e → Mn2+ + 4H2O Eθ

(Mn) = +1.51V (Reduction potential)

2MnO4 -+ 5Fe2+ + 8H+ → 5Fe3+ + Mn2++4H2O Eθ

cell = +0.75V

Eθ Cell calculation

Mg/Mg2+ ║ Sn2+/Sn

Reaction between Fe2+ + MnO4 -

Reaction between Fe2+ + CI - is spontaneous

Fe 2+ is strong RA on the right

• Undergo oxidation Fe2+ → Fe3+ + e

MnO4 - is strong OA on the left

• Undergo reduction MnO4 - + 8H+ + 5e → Mn2+ + 4H2O

• Z rule is followed and Eθ Cell is (+ve)

• 2MnO4 - + 5Fe2+ + 8H+ → 5Fe3+ + Mn2+ + 4H2O

Mg → Mg2+ + 2e Eθ(Mg) = +2.37V (Oxidation potential)

Sn 2+ + 2e → Sn Eθ(Sn) = -0.14V (Reduction potential)

Mg + Sn2+ → Mg2+ + Sn Eθcell = + 2.23V

Eθ Cell calculation

Fe2+/Fe3+ ║ MnO4 -/Mn2+

SPONTANEOUS

SPONTANEOUS

Is the reaction between Ag + AI3+ → Ag+ + AI possible ?

Reaction between Ag + AI3+ is NOT spontaneous

Ag is weak RA on the right

• Undergo oxidation Ag → Ag+ + e

Al3+ is weak OA on the left

• Undergo reduction AI3+ + 3e → Al

• Z rule is NOT X followed and EθCell is (-ve)

• Ag + AI3+ → Ag+ + AI

Reaction between Ag + Al3+

Eθ Cell calculation

Ag → Ag+ + e Eθ(Ag) = -0.80V (Oxidation potential)

AI3+ + 3e → AI Eθ(AI) = -1.66V (Reduction potential)

Ag + AI3+ → Ag+ + AI Eθcell = -2.46V

NON

SPONTANEOUS