# JEE Main & Advanced Chemistry Electrochemistry Electrolytic Conduction

Electrolytic Conduction

Category : JEE Main & Advanced

When a voltage is applied to the electrodes dipped into an electrolytic solution, ions of the electrolyte move and, therefore, electric current flows through the electrolytic solution. The power of the electrolytes to conduct electric current is termed conductance or conductivity.

(1) Ohm's law : This law states that the current flowing through a conductor is directly proportional to the potential difference across it, i.e., $I\propto V$

where I is the current strength (In Amperes) and V is the potential difference applied across the conductor (In Volts)

or  $I=\frac{V}{R}$ or $V=IR$

where R is the constant of proportionality and is known as resistance of the conductor. It is expressed in Ohm's and is represented as $\Omega .$The above equation is known as Ohm's law. Ohm's law may also be stated as,

the strength of current flowing through a conductor is directly proportional to the potential difference applied across the conductor and inversely proportional to the resistance of the conductor.”

(2) Resistance : It measures the obstruction to the flow of current. The resistance of any conductor is directly proportional to the length (l) and inversely proportional to the area of cross-section (a) so that

$R\propto \frac{l}{a}\,\,\,\,\text{or }R=\rho \frac{l}{a}$

where $\rho$(rho) is the constant of proportionality and is called specific resistance or resistivity. The resistance depends upon the nature of the material.

Units : The unit of resistance is ohm $(\Omega ).$In terms of SI, base unit is equal to $(kg{{m}^{2}})\,/\,({{s}^{3}}{{A}^{2}}).$

(3) Resistivity or specific resistance : We know that resistance R is

$R=\rho \frac{l}{a}$; Now, if $l=1\,cm,\,a=1\,c{{m}^{2}}$then $R=\rho$

Thus, resistivity is defined as the resistance of a conductor of 1 cm length and having area of cross-section equal to $1\,c{{m}^{2}}.$

Units : The units of resistivity are $\rho =R.\frac{a}{l}=Ohm\frac{c{{m}^{2}}}{cm}$ $=Ohm.\,cm$

Its SI units are Ohm metre $(\Omega \,m).$ But quite often Ohm centimetre $(\Omega \,cm)$ is also used.

(4) Conductance : It is a measure of the ease with which current flows through a conductor.  It is an additive property. It is expressed as G. It is reciprocal of the resistance, i.e.,

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

Units : The units of conductance are reciprocal Ohm$(oh{{m}^{-1}})$ or mho. Ohm is also abbreviated as $\Omega$ so that $Oh{{m}^{-1}}$ may be written as ${{\Omega }^{-1}}.$

According to SI system, the units of electrical conductance is Siemens, S (i.e., $1\text{S}=1\,{{\Omega }^{-1}}).$

(5) Conductivity : The inverse of resistivity is called conductivity (or specific conductance). It is represented by the symbol, $\kappa$ (Greek kappa). The IUPAC has recommended the use of term conductivity over specific conductance. It may be defined as, the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. In other words, conductivity is the conductance of one centimetre cube of a solution of an electrolyte.

Thus, $\kappa =\frac{1}{\rho }$

Units : The units of conductivity are

$\kappa =\frac{1}{Ohm.\,cm}=Oh{{m}^{-1}}$cm–1  or ${{\Omega }^{-1}}\,c{{m}^{-1}}$

In SI units, l is expressed in m area of cross-section in ${{m}^{2}}$ so that the units of conductivity are $\text{S }{{m}^{-1}}.$

(6) Molar conductivity or molar conductance : Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution.

It is denoted by $\Lambda$(lambda). Molar conductance is related to specific conductance ($\kappa$) as,

$\Lambda =\frac{\kappa }{M}$

where, M  is the molar concentration.

If M is in the units of molarity i.e., moles per litre $(mol\,{{L}^{-1}}),$the $\Lambda$may be expressed as,

$\Lambda =\frac{\kappa \times 1000}{M}$

For the solution containing 1 gm mole of electrolyte placed between two parallel electrodes of 1 sq. cm area of cross-section and one cm apart,

$\text{Conductance(}G\text{)}=\text{Conductivity}=\text{Molar}\,\text{conductivity}(\Lambda )$

But if solution contains 1 gm mole of the electrolyte therefore, the measured conductance will be the molar conductivity. Thus,

$\text{Molar conductivity}(\Lambda )=100\times \text{Conductivity}$

In other words, $(\Lambda )=\kappa \times V$

where V is the volume of the solution in $c{{m}^{3}}$containing one gram mole of the electrolyte.

If M is the concentration of the solution in mole per litre, then

M mole of electrolyte is present in $1000\,c{{m}^{3}}$

1 mole of electrolyte is present in $=\frac{1000}{M}c{{m}^{3}}$ of solution

Thus, $\Lambda =\kappa \times \text{Volume in }c{{m}^{3}}$containing 1 mole of electrolyte.

or $\Lambda =\frac{\kappa \times 1000}{M}$

Units of Molar Conductance : The units of molar conductance can be derived from the formula ,

$\Lambda =\frac{\kappa \times 1000}{M}$

The units of $\kappa$ are S$c{{m}^{-1}}$ and units of $\Lambda$are, $\Lambda =S c{{m}^{-1}}\times \frac{c{{m}^{3}}}{mol}=S\,c{{m}^{2}}\,mo{{l}^{-1}}=S c{{m}^{2}}mo{{l}^{-1}}$

According to SI system, molar conductance is expressed as $S\,{{m}^{2}}mo{{l}^{-1}},$if concentration is expressed as $mol\,{{m}^{-3}}.$

(7) Equivalent conductivity : It is defined as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte in solution.

It is expressed as ${{\Lambda }_{e}}$ and is related to specific conductance as

${{\Lambda }_{e}}=\frac{\kappa \times 1000}{C}=\kappa \times \frac{1000}{M}$  (M is Molarity of the solution)

where C is the concentration in gram equivalent per litre (or Normality). This term has earlier been quite frequently used. Now it is replaced by molar conductance. The units of equivalent conductance are $Oh{{m}^{-1}}\,c{{m}^{2}}{{(gm\,equiv)}^{-1}}.$

(8) Experimental measurement of conductance

(i) The conductance of a solution is reciprocal of the resistance, therefore, the experimental determination of the conductance of a solution involves the measurement of its resistance.

(ii) Calculation of conductivity :  We have seen that conductivity (k) is reciprocal of resistivity $(\rho )$, i.e.,

$\kappa =\frac{1}{\rho }$ and  $\rho =R\frac{a}{l}$

$\therefore$  $\kappa =\frac{1}{R}\left( \frac{l}{a} \right)\,\,\,\text{or}\,\,\,\kappa =G\left( \frac{l}{a} \right)$

where G  is the conductance of the cell, l is the distance of separation of two electrodes having cross section area $a\,c{{m}^{2}}.$

The quantity $\left( \frac{l}{a} \right)$ is called cell constant and is expressed in $c{{m}^{-1}}.$ Knowing the value of cell constant and conductance of the solution, the specific conductance can be  calculated as,

$\kappa =G\times \text{Cell constant}$

i.e., $\text{Conductivity}=\text{ Conductance }\times \text{Cell constant}$

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