Electrochemistry

Electrochemistry : 10.3 : Electrolysis cell

VOLTAN CELL VS ELECTRIOLYSIS CELL .

Voltaic cell :

use a spontaneous reaction to generate electric energy.

Electrolysis :

use electric energy to drive non- spontaneous energy.

VOLTAIC CELL.	ELECTROLYTIC
electrons generate at anode (-)	Electrons remove from anode (+)
electrons consumed at cathode (+)	Electrons supplied to cathode (-)
electrons flow from anode(-) to cathode (+)	Electron flow anode (+) to cathode (-)

ELECTROLYSIS

Splitting ('lysing') of a substance by input of electrical energy.

To decompose a compound into it element.

ELECTROLITE IN AN ELECTROLYTIC CELL.

1. Can be :

pure compound.

ex : H₂O , molten salt

Aqueous solution of salt.

ex : NaCI(aq) Na₂SO₄ (aq)

ELETROLYSIS OF PURE MOLTEN SALT

Example : molten NACI

Anode (oxidation) : 2Cl (l) ------ Cl₂(g) + 2e^-

Cathode (reduction) : 2Na⁺ (l) + 2e^------- 2Na (s)

Overall : 2Na⁺ (l) + 2Cl (l) ------ 2Na + Cl₂ (g)

*anion oxidised at anode!

*cation reduced at cathode!

ELECRTROLYSIS OF WATER

Anode (oxidation) 2H₂O(l) ------ O₂(g) + 4H⁺(aq) + 4e^-

Cathode (reduction) 4H₂O(l)+ 4e^- ------ 2H₂(g) + 4OH^-(aq)

Overall 2H₂O(l) ------ 2H₂(g) + O₂(g)

[Note : 4H⁺(aq) + 4OH^-(aq) ------ 4H₂O(l)]

volume of H₂ : O

= 2 : 1

electrode : inert (Pt, etc )

H₂O oxidized at cathode produce O₂

H₂O reduce at cathode to produce H₂

water can be oxidized and reduced.

ELECTROLYSIS OF AQUEOUS IONIC SOLUTION

Aqueous solution of salt are mixture of many species (ions and HO2)

So we have to compare various electrode potentials (E^◦) to predict.

PREDICTING ELECTROLYSIS PRODUCT

When two half- reaction are possible at an electrode.

Cathode : reduction with more positive E° occurs.

Anode : oxidation with more negative E° occurs.

CATION OF ACTIVE METALS (that cannot be reduced)

Cations of metal in group (1) and (2) and A 1

They are not reduced (E° more negative)

H₂O reduced to H₂ and OH

Example: Na₂SO₄(aq)

Species present in solution: Na⁺ , SO₄^2-, H₂O

At cathode(-):

Na⁺(aq) + e^- ------ Na(s) E= -2.71 V

2H₂O(l) + 2e^- ------ H₂(g) + 2OH^- (aq) E= -0.83 V

ANION (OXOANIONS AND F^-) (cannot be oxidised)

Common oxoanion such as SO4^2-, CO3^2-, HO^3-, AND PO4^3-, (AND F) are not oxidized.

Because the central atom already at it highest oxidation state.

H₂O oxidixed to O₂ and H⁺

EXAMPLE :Na₂SO₄(aq)

Special present in the solution: Na⁺, SO₄^2-, H₂O

At anode(+):

2H₂O ------ O(g) + 4H⁺(aq) + 4e^-

CATION OF LESS ACTIVE METALS (can be reduce)

Cation of Au, Ag, Cu, Cr, Pb and Cd

They are reduce at cathode (E° more +ve)

Example: AgNO(aq)

Species present in the solution: Ag⁺ , NO^-, H₂O

At cathode(-):

Ag⁺(aq) + e^- ------ Ag(s)

2H₂O(l) + 2e^- ------ H(g) + 2OH^-(aq)

HALIDES THAT CAN BE OXIDIZER

I^-, Br^-, Cl^-, except F

The concentration must be high

Example : NaCl(aq)

Species present in the solution : Na⁺, Cl^-, H₂O

At cathode(-):

2H₂O(l) + 2e^- ------ H₂(g) + 2OH^-(aq)

At anode(+):

2Cl^-(aq) ------ Cl₂(g) + 2e^-

SUMMARY ON PREDICTING ELECTROLYSIS PRODUCT

Which species will be reduced : Au³⁺(aq) or H₂O

ans : Au³⁺

Au³⁺(aq) + e^------- Au(s)

Au³⁺ is ion of less active metal (E more positive).

Which species will be reduces : NA⁺(aq) or H₂O

ans : H₂O

2H₂O(l) + 2e^- ------ H₂(g) + 2OH^-(aq)

Na+ is ion of active metal (E more negative).

Which species will be oxidised:

ans : H₂O

2H₂O(l) + O(g) ------ 4H⁺(aq) + 4e^-

SO₄^2- is an oxoanion (cannot be oxidised because the central atom alreary at the highest oxidation states).

Which species will be oxidized: CL (aq) or H₂o

ans : Cl

2Cl^-(aq) ------ Cl₂(g) + 2e^-

hallides (except F^-) can be oxidized (due to overvoltage).

EFFECT OF CONCENTRATION

Ex: Concentration NaCl(aq)

At cathode (-) : 2H₂O(l) + 2e^- ------ H₂(g) + 2OH^-(aq)

At anode (+) : 2Cl^-(aq) ------ Cl₂(g) + 2e^-

EXAMPLE : dilute NaCl(aq)

At cathode (-) 4H₂O(l) + 4e^- ------ 2H₂(g) + 4OH^-(aq)

At anode (+) 2H₂O(l) ------ O₂(g) + 4H⁺(aq) + 4e^-

Overall 2H₂O(l) ------ 2H₂(g) + O₂(g)

TYPE OF ELECTRODE

Inactive electrode such as graphite and Pt(s) are normally use in electrolysis.
-They do not involve in the reaction

Active electrodes such as metal (anode) dissolve to form metallic ions

EXAMPLE : Electrolysis of using inert electrode

Cathode : Cu²⁺(aq) + 4e^- ------ 2Cu (s)

Anode : 2H₂O(l) ------ O₂(g) + 4 H⁺ (aq) + 4e^-

Overall : 2H₂O(l) + 2Cu²⁺(aq) ------ 2Cu(S) + O₂(g) + 4 H⁺ (aq)

FARADAY LAW

Amount of substance produce of each electrode is directly proportional to quantity of charge flowing through the cell

Also called Faraday's First Law of electrolysis.

CALCULATING USING FARADAY'S LAW

Main steps
- Balance half reaction to find number of moles of electrons.

ELECTRIC CHARGE (Q)

Charge ( Q ) = Current ( I ) × time ( t )

Unit Coulomb , C Ampere , A Second , s

- 10.2 Nernst Equation
Electrochemistry : 10.2 Nernst Equation NERNST EQUATION Cell potential (E°cell) under any condition   Ecell = E°cell – (RT/nF) ln Q R: universal gas constant.            Q...

- Standard Hydrogen Potential
Standard Reduction PotentialsStandard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Standard hydrogen electrode (SHE) Reduction Reaction                                       ...

- Galvanic Cell
                           Oxidation is defined as the addition of oxygen to a substancethe loss of hydrogen from a substancethe...

- Electrochemistry : 10.2 Nernst Equation
NERNST EQUATION Cell potential (E°cell) under any condition   Ecell = E°cell – (RT/nF) ln Q R: universal gas constant.            Q...

- Thermochemistry : 9.3 Born-haber Cycle
First Ionization Energy (IE1)Energy required for 1 mol of gaseous atom to lose 1 mol of electrons. Affinity Electron (EA) Energy change that occurs when 1 mol of gaseous atom gains 1 mol of electrons. Lattice EnergyEnergy change when 1 mol of solid...

Electrochemistry