UNIT-3

 Electrochemistry

Q1-Describe Gibbs Free Energy. Prove it.

A-  Gibbs free energy of the system is the difference of enthalpy of the system with the product of temperature times the entropy of the system.

  G=H-TS

Gibbs free energy of the system is defined in term of the thermodynamics which are state in function. Any change in the Gibbs Free Energy System is directly proportional to the difference of change in the enthalpy of the system with the products of temperature times the entropy of the system.

                   G=     H-     (TS)

While at constant temperature this reaction transform into:

                  G=      H-T     S

The Nernst Equation is derived from the Gibbs free energy under standard conditions.

  E*=E*reduction-E*oxidation   ………..(i)

  G=-nFE      ………..(ii)

Where,

 n=no. of transferred electrons in the reaction

F= Faraday constant

E=Potential Difference.

While when we see in the standard condition then, equation (ii) becomes

  G*=-nFE*     ………….(iii)

Hence,

Reaction is Spontaneous when E* is positive while non- spontaneous in vice-versa.

 G=     G*+RT lnQ     ………….(iv)

Now, Substituting       G=−nFE and     G*=−nFE* into Equation 4, we have:

−nFE=−nFEo+RTlnQ  …………….(v)

On Dividing both sides of the Equation above by −nF,

E=E*−RTnFlnQ(6)   ……….(vi)

Equation (vi) in the form of log10:

E=E*−2.303RT/nF log10Q    …….(vii)

At standard temperature T = 298 K, the 2.303RT/F term equals 0.0592 V and Equation

(vii) can be rewritten:

E=E*−0.0592V/n log10Q   ……..(viii)

The equation (viii) clearly indicates that electric potential of cell depends on reaction quotient of reaction. The product formation leads to the increase in the concentration of the products. This tends to decrease the the potential of the cell until it reaches at the stage of equilibrium where,      G=0 and       G=-nFE Q=K so E=0

Then on substituting the these values to Nernst Equation we get,

  0=E*-RT/nF In K       …….(ix)

At room temperature it becomes;

  0=E*-0.0592V/n Log10K

  LogK=nE*/0.0592V

The above equation clearly indicates the equilibrium constant K is proportional to the standard potential.

Q2- Describe the chemistry of battery.

Negative plate reaction:

 PbSO4(s) + H+(aq) + 2e– → Pb(s) + HSO4–(aq)

Positive plate reaction:

PbSO4(s) + 2H2O(l) → PbO2(s) + HSO4–(aq) + 3H+(aq) + 2e–

Combining these two reactions, the overall reaction is the reverse of the discharge reaction:

2PbSO4(s) + 2H2O(l) → Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4–(aq)

Discharge Chemistry of the Battery:

The positive and negative plate of the batteries becomes lead sulphate. Due to the loss of sulfuric acid from electrolytes it becomes the water.

Negative plate reaction:

 Pb(s) + HSO4–(aq) → PbSO4(s) + H+(aq) + 2e–

Positive plate reaction:

 PbO2(s) + HSO4–(aq) + 3H+(aq) + 2e– → PbSO4(s) + 2H2O(l)

Combining these two reactions, one can determine the overall reaction:

Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4–(aq) → 2PbSO4(s) + 2H2O(l)

Q3- What is Corrosion?

A-  Corrosion is defined as the phenomenon in which metal get a coated covering over its whole body due to the chemical reaction from its surrounding that result in the conversion of metal into the oxide, salt or any other compound.

Q4-What do you understand by Direct corrosion?

A-  The Direct Corrosion is the simplest corrosion which occurs directly in the environment temperature. Under this there is a chemical attack which includes oxidation under which the oxygen present in the environment combines with all the possible part of the surface of material.

Fe + O + 2CO2 + H2O → Fe(HCO3) 2

2Fe(HCO3)2 + O → 2Fe(OH)CO3 + 2CO2 + H2O

Fe(OH)CO3 + H2O  → Fe(OH)3 + CO2

Q5-What is High Temperature Oxidation? What are its causes and preventions.

A-  The formation of scales and oxides in iron rusting took place for rusting of ferrous alloys at high temperature while the other form of the corrosion may be noted when liquid metal flow through other metals. The corrosion is caused due to the tendency of the solid to dissolve into the liquid metal unto the limit of solubility.

 Causes:

1-     Corrosion took place due to the presence of salt.

2-     Corrosion took place due to the inadequate design procedure.

3-     Presence of excess water-cement ratio causes corrosion.

4-     Internal structure of metal.

Prevention:

1-     Maintaining a high degree of workmanship.

2-     Use of high quality and impermeable concrete.

3-     Using the correct cement-water ratio.

Q6-Describe the applications of water system.

A-  In water there is only one component i.e., water and its phases: ice, water, steam that is solid, liquid and gaseous form.                    

Q7- Complete the following reactions:

 Fe + O + 2CO2 + H2O →

2Fe(HCO3)2 + O →

Fe(OH)CO3 + H2O  →

A-  Fe + O + 2CO2 + H2O → Fe(HCO3) 2

2Fe(HCO3)2 + O → 2Fe(OH)CO3 + 2CO2 + H2O

Fe(OH)CO3 + H2O  → Fe(OH)3 + CO2

Q8- Mention the applications of Gibbs Free Energy.

A-     

1-     This is used in the solubility product and potentio-metric titration.

2-     It is used to calculate the potential of ion charge.

3-     It is used in oxygen and aquatic environment.

Q9-Explain Homogenous phase and Heterogenous phase.

A-  Air constitutes a single phase only as it contains a mixture of nitrogen, oxygen, carbon dioxide, water vapour etc.  A system consisting of only one phase is said to be homogeneous. A mixture of two  immiscible  liquids  such  as water and benzene, will exist in two distinct  liquid phases and in addition there will be a vapour phase. Thus there will be three phases each separated from the other by a well-defined bounding surface while the system consist of more than one phase is called as the heterogenous phase

Q10- What is Phase rule?

    F=C-P+2