Chemistry

Session

Electrochemistry - 2

Session Objectives

• Electrolysis

• Faradays Laws of electrolysis

• Electrode Potential

• Electromotive force

• Electrochemical cells

Electrolysis

The process of decomposition of an electrolyte by the passage of electricity is called electrolysis. In electrolysis electrical energy is used to cause a chemical reaction.

Electrolysis

+NaCl ( ) Na ( ) Cl ( )

Na e Na Reduction At cathode

At anode21

Cl Cl (g) e Oxidation2

For example,

Electrolysis of molten sodium chloride

Electrolysis of aqueous sodium chloride

+NaCl (aq.) Na (aq.) + Cl (aq.)

2H O ( ) H (aq.) + OH (aq.)

H+ ions are discharged at cathode because discharge potential of H+ ion is much lower than Na+ ion

21

H + e H (gas)2

At anode, Cl– is discharge as Cl2 (gas) because discharge potential of Cl– is much lower than that of OH– ion.

21

Cl Cl g e2

Electrolysis of aqueous copper sulphate

Electrolysis of aqueous copper sulphate solution using inert electrodes

2+ 24 4CuSO (aq.) Cu (aq.) + SO (aq.)

+2H O ( ) H (aq.) + OH (aq.)

At cathode, Cu2+ ions are discharged in preference to H+ ions because discharge potential of Cu2+ is much lower than H+ ions.

2Cu aq. 2e Cu aq.

At anode, OH– ions are discharged in preference to SO42– ions because

discharge potential of OH– is much lower than SO42– ions.

2 24OH 2H O ( ) O (g) 4e

Faraday’s law

If W grams of the substance is deposited by Q coulombs of electricity, then

W Q

But Q = it, Hence W i t

or W = Z it

I = current in amperes

t = time in seconds.

Z = constant of proportionality (electrochemical equivalent.)

Faraday’s LawE

By definition Z96500

I . t . EW

96500

E=Equivalent mass of the substance

1 Faraday=96500 coulomb

Na e Na E 23g1F 23g

3Al 3e Al E 9g27g3F

2Cu 2e Cu E 31.75g2F 63.5g

Illustrative Example

Find the total charge in coulombson 1 g ion of N3– .

Electronic charge on one mole ion N3–

= 3 ×1.602 ×10–19 × 6.023 × 1023 coulombs

= 2.89 ×105 coulombs

Solution :

No. of moles in 1g N3- ion = 1/14 =0.0714286.

Therefore, charge on 1g N3– ion=0.0714286 x 2.89 ×105 coulombs=2.06 x 104 Coulombs

Illustrative Example

On passing 0.1 Faraday of electricity through aluminium chloride what will be the amount of aluminium metal deposited on cathode (Al = 27)

27w 0.1 0.9 g

3

Solution:

Illustrative Example

How many atoms of calcium will be deposited from a solution of CaCl2 by a current of 25 milliamperes flowing for 60 s?

Number of Faraday of electricity passed

325 10 6096500

325 10 6096500 2

32325 10 60

6.023 1096500 2

= 4.68 × 1018 atoms of calcium.

Solution:

moles of Ca atoms

atoms of Ca

Illustrative Example

What current strength in amperes will be required to liberate 10 g of bromine from KBr solution in half an hour?

Bromine is liberated by the reaction at anode as follows

22Br Br 2e

Solution:

E itW

96500

160 i 30 6010

2 96500

i = 6.701 ampere

Faraday’s second law of electrolysis

When the same quantity of electricity is passed through different electrolytes the masses of different ions liberated at the electrodes are directly proportional to their chemical equivalence

1 1 1 1

2 2 2 2

1 1

2 2

W E Z It Eor

W E Z It E

Z EHence,

Z E

Illustrative Example

The electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. What will be the ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells ?

Solution

Ratio of iron deposited at cathode will be in their ratio of equivalents

Equivalent of iron from ferrous saltEquivalent ofi ron from ferric salt

M32=

M 23

Illustrative Example

A 100 W, 110 V incandescent lamp is connected in series with an electrolyte cell containing cadmium sulphate solution. What mass of cadmium will be deposited by the current flowing for 10 hr? (Gram atomic mass of Cd = 112.4 g)

SolutionWatt = Volt × Current

100 = 110 × Current

100 10current ampere

110 11

But we know, Q = i × t

1010 60 60 coulombs

11

Number of equivalent =10 10 60 6011 96500

= 0.339

Mass of cadmium deposited

0.339 112.42

= 19.06 g

Electrode Potential

The electrode potential depends upon:

• Nature of the metal

• Concentration of the metallic ions in solution

• Temperature of the solution.

Zn (s) Zn++ (aq) + 2e–

Zn++ (aq.) + 2e– Zn (s)

Oxidation electrode potential

Or

Reduction electrode potential

Standard Hydrogen Electrode

21

H g H aq. e2

21

H aq. e H g2

OR

Pt, H2 (g) (1 atm)/ H+ (aq) (c = 1 M)

When elements are arranged in increasing order of standard electrode potential as compared to that of standard hydrogen electrode,

It is called electrochemical series.

Electrochemical series

0E SRP of cationnM / M

0E SRP of anion1X / X22

Higher the RP , more tendency to get reduced (strong oxidising agent) and vice versa.

Note: Oxidation-Reduction-Potential of an element havesame magnitude and different sign

(a) To compare the relative oxidizing and reducing powers.

(b) To compare the relative activities of metals.

(c) To calculate the standard EMF of any electrochemical cell.

(d) To predict whether a redox reaction is spontaneous.

Applications of electrochemical series:

Applications of Electrochemical Series

Higher the standard reduction potential, the lesser will be its reducing strength.

Li is strongest reducing agent in aqueous solution.

The lesser the standard reduction potential of an element, greater will be its activity.

A more active metal displaces a less active one from its salt solution.

Those metals which have positive oxidation potential will displace hydrogen from acids.

The metals above hydrogen are easily rusted and those situated below are not rusted.

E0 values of Mg+2/Mg is –2.37V, Zn+2/Zn is –0.76V, Fe+2/Fe is –0.44V.Using this information predict whether Zn will reduce iron or not?

Illustrative Example

Zinc has lower reduction potential than iron. Therefore, it can reduce iron.

Solution:

Standard electrode potential

When the ions are at unit activityand the temperature is 25°C (298 K),the potential difference is called the standard electrode potential (E°).

Electrochemical cell

Daniel cell

Chemical energy electrical energy.

Galvanic cell:-

–(s) (aq)Zn Zn e 2 2

Oxidation at anode

Reduction at cathode

–(aq) (s)Cu e Cu 2 2

Cell reaction

(s) (aq) (aq) (s)Zn Cu Zn Cu 2 2

Electromotive force(EMF)

E0cell=E0

cathode-E0anode

Reactions involved

Electromotive force(EMF) of a cell

As per IUPAC convention if we consider standard reduction potentialof both the electrodes then

Emf = ER.H.Electrode - EL.H.Electrode

Emf E Ecathode anode

Salt Bridge

Salt bridge Agar—agar mixed with KCl, KNO3, NH4NO3) etc.

• Eliminates liquid - liquid junction potential

• Maintains the electrical neutrality of solutions.

• Completes the circuit.

Illustrative Example

Calculate the emf of the cell

2+ 3+Ni/Ni 1.0 M || Au 1.0 M | Au

= –0.25 V for and 1.5 V Au3+/Au.

oE 2Ni | Ni

o o ocell cathode anodeE E E

= 1.5 – (–0.25) = 1.75 V

Solution:

Illustrative Example

The standard oxidation potential Eo for the half reaction are given below.

o

2 o

Zn 2e Zn E –0.76 V

Fe 2e Fe E –0.41 V

Calculate E0 for the following cell reaction Zn + Fe++ Zn++ + Fe

Solution

o o ocell cathode anodeE E E

= –0.41 – (–0.76)

= +0.35 V

Thank you