Question

For $Na^+$ , the value of the symbol ${{\lambda}_m}^0$ is $50.0{{\Omega }^{-1}}{{cm}^{2}}mo{{l}^{-1}}$ . The speed of $Na^+$ ion in the solution, if in the cell, electrodes are $5\;cm$ apart and to which a potential of $19.3\;volt$ is applied is
A. $2\times {{10}^{-3}}cm\;{{s}^{-1}}$
B. $1\times {{10}^{-3}}cm\;{{s}^{-1}}$
C. $2\times {{10}^{-4}}cm\;{{s}^{-1}}$
D. $2\times {{10}^{-2}}cm\;{{s}^{-1}}$

Answer 1

Hint: The value given here for the symbol ${{\lambda}_m}^0$ is known as the molar ionic conductivity. Using this value we can find the value of the ionic mobility, which can be defined as the molar conductivity of the solution per charge flowing by one mole of electrons (also known as Faraday). The ionic mobility can also be defined as the drift velocity per unit electric field. From this, the value of the velocity of ions can be obtained.

Complete step by step answer:
Let us note down the given data and data we need to find;
Molar conductivity of $Na^+$ ion $50.0{{\Omega }^{-1}}{{cm}^{2}}mo{{l}^{-1}}$
Distance between the electrodes $d=5cm$
Potential difference applied between two ends $V=19.3volt$
Mobility of $Na^+$ ion ${{\mu }_{m}}^{0}=?$
Velocity or speed of $Na^+$ ion $v=?$

The molar conductivity can be defined as conductivity of the ion in the solution with volume $V$ . The ionic mobility is found to be proportional to the molar conductivity of the solution, and the proportionality constant is the charge of one mole of electrons (also known as Faraday) which is mathematically expressed as
${{\lambda }_{m}}^{0}=F{{\mu }_{m}}^{0}$
From the above equation, mobility is expressed as
${{\mu }_{m}}^{0}=\dfrac{{{\lambda }_{m}}^{0}}{F}$
Substituting the value of molar conductivity and $Faraday=96500\;Coulomb$
${{\mu }_{m}}^{0}=\dfrac{50oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}}{96500A\sec }$
For the units, from Ohm’s Law, $\Omega A=Volt$
${{\mu }_{m}}^{0}=5.18\times {{10}^{-4}}vol{{t}^{-1}}c{{m}^{2}}{{\sec }^{-1}}$

The ionic mobility can be defined as the velocity achieved by an ion moving through a gas under an electric field. Hence, the ionic mobility can be mathematically expressed as
${{\mu }_{m}}^{0}=\dfrac{\text{ionic velocity (cm se}{{\text{c}}^{-1}})}{\dfrac{\text{Potential difference (volt)}}{\text{Distance between electrodes (cm)}}}$
From the above formula, the ionic velocity is expressed as
$v={{\mu }_{m}}^{0}\times \dfrac{V}{d}$
Substituting the given values,
$v=5.18\times {{10}^{-4}}\times \dfrac{19.3}{5}$
$\therefore v=2\times {{10}^{-3}}cm\;{{s}^{-1}}$

Hence, the correct answer is option A.

Note: Here, as the name of the symbol ${{\lambda}_m}^0$ is not given, we should be able to remember it as the molar conductivity of the solution, which is defined as the conductivity of volume $V$ having $1\;mole$ solution. The unit of faraday is $A\;sec$ or $coulomb\;$ but, it shows the charge of $1\;mole$ of electrons. Hence, the unit of $\;mol$ is present but not written for Faraday and due to this, the unit of $\;mol$ is common from the numerator and denominator, and thus canceled.