Question 3.6:
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.
Answer:
= 1.104 V
We know that,
= −2 × 96487 × 1.04
= −213043.296 J
= −213.04 kJ
Question 3.7:
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:
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 ρ 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 with for strong and weak electrolytes is shown in the following plot:
Question 3.8:
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm−1. Calculate its molar conductivity.
Answer:
Given,
κ = 0.0248 S cm−1
c = 0.20 M
Molar conductivity,
= 124 Scm2mol−1
Question 3.9:
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 S cm−1.
Answer:
Given,
Conductivity, κ = 0.146 × 10−3 S cm−1
Resistance, R = 1500 Ω
Cell constant = κ × R
= 0.146 × 10−3 × 1500
= 0.219 cm−1
Question 3.10:
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 .
Answer:
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 interrupts at 124.0 S cm2 mol−1, = 124.0 S cm2 mol−1.
Question 3.11:
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?
Answer:
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
Question 3.12:
How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al.
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of to Mn2+.
Answer:
(i)
Required charge = 3 F
= 3 × 96487 C
= 289461 C
(ii)
Required charge = 2 F
= 2 × 96487 C
= 192974 C
(iii)
i.e.,
Required charge = 5 F
= 5 × 96487 C
= 482435 C
Question 3.13:
How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl2.
(ii) 40.0 g of Al from molten Al2O3.
Answer:
(i) 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
(ii) 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
Question 3.14:
How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2.
(ii) 1 mol of FeO to Fe2O3.
Answer:
(i) According to the Question,
Now, we can write:
Electricity required for the oxidation of 1 mol of H2O to O2 = 2 F
= 2 × 96487 C
= 192974 C
(ii) According to the Question,
Electricity required for the oxidation of 1 mol of FeO to Fe2O3 = 1 F
= 96487 C
Question 3.15:
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Answer:
Given,
Current = 5A
Time = 20 × 60 = 1200 s
Charge = current × time
= 5 × 1200
= 6000 C
According to the reaction,
Nickel deposited by 2 × 96487 C = 58.71 g
Therefore, nickel deposited by 6000 C
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.
Question 3.16:
Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Answer:
According to the reaction:
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by =
= 1295.43 C
Given,
Current = 1.5 A
Time
= 863.6 s
= 864 s
= 14.40 min
Again,
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit
= 0.426 g of Cu
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit
= 0.439 g of Zn
Question 3.17:
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe3+(aq) and I−(aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br− (aq)
(iv) Ag(s) and Fe3+ (aq)
(v) Br2 (aq) and Fe2+ (aq).
Answer:
Since for the overall reaction is positive, the reaction between Fe3+(aq) and I−(aq) is feasible.
Since for the overall reaction is positive, the reaction between Ag+ (aq) and Cu(s) is feasible.
Since for the overall reaction is negative, the reaction between Fe3+(aq) and Br−(aq) is not feasible.
Since E for the overall reaction is negative, the reaction between Ag (s) and Fe3+(aq) is not feasible.
Since for the overall reaction is positive, the reaction between Br2(aq) and Fe2+(aq) is feasible.