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NEET-XII-Chemistry

03: Electrochemistry

 page 3

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  • Qstn #5-iii
    Sn(s) | Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) | Pt(s)
    Ans : For the given reaction, the Nernst equation can be given as:



    = 0.14 - 0.0295 × log125

    = 0.14 - 0.062

    = 0.078 V

    = 0.08 V (approximately)
  • Qstn #5-iv
    Pt(s) | Br2(l) | Br-(0.010 M) || H+(0.030 M) | H2(g) (1 bar) | Pt(s).
    Ans : For the given reaction, the Nernst equation can be given as:


  • Qstn #6
    In the button cells widely used in watches and other devices the following reaction takes place:

    Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH-(aq)

    Determine

    and for the reaction.
    Ans :

    = 1.104 V

    We know that,



    = -2 × 96487 × 1.04

    = -213043.296 J

    = -213.04 kJ
  • Qstn #7
    Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
    Ans : Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbolκ. If ``\rho`` is resistivity, then we can write:



    The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.

    i.e.,

    (Since a = 1, l = 1)

    Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

    Molar conductivity:

    Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.



    Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).



    Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.

    The variation of withfor strong and weak electrolytes is shown in the following plot:


  • Qstn #8
    The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm-1. Calculate its molar conductivity.
    Ans : Given,

    κ = 0.0248 S cm-1

    c = 0.20 M

    Molar conductivity,



    = 124 Scm2mol-1
  • Qstn #9
    The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 ``\Omega``. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10-3 S cm-1.
    Ans : Given,

    Conductivity, κ = 0.146 × 10-3 S cm-1

    Resistance, R = 1500 ``\Omega``

    Cell constant = κ × R

    = 0.146 × 10-3 × 1500

    = 0.219 cm-1
  • Qstn #10
    The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

    Concentration/M 0.001 0.010 0.020 0.050 0.100

    102 × κ/S m-1 1.237 11.85 23.15 55.53 106.74

    Calculate for all concentrations and draw a plot between and c½. Find the value of.
    Ans : Given,

    κ = 1.237 × 10-2 S m-1, c = 0.001 M

    Then, κ = 1.237 × 10-4 S cm-1, c½ = 0.0316 M1/2





    = 123.7 S cm2 mol-1

    Given,

    κ = 11.85 × 10-2 S m-1, c = 0.010M

    Then, κ = 11.85 × 10-4 S cm-1, c½ = 0.1 M1/2





    = 118.5 S cm2 mol-1

    Given,

    κ = 23.15 × 10-2 S m-1, c = 0.020 M

    Then, κ = 23.15 × 10-4 S cm-1, c1/2 = 0.1414 M1/2





    = 115.8 S cm2 mol-1

    Given,

    κ = 55.53 × 10-2 S m-1, c = 0.050 M

    Then, κ = 55.53 × 10-4 S cm-1, c1/2 = 0.2236 M1/2





    = 111.1 1 S cm2 mol-1

    Given,

    κ = 106.74 × 10-2 S m-1, c = 0.100 M

    Then, κ = 106.74 × 10-4 S cm-1, c1/2 = 0.3162 M1/2





    = 106.74 S cm2 mol-1

    Now, we have the following data:





    		 0.0316
    		 0.1
    		 0.1414
    		 0.2236
    		 0.3162

    		123.7
    		 118.5
    		 115.8
    		 111.1
    		 106.74





    Since the line interruptsat 124.0 S cm2 mol-1, = 124.0 S cm2 mol-1.
  • Qstn #11
    Conductivity of 0.00241 M acetic acid is 7.896 × 10-5 S cm-1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant?
    Ans : Given, κ = 7.896 × 10-5 S m-1

    c = 0.00241 mol L-1

    Then, molar conductivity,



    = 32.76S cm2 mol-1

    Again, = 390.5 S cm2 mol-1

    Now,

    = 0.084

    Dissociation constant,



    = 1.86 × 10-5 mol L-1
  • Qstn #12
    How much charge is required for the following reductions:
  • Qstn #12-i
    1 mol of Al3+ to Al.
    Ans :

    Required charge = 3 F

    = 3 × 96487 C

    = 289461 C
  • Qstn #12-ii
    1 mol of Cu2+ to Cu.
    Ans :

    Required charge = 2 F

    = 2 × 96487 C

    = 192974 C
  • Qstn #12-iii
    1 mol of to Mn2+.
    Ans :

    i.e.,

    Required charge = 5 F

    = 5 × 96487 C

    = 482435 C
  • Qstn #13
    How much electricity in terms of Faraday is required to produce
  • Qstn #13-i
    20.0 g of Ca from molten CaCl2.
    Ans : According to the question,



    Electricity required to produce 40 g of calcium = 2 F

    Therefore, electricity required to produce 20 g of calcium

    = 1 F
  • Qstn #13-ii
    40.0 g of Al from molten Al2O3.
    Ans : According to the question,



    Electricity required to produce 27 g of Al = 3 F

    Therefore, electricity required to produce 40 g of Al

    = 4.44 F
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