Electrochemistry: Chemical Change and Electrical Work

ELECTROCHEMISTRYA study of chemical properties & reaction of ion in a solution where:

i. chemical energy is converted to electrical energy

ii. electrical energy is converted to chemical energy

Types of electrochemical cell:

i. Galvanic /Voltaic Cell

(chemical energy is converted to electrical energy)

ii. Electrolysis Cell

(electrical energy is converted to chemical energy)

Reactions in voltaic/galvanic cells

Two types of reactions in electrochemical cell:

i. Reduction

ii. Oxidation ~ RedOx reaction

Key Points About Redox Reactions

Oxidation (electron loss) always accompanies reduction (electron gain).

The oxidising agent is reduced, and the reducing agent is oxidised.

The number of electrons gained by the oxidising agent always equals the number lost by the reducing agent.

Galvanic Cell(electric cells)V

Salt bridge(Saturated KCl/KNO3)

(+)(-)

M+ N+

eElectrode M(anode)

Electrode N(cathode)

*Electron moves from the anode to the cathode

*Anode to lose mass and the cathode to gain mass.*Function of salt bridge :–i. To complete the electrical circuit. ii. To balance the neutrality of electrical charge in electrolytes

Electrolyte

Reduction:

Occurs at cathode (+ve electrode)

Electrode N (cathode) accepts electrons & the electrolyte is reduced to metal N, as a result the mass of cathode increases.

Half cell reaction at cathode: Nn+

(aq) + ne N(s)

Oxidation:

Occurs at anode (-ve electrode)

Electrode M (anode) releases electrons & itself oxidised to Mn+ ion

Half cell reaction at anode:M(s) Mn+

(aq) + ne

So, overall cell reaction:Nn+

(aq) + M(s) Mn+(aq) + N(s)

Electrode potential: The metal atoms give up electrons, becoming metal

ions.The metal ions dissolve in the water leaving the electrons on the metal.

At equilibrium there is a potential difference between the positive solution and the negative metal.

MMn+ + ne Mn+ + ne - M A small electric potential difference is produced

between electrode surface (M) and solution (Mn+). This difference is known as electrode potential of

metal M in a solution of its ions..

M (electrode)

Mn+ +--- + --- +--- + --- + +--- --- + +--- + +

Standard Hydrogen Electrode (SHE)

The hydrogen electrode is a platinum strip, coated with black platinum, immersed in 1M solution of hydrogen ions, at 298K and the pressure of hydrogen gas is 1 atm.

This electrode compares the ability of donating electrons by various reactions to aqueous solution of hydrogen ions.

Standard Electrode Potentials

Standard 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.

E0 = 0 V

Standard hydrogen electrode (SHE)

2e- + 2H+ (1 M) H2 (1 atm)

Reduction Reaction

Hydrogen electrode is used as the reference electrode to determine electrode reduction potential of the other species.

Standard electrode potential (Eo).of a metal is the potential difference between the metal and a 1.00M aqueous solution of its ion, measured relative to a SHE

Standard reduction potential is not dependent on the volume of solution or the size of electrode.

Standard Electrode/reduction Potential

E0 = 0.76 Vcell

Standard emf (E0 )cell

0.76 V = 0 - EZn /Zn 0

2+

EZn /Zn = -0.76 V02+

Zn2+ (1 M) + 2e- Zn E0 = -0.76 V

E0 = EH /H - EZn /Zn cell0 0

+ 2+2

Standard Electrode Potentials

E0 = Ecathode - Eanodecell0 0

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

Standard Electrode Potentials

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)

2e- + Cu2+ (1 M) Cu (s)

H2 (1 atm) 2H+ (1 M) + 2e-Anode (oxidation):

Cathode (reduction):

H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)

E0 = Ecathode - Eanodecell0 0

E0 = 0.34 Vcell

Ecell = ECu /Cu – EH /H 2+ +2

0 0 0

0.34 = ECu /Cu - 00 2+

ECu /Cu = 0.34 V2+0

Table: Selected Standard Electrode Potentials (298K, 1atm)

Half-Reaction E0(V)

2H+(aq) + 2e- H2(g)

F2(g) + 2e- 2F-(aq)

Cl2(g) + 2e- 2Cl-(aq)

MnO2(g) + 4H+(aq) + 2e- Mn2+(aq) + 2H2O(l)

NO3-(aq) + 4H+(aq) + 3e- NO(g) + 2H2O(l)

Ag+(aq) + e- Ag(s)

Fe3+(g) + e- Fe2+(aq)

O2(g) + 2H2O(l) + 4e- 4OH-(aq)

Cu2+(aq) + 2e- Cu(s)

N2(g) + 5H+(aq) + 4e- N2H5+(aq)

Fe2+(aq) + 2e- Fe(s)

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

Na+(aq) + e- Na(s)

Li+(aq) + e- Li(s)

+2.87

-3.05

+1.36

+1.23

+0.96

+0.80

+0.77

+0.40

+0.34

0.00

-0.23

-0.44

-0.83

-2.71

• E0 is for the reaction as written

• The more positive E0 the greater the tendency for the substance to be reduced

• The half-cell reactions are reversible

• The sign of E0 changes when the reaction is reversed

• Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

Important:

1.Half cell reaction equation in standard reduction potential table is in reduction form.Ex:

Eo Ag+/Ag = +0.80V

2. If we write the half cell reaction equation in oxidation form, the sign of Eo has to be changed.Ex:

Ag(s) Ag+(s) + e Eo

Ag/Ag+ = -0.80V

Ag+(aq) + e- Ag(s)

3. Eo value is independent of stoichiometric coefficients

Ex:

Cr3+(aq) + 3e Cr(s) Eo = -0.74V

2Cr3+(aq) + 6e 2Cr(s) Eo = -0.74V

1/3Cr3+(aq) + e 1/3Cr(s) Eo = -0.74V

Strength of [O] agent : Br2 > Fe3+

Strength of [R] agent : Fe2+ > Br-

Oxidising agent Reducing agent

Fe3+ Fe2+

Br2 Br-

4. Oxidising agent and reducing agent Fe3+

(aq) + e Fe2+(aq) Eo = +0.77V

Br2(l) + 2e 2Br-(aq) Eo = +1.07V

5. Selecting electrode to be an anode or a cathode :

The element with more negative Eo is the anode. The element with more positive Eo is the

cathode.

Ex:

Cr3+(aq) + 3e Cr(s) Eo = -0.74V

Fe3+(aq) + e Fe2+

(aq) Eo = +0.77V

(anode)

(cathode)

6. Cell notation

Zn(s)/ Zn 2+(aq) // Cu2+

(aq)/Cu(s) Eo cell = 1.10V

Anode // cathode

// - represents salt bridge

/ - represents the boundary of electrode- electrolyte @ different phase

Eg:

Zn2+(aq) + 2e → Zn(s) Eo = -0.76 (anode)

Cu2+(aq) + 2e → Cu(s) Eo = +0.34 (cathode)

Calculate a standard cell potential by combining two standard

electrode potential

V

Salt bridge(Saturated KCl/KNO3)

e

Electrode Zn(anode)

Electrode Cu(cathode)

e

Zn2+

Cu2+

Example:Zn2+

(aq) + 2e → Zn(s) Eo = -0.76 (anode) Cu2+

(aq) + 2e → Cu(s) Eo = +0.34 (cathode)

1. Half cell reaction:a) At the anode~

Zn(s) → Zn2+(aq) + 2e Eo = +0.76

b) At the cathode~

Cu2+(aq) + 2e → Cu(s) Eo = +0.34

2. Cell reaction:

Zn(s) + Cu2+(aq) → Zn2+

(aq) + Cu(s) Eo cell = +1.10V

3. Cell notation:

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

(aq)/ Cu(s)

4. Eocell = Eo

red(cathode) – Eo red(anode)

= +0.34 – (-0.76)V

=1.10V ~ e.m.f

Cell potential: Ecell is the emf (electromotive force) of a cell.

For 1M solution at 25 C (standard conditions), the standard emf (standard cell potential) is denoted as Ecell.

Eo Zn2+/Zn = -0.76V

Eo Cu2+/Cu = +0.34

To measure the SEP of ions of the same element in different ox. state.• Draw the diagram

• Example: Fe3+/ Fe2+ (Half-cell )• E.m.f of the cell = +0.77V

• 2. MnO-4 + 8H+ + 5e- <= > Mn2+ 4H2O Eo = +

1.52 V• Hall- cell = ?•

To measure the SEP consisting of gases and their aqueous ion

• Example: chlorine: half –life• Diagram:

• The result are:• A)the emf of the cell= 1.35v• B) The chlorine half-cell is the positive terminal• SEP for the chlorine half-cell = +1.35v.

To predict the feasibility of redox reaction

We used Eocell to predict spontaneity of reaction

Eocell Reaction under standard conditions

+ve Spontaneous/feasible

0 In equilibrium

-ve Non-spontaneous/not feasible

• Example

• Predict whether aqueous hydrogen peroxide is able to oxide bromide ions to bromine.

Variation of electrode potentials with concentration

• Electrod potentials vary with conc. of the ions if the conc. of metal ions Mn+, in the equilibrium

• Mn+ + ne <=> M• is increased, the equilibrium goes to the

right, more Mn+ combines with electrons to produce more metal M. Hence more electrons are removed from the electrod and it becomes more positive

Fuel Cells

• A fuel cell is an electrochemical cell that converts the chemical energy of a continuous supply of reactants into electrical energy.

• A fuel (e.g. hydrogen,methane) is supplied to one electrode and an oxidising agent.(usually O2) to the other electrode

• Cell notation:• Ni(s)/H2(g); OH-(aq) // O2(g); OH-(aq)/ Ni(s)• 1)Write the half cell at cathode and anode• 2)calculate the E o cell

Worked Example 1

Given below are the standard reduction potentials of silver ion and iron (II) ions

Ag+ (aq) + e Ag (s) Eo = +0.80V

Fe2+ (aq) + 2e Fe (s) Eo = -0.44V

Determine the Eocell of the electric cell.

SolutionDetermine which element will be an anode or cathode by referring to the Eo : remember that the more +ve REDCAT So, Fe is anode & Ag is cathode

Write down the Eocell equation

Eocell = Eo (cathode) – Eo (anode)

Eocell = +0.80 – (-0.44) = 1.24V

Worked Example 2

Given the following standard reduction potentials

Zn2+ (aq) + 2e Zn (s) Eo = -0.76V

Fe3+ (aq) + e Fe2+ (aq) Eo = +0.77V

Will Zinc reduced Fe3+ ion to Fe2+ ions?SolutionFe3+ reduced to Fe2+, that means Zn is oxidised to Zn2+

Write the half-equation

2Fe3+ + 2e 2Fe2+ Eo = +0.77

Zn Zn2+ + 2e Eo = +0.76

Overal: 2Fe3+ + Zn 2Fe2+ + Zn2+ Eocell = +1.53

Since Eocell is +ve, Zn will reduce Fe3+ to Fe2+

Worked Example 3

Given below is the standard reduction potential of copper

Cu2+ (aq) + 2e Cu (s) Eo = + 0.34V

Is the following reaction spontaneous under standard conditions ? Cu2+ (aq) + H2 (g) Cu (s) + 2H+ (aq)

SolutionBy looking at the eqn, Cu2+ is reduced, Cu as cathode.

Write down the Eocell eqn:

Eocell = Eo (cathode) – Eo (anode)

Eocell = Eo Cu – Eo H2

Eocell = 0.34 – 0 = + 0.34 V

Since Eocell is +ve, the above reaction is spontaneous

Exercise:

1. Calculate the e.m.f of the following cells at standard conditions

a. Cr(s)/ Cr3+(aq) //Br2(l)/Br-

(aq)/Pt(s)

b. Al(s)/ Al3+(aq) // Pb 2+

(aq)/ Pb(s)

c. Cu(s)/Cu2+(aq)//Cl2(g)/Cl-(aq)/ Pt(s)

d. Pt/Sn2+(aq), Sn4+

(aq)//Fe3+(aq), Fe2+

(aq)/Pt

Exercise:

2. Calculate standard reduction potential for Cd2+/Cd electrode, if we are given:

Cd(s)/ Cd2+(aq) // Ag+

(aq) / Ag(s) Eocell = +1.20V

Ag+ (aq) +e Ag (s) Eo = +0.80V

Exercise:

3 a. Determine Eocell for this reaction:

5Fe2+(aq) + MnO4

-(aq) + 8H+

(aq)

5Fe3+(aq) + Mn2+

(aq) + 4H2O(l)

b. Write the cell notation for the above reaction

Answer:

a. Eocell = +0.75V

b. Pt/Fe2+(aq), Fe3+

(aq)//MnO4-(aq),Mn2+/Pt

Exercise4. Aluminium electrode is immersed in 1M Al(NO3)3

solution, while lead electrode is immersed in 1.0M Pb(NO3)2.

Eo Al3+/Al = -1.66V

Eo Pb2+/Pb = -0.13V

a. Write reaction equation at the anode and cathode

b. Draw the cell diagram

c. Calculate Eocell

d. Calculate Ecell if [Al(NO3)3] is changed to 0.005M