Chapter 3 Electrochemistry Solutions
Question - 21 : - In the button cells widely usedin watches and other devices the following reaction takes place:
Answer - 21 : -
Zn(s) + Ag2O(s) + H2O(l)→ Zn2+(aq) + 2Ag(s) + 2OH−(aq)
Determine 
and 
for the reaction.
Answer
= −2 × 96487 × 1.04
= −213043.296 J
= −213.04 kJ
Define conductivity and molarconductivity for the solution of an electrolyte. Discuss their variation withconcentration.
Answer - 22 : -
Conductivity of a solution isdefined as the conductance of a solution of 1 cm in length and area ofcross-section 1 sq. cm. The inverse of resistivity is called conductivity orspecific conductance. It is represented by the symbolκ. If ρ isresistivity, then we can write:
The conductivity of a solution atany given concentration is the conductance (G) of one unit volume ofsolution kept between two platinum electrodes with the unit area ofcross-section and at a distance of unit length.
i.e., 
(Since a =1, l = 1)
Conductivity always decreaseswith a decrease in concentration, both for weak and strong electrolytes. Thisis because the number of ions per unit volume that carry the current in asolution decreases with a decrease in concentration.
Molar conductivity:
Molarconductivity of a solution at a given concentration is the conductance ofvolume V of a solution containing 1 mole of the electrolyte kept between twoelectrodes with the area of cross-section A and distance ofunit length.
Now, l = 1and A = V (volume containing 1 mole of the electrolyte).
Molar conductivity increases witha decrease in concentration. This is because the total volume V of the solutioncontaining one mole of the electrolyte increases on dilution.
Thevariation of 
with
for strong and weak electrolytes is shown in thefollowing plot:
Question - 23 : - The conductivity of 0.20 Msolution of KCl at 298 K is 0.0248 Scm−1. Calculate its molarconductivity.
Answer - 23 : -
Given,
κ =0.0248 S cm−1
c = 0.20 M
Molar conductivity, 
= 124 Scm2mol−1
Question - 24 : - The resistance of a conductivitycell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cellconstant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 Scm−1.
Answer - 24 : -
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 - 25 : - The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:
Answer - 25 : -
Concentration/M 0.001 0.010 0.0200.050 0.100
102 × κ/S m−1 1.23711.85 23.15 55.53 106.74
Calculatefor all concentrations and draw a plot between
and c½. Find the value of
Answer
Given,
κ = 1.237 × 10−2 Sm−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 Sm−1, c = 0.010M
Then, κ = 11.85 × 10−4 Scm−1, c½ = 0.1 M1/2
= 118.5 S cm2 mol−1
Given,
κ = 23.15 × 10−2 Sm−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 Sm−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 Sm−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:
Since the line interruptsat 124.0 S cm2 mol−1, = 124.0 S cm2 mol−1.
Answer - 26 : -
Given, κ = 7.896× 10−5 S m−1
c = 0.00241 mol L−1
Then,molar conductivity, 
Now,
= 0.084
Dissociation constant, 
= 1.86 × 10−5 molL−1
Answer - 27 : -
(i) 1 mol ofAl3+ to Al.
(ii) 1 mol ofCu2+ to Cu.
(iii) 1mol of 
to Mn2+.
Answer
Question - 28 : - How much electricity in terms ofFaraday is required to produce
Answer - 28 : -
(i) 20.0 g of Ca from molten CaCl_2.
(ii) 40.0 g of Al from molten Al_2O_3.
Answer
(i) Accordingto the question,
How much electricity is requiredin coulomb for the oxidation of
Answer - 29 : -
(i) 1 mol of H_2O to O_2.
(ii) 1 mol of FeO to Fe_2O_3.
Answer
(i) Accordingto the question,
Electricity required for theoxidation of 1 mol of FeO to Fe2O3 = 1 F
= 96487 C
Question - 30 : - A solution of Ni(NO3)2 iselectrolysed between platinum electrodes using a current of 5 amperes for 20minutes. What mass of Ni is deposited at the cathode?
Answer - 30 : -
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 bedeposited at the cathode.