Question

Cell constant has unit:
A) $\text{ }s{{m}^{-1}}\text{ }$
B) $\text{ }m{{s}^{-1}}\text{ }$
C) $\text{ }{{m}^{-1}}\text{ }$
D) $\text{ }{{m}^{-2}}\text{ }$

Answer 1

For a conductivity cell, the specific conductance is only when the electrodes are 1 m apart and exactly $\text{ }1\text{ }{{\text{m}}^{\text{2}}}$ in area. However, this is not a practical case. Thus we have the multiplying factor known as cell constant. The cell constant is equal to the ratio of the distance between the electrodes and their cross-sectional area.
$\text{ }{{\text{K}}_{\text{cell}}}\text{ = }\left( \dfrac{l}{\text{a}} \right)$

Complete step by step answer:
We know that the specific conductance is defined as the conductance of the specimen which is 1 m in length and $\text{ }1\text{ }{{\text{m}}^{\text{2}}}$ cross-section. It is a reciprocal of specific resistance i.e.$\text{ }\dfrac{\text{1}}{\text{ }\\!\\!\rho\\!\\!\text{ }}\text{ }$ and generally represented by the kappa $\text{ }\\!\\!\kappa\\!\\!\text{ }$ . The specific conductance is given as follows: $\text{ }\\!\\!\kappa\\!\\!\text{ = }\dfrac{\text{1}}{\text{ }\\!\\!\rho\\!\\!\text{ }}\text{ = }\left( \dfrac{\text{l}}{\text{a}} \right)\text{ }\\!\\!\times\\!\\!\text{ Conductance}$
For the specific conductance as mentioned above, it is the conductance for a one-meter cube of the solution. Therefore, conductance measured by using a conductivity cell will be specific conductance only if the electrodes are exactly $\text{ }1\text{ }{{\text{m}}^{\text{2}}}$ in the area and 1 m apart. But this is not the usual case. The conductance obtained will have to be multiplied by a certain factor to get the specific conductance. This factor is called the cell constant.
The cell constant is represented by$\text{ }{{\text{K}}_{\text{cell}}}\text{ }$.
We know that specific conductance is,
$\text{ Kappa ( }\\!\\!\kappa\\!\\!\text{ ) =}\left( \dfrac{l}{\text{a}} \right)\text{ }\\!\\!\times\\!\\!\text{ Conductance}$
Hence the conductance measured by the cell is multiplied by the factor $\left( \dfrac{l}{\text{a}} \right)$ to get the specific conductance. Thus, we can be called the factor $\left( \dfrac{l}{\text{a}} \right)$ as the cell constant.
Where ‘l’ is the distance in the ‘m’ between the electrodes and ‘a’ is the cross-sectional area of the electrodes in ${{\text{m}}^{\text{2}}}$.
Let's find out the units of cell constant.
We know that cell constant is written as follows,
$\text{ }{{\text{K}}_{\text{cell}}}\text{ = }\left( \dfrac{l}{\text{a}} \right)$
‘l’ has dimensions of meter and ‘a’ have dimensions of ${{\text{m}}^{\text{2}}}$. Then cell constant unit would be,
$\text{ }{{\text{K}}_{\text{cell}}}\text{ = }\left( \dfrac{l}{\text{a}} \right)\text{ = }\dfrac{\text{m}}{{{\text{m}}^{\text{2}}}}\text{ = }\dfrac{1}{\text{m}}\text{ = }{{\text{m}}^{-1}}$
Therefore, the cell constant $\text{ }{{\text{K}}_{\text{cell}}}\text{ }$has the unit${{\text{m}}^{-1}}$.

Hence, (C) is the correct option.

Note: The measurements of distance and the area of the electrode is very inconvenient and also unreliable. The cell constant value is determined by measuring the resistance of a solution whose conductivity is known. We use $\text{ KCl }$ solutions whose conductivity is accurately known to us at various concentrations and temperatures. The cell constant can be also written as,
$\text{Cell constant = }{{\text{G}}^{\text{*}}}\text{= R}\times \text{ }\\!\\!\kappa\\!\\!\text{ }$
Where R is resistance and kappa is$\text{ }\dfrac{\text{1}}{\text{ }\\!\\!\rho\\!\\!\text{ }}\text{ }$.