chemistry. session electrochemistry - 3 session objective nernst equation equilibrium constant and...

Chemistry

Session

Electrochemistry - 3

Session Objective

• Nernst equation

• Equilibrium constant and Nernst equation

• Primary cell(Batteries)

• Secondary cell(Batteries)

• Fuel Cell

• Corrosion and its prevention.

Nernst Equation

For a general reduction reaction,

nM ne M(s)

0n+ n+M / M M / M

RT 1E = E - 2.303 log n+nF [M ]

The Nernst equation can be written as

0.059 1oE = E - logn+ n+ n+M /M M /M n M

Where n = Number of electrons involved

[Mn+] = molar concentrations at 298K

(At 298K)

Illustrative ExampleCalculate the electrode potential at a copper electrode dipped in a 0.1M solution of copper sulphate at 250C . The standard potential of Cu2+/Cu system is 0.34 volt at 298 K.

Solution:

0 2redPutting the values of E 0.34 V,n 2 and [Cu ] 0.1 M

red 100.0591

E 0.34 log [0.1]2

0.34 0.02955 ( 1)

0.31045 volt

Cu2+ + 2e- Cu

2 2

0 210Cu / Cu Cu / Cu

0.0591We know that E E log [Cu ]

n

Equilibrium constant from Nernst equation

Consider the following cell reaction in equilibrium then

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

2+0

2+

2+0

2+

2.303RT [Zn ]E = 0 = E (cell) - log

2F [Cu ]

2.303RT [Zn ]E (cell) = log

2F [Cu ]

Then Nernst equation is:

2+

c2+

0c

0

[Zn ]At equilibrium = K

[Cu ]

2.303RTTherefore, E (cell) = logK

2F2.303RT

Generally, E (cell) = logK2F

Progress of reaction

Ele

ctro

de p

ote

nti

al

Reduction half reaction

Oxidation half reaction

Equilibrium point

Illustrative Example

Calculate the equilibrium constant of the reaction:

Cu(s)+2Ag+(aq.)Cu2+(aq.)+Cu(s) E0=0.46 V

Solution :

0c

c

15c

0.059E (cell) = log K

20.46×2

log K = =15.60.059

K = 4×10

Electrochemical cell and Gibbs Energy of the reaction

r

2+ 2+

r

Δ G = -nFE

e.g. for cell reaction

Zn(s) +Cu (aq.) Zn (aq.) +Cu(s)

Δ G = -2FE

0 0r r

0 0r

0r

I f the reactants are in standard state then,

Δ G = Δ G , E = E

Δ G = -nFE

RT= -nF × lnK

nF

Δ G = -RTlnK

Illustrative Example

2 2cell Cu /Cu Zn / ZnE E E

Solution

= + 0.34 V – (–0.76 V) = 1.10 V

G° = –nFE°

= 2×96500×1.10

= –212.3 kJ mol–1

o

Zn /Zn2+E 02+Cu /Cu

E

Calculate G° for Zn-Cu cell at standard state conditions

[Given = –0.76 V, = +0.34 V ]

Commercial Cells

Primary Cell

Dry Cell

Commercial Cells

2

2 2 2 3Zn(s) 2MnO (s) H O Zn Mn O 2OH

The oxidation taking place at the negative zinc electrode.

The reduction takes place at positive electrode

The net cell reaction is

The emf of the cell is about 1.45 V.

2+ -Zn(s) Zn (aq)+2eAnode:

- -

2 2 2 32MnO +H O +2e Mn O +2OHCathode:

Net cell reaction is reversible. Hence, it can be recharged.

Secondary Cells

Lead storage battery

Lead storage battery

At anode:

2- -4 4Pb s + SO aq. PbSO s + 2e

2- + -2 4 4 2PbO + SO aq. + 4H aq. + 2e PbSO s +2H O

+ 2-2 4 4 2Pb s +PbO s +4H aq. +2SO aq. 2PbSO s +2H O

At cathode:

Overall reaction:

Secondary Cell

In the above equation H2SO4 is used up during the discharge.During recharging the reactions are the reverse of those that occurs during discharge.

At cathode:

24 4PbSO s 2e Pb s SO aq.

24 2 2 4PbSO s 2H O PbO SO aq. 4H aq. 2e

At anode:

24 2 2 42PbSO s 2H O Pb s PbO s 4H aq. 2SO aq.

Overall reaction:

Fuel Cells

Galvanic cells which converts energy of combustion of fuel like hydrogen, methane and methanol etc. directly into electrical energy are called fuel cells.

One of the most successful fuel cell uses hydrogen and oxygen reaction to form water.

At cathode: O2(g)+2H2O(l)+4e- 4OH-(aq.)

At anode: 4H2O(l)+4e- 2H2 + 4OH-(aq.)

Overall cell reaction is:

2H2(g)+O2(g) 2H2O(l)

Efficiency of fuel cell is 70% much more as compared thermal plants(40%).

Example

Corrosion

Process of slowly eating away of the metal due to attack of atmospheric gases on the surface of the metal.

• Rusting of iron

• Tarnishing of silver

• Development of green coating on copper and bronze, etc.

Examples of corrosion

Corrosion

• Barrier protection

• Using anti rust solutions

• Sacrificial protection

Methods of preventing corrosion

For example iron surface is covered with a metal which has higher tendency to get oxidized (larger negative value of standard reduction potential) than iron.

Zinc is used for covering iron and the process is called galvanization.

The standard reduction potentials of Sn+2/Sn and Zn+2/Zn are respectively –0.14V, -0.76V. Predict whether the corrosion of tin can be prevented by coating with zinc or not.

Illustrative Example

Zinc lies above tin in the electrochemical series, therefore it has a lower reduction potential than tin.This property is employed to prevent corrosion of tin by coating it with zinc as zinc acts as a sacrificial electrode.

Solution :

Thank you