March 23, 2012

Electrochemistry – I Three Marks

1. Define electrochemical equivalent. What is its unit?
The amount of a substance deposited by 1 ampere current passing for 1 second (i.e., one coulomb)
Unit: kg.coulomb–1
2. Define equivalent conductance. Give the equation for it.
Equivalent conductance c) is defined as the conductance of an electrolyte solution containing one gram equivalent of the electrolyte. It is equal to the product of specific conductance (κ) of the solution and the volume (V) of the solution that contains one gram equivalent of the electrolyte.
c) = κ x V
3. State Faraday's first and second laws of electrolysis.
Faraday’s First law:
The mass of the substance (m) liberated at the electrodes during the electrolysis is directly proportional to the quantity of electricity (Q) that passes through the electrolyte.         m µ Q
Faraday’s Second law:
When the same quantity of electricity passes through solutions of different electrolytes, the amounts of the substances liberated at the electrodes are directly proportional to their chemical equivalents.
4. State Kohlrausch's law.
At infinite dilution wherein the ionisation of all electrolytes is complete, each ion migrates independently and contributes a definite value to the total equivalent conductance of the electrolyte
5. State Ostwald's dilution law.
Ostwald’s dilution law relates the dissociation constant of the weak electrolyte with the degree of dissociation and the concentration of the weak electrolyte.
Or
Ka = α2C / (1 - α) is known as the Ostwald’s dilution law.
Where,
Ka = Dissociation constant of the weak electrolyte
α = Degree of dissociation
C = Concentration of the weak electrolyte
6. The mass of the substance deposited by the passage of 10 ampere of current for 2 hours 40 minutes and 50 seconds is 9.65 g. Calculate the electrochemical equivalent.
Mass of the substance deposited m = 9.65 g
Current passing I = 10 amp
 Time t = 2 hours 40 minutes and 50 seconds
      = [(2 x 60 x 60) + (40 x 60) + 50] seconds
      = [7200 + 2400 + 50]
      = 9650 seconds
Electrochemical equivalent Z = ?
          Z = m / I ´ t
Quantity of electricity used is Q = I x t = 10 ´ 9650 = 96500 coulombs
Amount of the substance deposited by 96500 coulombs = 9.65 g
Amount of the substance deposited by 1 coulomb = Electrochemical equivalent of the substance
                Z = 9.65 g / 96500 coulombs
                    = 0.0001g.c–1 
                    = 1 x 10– 4 g.c–1
                    = 1 x 10–7 kg.c–1
7. What do you understand / is meant by Buffer solution? Mention its types. Give an example.
A buffer solution is one which maintains its pH fairly constant even upon the addition of small amounts of acid or base.
In other words, a buffer solution resists (or buffers) a change in its pH. That is, we can add a small amount of an acid or base to a buffer solution and the pH will change very little.
Two common types of buffer solutions are:
1. A weak acid and its salt with a strong base. These are called Acid buffers.
Example: CH3COOH + CH3COONa.
2. A weak base and its salt with a strong acid. These are called Basic buffers.
Example: NH4OH + NH4Cl.
8. What is common ion effect? Give an example. Or What is meant by common ion effect?
The reduction of the degree of dissociation of a salt by the addition of a common-ion is called the Common-ion effect.
Example:
When solid NH4Cl is added to NH4OH solution
NH4OH NH4+ + OH
 the equilibrium shifts to the left. Thereby the equilibrium concentration of OH decreases.
9. Why does the metallic conduction decrease with increase in temperature?
Conductivity of metal decreases with increase in temperature due to the enhanced thermal vibration of metal atoms disrupting the movement of electrons passing through them.
10. Write three significances of Henderson equation

With the help of Henderson - Hasselbalch equation
1. The pH of a buffer solution can be calculated from the initial concentrations of the weak acid and the salt provided Ka is given.
However, the Henderson - Hasselbalch equation for a basic buffer will give pOH and its pH can be calculated as (14 – pOH).
2. The dissociation constant of a weak acid (or weak base) can be determined by measuring the pH of a buffer solution containing equimolar concentrations of the acid (or base) and the salt.
Likewise we can find the pKb of a weak base by determining the pOH of equimolar basic buffer.
3. A buffer solution of desired pH can be prepared by adjusting the concentrations of the salt and the acid added for the buffer.
 
ONE MARKS  FIVE MARKS

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