Electrolytic Conduction: Electrolytic Conduction, Conductivity Cell, Determination of Electrolytic Conduction, practice problems and FAQs:

Have you ever seen the batteries used in an inverter?

You must have observed that after a certain months of usage we have to fill the cells of the batteries with distilled water.

Till now we have discussed a lot about electrical conduction on metals and semiconductor.

Let’s understand how the electrical conduction is happening in an electrolytic solution.

Table of content

Electrical Resistance and Conductance
Resistivity or Specific Conductance
Conductance (G)
Conductivity/ Specific Conductance
Determination of Cell Constant
Determination of Electrolytic Conductance
Practice Problems
Frequently Asked Questions (FAQs)

Electrical Resistance and Conductance

Every substance provides some level of resistance to the flow of electricity. Ohm's law is the formula that describes resistance in terms of current and potential gradient.

Lets consider a wire

In which I current is flowing

Where as E=voltage across the ends of the wire

Then,

According to the Ohm’s law

$R = \frac{E}{I}$

Where R=Resistance of the wire

Voltage is often measured in volts, whereas current is typically measured in amperes. Thus, according to Ohm's law,

V=IR

Where I=current flowing through the conductor

V= potential difference across the conductor

R= resistance offered by the conductor

Resistivity or Specific Conductance

The resistance provided by a substance per unit length for a unit cross-section is known as resistivity. Temperature increases resistivity in a linear fashion.

Mathematically,

$R = ρ \frac{l}{A} o r ρ = R \frac{A}{l}$

Where R= resistance offered by a substance

=resistivity of the substance or material

A= area of cross section

l=length of the substance

Unit of resistivity

$ρ = o h m \frac{c m^{2}}{c m} = o h m c m$ or ohm m

S.I unit is ohm m but in electrochemistry most of the cases are dealt with ohm cm

Conductance (G)

It represents the ease by which current can flow through the conductor. It is a measure of degree through which conductor can conduct electricity. Greater the value of conductance, greater the conduction.

It is generally reverse of resistance.

Mathematically, we can write,

$G = \frac{1}{R}$

Unit of conductance is simen (S)

$1 s i m e n (S) = 1 o h m^{- 1} o r 1 Ω^{- 1} o r 1 m h o$

Conductivity/ Specific Conductance

Mathematically

$κ \propto \frac{1}{ρ}$

$κ = \frac{1}{R} \times \frac{l}{A}$

$κ = G \frac{l}{A}$

Where

G= conductance of 1 (unit)3 of conductors or 1 (unit)3of solution.

$\frac{l}{A} = c e l l c o n s t a n t$

l= distance between electrodes

A= area of cross section of electrodes

S.I unit of

$1 S m^{- 1} = 1 o h m^{- 1} m^{- 1} o r 1 m h o c m^{- 1}$

Common unit : $1 S c m^{- 1} = 1 o h m^{- 1} c m^{- 1} = 1 m h o c m^{- 1}$

Determination of Cell Constant

It is represented by $\frac{K}{G} o r K R$

Cell constant can be determined by above three formulas but most commonly it is determined through either $\frac{K}{G} o r K R$

Determination of Electrolytic Conductance

Wheatstone bridge can be used to assess electrolytic conductivity.

Utilizing a wheatstone bridge configuration, the conductivity of an electrolytic solution is determined by substituting one resistance with a conductivity cell that is filled with the electrolytic solution's unknown conductivity.

The tools utilised in this procedure are Rheostat, resistance, conductivity cell, detector,and oscillator.

Rheostat

A rheostat is a variable resistor that regulates current flow by changing the resistance in a circuit..

Detector

Here galvanometer is used to detect the small amount of current flowing through the circuit.

Conductivity cell

It is a device in which the conductivity of the electrolyte is unknow. The conductivity cell has two platinised electrodes which are connected to the terminals. These terminal ends are connected with the arms of wheatstone bridge.

When the detector detects null point. At this point

$\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$

Where R1= resistance of the rheostat

R2=resistance of the conductivity cell

R3=resistance of resistor in arm 3

R4=resistance of resistor in arm 4

From the above equation we can calculate the resistance of the conductivity cell and thus conductivity of the electrolyte can be calculated.

Practice problems

Q1. What is the S.I unit of conductivity?

$o h m^{- 1} m^{- 1}$
ohm m^-1
mho^-1 m^-1
ohm^-2 m^-1

Answer: (A)

Solution: we know that,

$κ = G \frac{l}{A}$

=ohm-1mm2=ohm-1 m-1

Q2. Two platinum electrodes are separated by 3.0 cm in a conductivity cell, and each electrode has a cross sectional area of 10 sq cm. The resistance of a 0.8 N electrolytic solution was measured using this cell and found to be 30. Calculate the solution's conductivity?

10 ohm^-1 cm^-1
0.1 ohm^-1 cm^-1
0.01 ohm^-1 cm^-1
100 ohm^-1 cm^-1

Answer: (C)

Given data :

l=3.0 cm

A=10 cm2

R=30 ohm

we know that,

$G = \frac{1}{R}$

$κ = G \frac{l}{A} = κ = \frac{1}{30 o h m} \times \frac{3.0 c m}{10 c m^{2}} = 0.01 o h m^{- 1} c m^{- 1}$

Q3. At 298 K, the resistance and conductivity of a cell holding 0.01 M KCl solution are 2000 ohm and 210-4 S cm-1, respectively. What will be the value of the cell constant?

0.5 cm^-1
0.05 cm^-1
0.04 cm^-1
0.4 cm^-1

Answer: (D)

Given data

R=200 ohm

$κ = 2 \times 10^{- 4} S c m^{- 1}$

$c e l l c o n s t a n t \frac{l}{A} = ?$

We know that

$κ = G \frac{l}{A} = \frac{1}{R} \times \frac{l}{A}$

Hence $, \frac{l}{A} = R \times κ = 2000 o h m \times 2 \times 10^{- 4} S c m^{- 1} = 0.4 c m^{- 1}$

Q4. At 293 K, the conductivity of 0.1N KCl solution is 0.0250 S cm-1 and the resistance of a cell containing this solution is 40 ohm. The value of the cell constant is:

0.1 cm^-1
0.01 cm^-1
1 cm^-1
10 cm^-1

Answer: (C)

Solution: Given data

R=40 ohm

= 0.0250 S cm-1

$c e l l c o n s t a n t \frac{l}{A} = ?$

We know that

$κ = G \frac{l}{A} = \frac{1}{R} \times \frac{l}{A}$

Hence, $\frac{l}{A} = R \times κ = 40 o h m \times 0.0250 S c m^{- 1} = 1 c m^{- 1}$

Frequently Asked Questions (FAQs)

Q1. In how many ways we can influence the value of electrolytic conductance?
Answer: The value of electrolytic conductance can be influenced by the following factors

Electrolytic conductivity happens only by the ions the electrolyte releases into the solution. Naturally then, the electrolytic conductivity is directly proportional to the concentration of the ions present in the solution.
The type of electrolytes has an influence the electrolytic conduction. Weak electrolyte dissociates only slightly and hence the number of ions and the conductivity will be lesser. On the other hand the strong electrolytes that almost completely dissociates have higher number of ions and conductivity. .
Increase of temperature increases the solubility of salts. Higher the electrolyte concentration higher will be conductivity. So increase of temperature increases the conductivity..

Q2. What is the difference between metallic and electrolytic conduction?
Answer: The passage of electrons through a metal is referred to as metallic conductance. The transport of ions in a pure liquid or solution is known as electrolytic conduction. The main distinction between them is that one involves electron mobility and the other involves ion movement.

Q3. Can we observe any change in the value of cell constant if we change the electrolyte of the solution?
Answer: When the temperature of the solution is changed, the value of the cell constant does not change. Because the distance between the electrode and the electrode's surface area were both kept constant.

Q.4 What is the impact of viscosity on electrolytic conduction?
Answer: The viscosity of the medium has an inverse relationship with conductance. This means that conductivity rises as viscosity falls.

Related topics