Electrochemistry 2 Question 3
3. Consider the statements $S_{1}$ and $S_{2}$ :
$\mathbf{S}_{1}$ : Conductivity always increases with decrease in the concentration of electrolyte.
$\mathbf{S}_{\mathbf{2}}$ : Molar conductivity always increases with decrease in the concentration of electrolyte.
The correct option among the following is
(2019 Main, 10 April I)
(a) $S_{1}$ is correct and $S_{2}$ is wrong
(b) $\mathrm{S}{1}$ is wrong and $\mathrm{S}{2}$ is correct
(c) Both $\mathrm{S}{1}$ and $\mathrm{S}{2}$ are wrong
(d) Both $\mathrm{S}{1}$ and $\mathrm{S}{2}$ are correct
Show Answer
Solution:
- The explanation of statements $\left(\mathrm{S}{1}\right.$ and $\left.\mathrm{S}{2}\right)$ are as follows :
In conductivity cell, conductivity $(\kappa)$ is equal to the sum of ionic conductances (c), of an electrolytic solution present is unit volume of the solution enclosed by two electrodes of unit area $(a \neq 1)$ separated by a unit length $(l=1)$.
$$ \kappa=c \times \frac{l}{a} \Rightarrow \kappa=c \text { when } l=1, a=1 $$
So, with decrease in the concentration of electrolyte, number of ions in the given unit volume also decreases, i.e. $\kappa$ [conductivity] also decreases.
Thus, statement $S_{1}$ is wrong. $S_{2}$ : Molar conductivity $\left(\lambda_{m}\right)$ is defined as the conducting power of all the ions present in a solution containing 1 mole of an electrolyte.
$$ \lambda_{m}=\kappa \times V_{\mathrm{mL}}=\kappa \times \frac{1000}{M} $$
where, $V_{\mathrm{mL}}=$ volume in $\mathrm{mL}$ containing 1 mole of electrolyte $m=$ molar concentration $(\mathrm{mol} / \mathrm{L})$ So, in a conductivity cell
$$ \lambda_{m} \propto \frac{1}{M} $$
i.e. molar conductivity increases with decrease in the concentration $(M)$ of electrolyte.
Thus, statement $\mathrm{S}_{2}$ is correct.
