Question

(A) Define the following terms:
(i) Limiting molar conductivity
(ii) Fuel cell
(B) Resistance of a conductivity cell filled with $0.1mol{L^{ - 1}}$ KCl solution is $100\Omega $. If the resistance of the same cell when filled with $0.2mol{L^{ - 1}}KCl$ solution is $540\Omega $, calculate the conductivity and molar conductivity of $0.02mol{L^{ - 1}}KCl$ solution. The conductivity of $0.1mol{L^{ - 1}}KCl$ solution is $1.29 \times {10^{ - 2}}{\Omega ^{ - 1}}c{m^{ - 1}}$.

Answer 1

Hint: We have learnt that conductance is the property of a conductor that supports the flow of currents. It depends upon temperature and concentration and fuel cells work like a battery by producing electricity and heat. Resistance is basically those properties which oppose the flow of current.

Complete step by step answer:
We just got to know that conductivity is the conductance of $1c{m^3}$ of a conductor, also called specific conductance and molar conductance is the conductance offered by 1 mole of an electrolyte which is placed between two electrodes separated by a distance of 1cm. Molar conductivity can be given by the formula: ${\lambda _m} = \dfrac{{K \times 1000}}{M}$ where K is the conductivity and M is molarity. When the electrolyte concentration reaches to zero the molar conductivity becomes limiting molar conductivity and it increases when the concentration is decreased as the volume containing 1 mole of electrolytes increases.
Now we can define the fuel cells which are those in which energy produced from the combustion of fuels like hydrogen, carbon monoxide, methane etc is converted into electric current through a series of redox reactions. As long as sufficient fuel and oxygen are provided to the fuel cells, they will keep on producing electricity. A fuel cell consists of a cathode, an anode and an electrolyte. At anode, oxidation reaction occurs that produces positive ions and electrons which then moves from anode to cathode via an electrolyte, at cathode another catalyst causes the formation of water and various other products by reaction of electrons, oxygen and ions. While ions and electrons move from anode to cathode, a direct current electricity is produced at the same time.
(B) To answer these types of numericals, we should first write the given point. So, let us write them-
For $0.1mol{L^{ - 1}}KCl$ solution,
Conductivity K is given as: $k = 1.29 \times {10^{ - 2}}{\Omega ^{ - 1}}c{m^{ - 1}}$Resistance is given as: $R = 100\Omega $. Using formula:
$Conductivity = \dfrac{{Cell{\text{ constant}}}}{{Resistance}} \\
Cell{\text{ constant = Conductivity }} \times {\text{Resistance}} \\
= 1{\text{.29}} \times {\text{1}}{{\text{0}}^{ - 2}} \times 100 \\
= 1{\text{.29 c}}{{\text{m}}^{ - 1}}$
Similarly, for $0.2mol{L^{ - 1}}KCl$ solution, we can calculate the conductivity using calculated value of cell constant and given resistance, we get:
 $Conductivity = \dfrac{{Cell{\text{ constant}}}}{{Resis\tan ce}} \\
= \dfrac{{1.29}}{{540}} \\
= 0{\text{.0024}}{\Omega ^{ - 1}}{\text{c}}{{\text{m}}^{ - 1}}$
Now we can easily calculate the molar conductivity of $0.2mol{L^{ - 1}}KCl$ solution by using the formula of molar conductivity we discussed above:
${\lambda _m} = \dfrac{{k \times 1000}}{M} \\
=\dfrac{{0.0024 \times 1000}}{{0.02}} \\
= 120{\Omega ^{ - 1}}c{m^{ - 1}}mo{l^{ - 1}}$

Hence, the conductivity is $0.0024{\Omega ^{ - 1}}c{m^{ - 1}}$ and molar conductivity is $120{\Omega ^{ - 1}}c{m^{ - 1}}mo{l^{ - 1}}$.

Note: Conductance of a solid conductor depends only on the temperature and conductance of a liquid conductor depends upon the concentration of the electrolyte as well as the temperature. When temperature and concentration are increased, conductance increases.