Question

Transport number $\text{C}{{\text{l}}^{\text{-}}}$ is least in:

A) $\text{HCl}$ 

B) $\text{NaCl}$ 

C) $\text{KCl}$ 

D) $\text{CsCl}$

Answer 1

Hint: The total current by the ion in the solution is called the transport number. The transport number depends on the size of the ion. Larger the size of the ion, the less is the velocity or ionic mobility of the ion under the influence of electricity. For an electrolyte, the transport number for chloride is represented as, 

\[\dfrac{\text{Current carried by the anion }}{\text{Total current}}\text{=}\dfrac{{{\text{u}}_{-}}}{{{\text{u}}_{\text{+}}}\text{ + }{{\text{u}}_{\text{-}}}}\].

Here higher the ionic mobility of cation less is the velocity of the anion and so the transport number.

Complete step by step answer:

According to the faraday, first law of electrolysis, the number of ions discharged at the electrode is found to be proportional to the total quantity of electricity passed through the solution, hence it follows that,

$\text{Total quantity of electricity that passes through the solution}\, \propto \, \text{ sum of the mobilities or speed of the ions}$ 

$\text{Total quantity of electricity that carried by the particular ion}\, \propto \, \text{   the mobility of that particular ion}$

The number of ions discharged or produced at the electrode depends on the speed of the ions or the mobility of ion to move towards the oppositely charged electrode.

Thus the total current carried by each ion is called the transport number. It is denoted by the ‘t’.Thus if the ${{\text{u}}_{\text{+}}}$ is the mobility of the cation and the mobility of anion is ${{\text{u}}_{\text{-}}}$, then

\[\text{Transport number of cation }{{\text{t}}_{\text{+}}}\text{=}\dfrac{\text{Current carried by the cation }}{\text{Total current}}\text{=}\dfrac{{{\text{u}}_{\text{+}}}}{{{\text{u}}_{\text{+}}}\text{ + }{{\text{u}}_{\text{-}}}}\] 

Similarly, the transport number for anion is \[\dfrac{\text{Current carried by the anion }}{\text{Total current}}\text{=}\dfrac{{{\text{u}}_{-}}}{{{\text{u}}_{\text{+}}}\text{ + }{{\text{u}}_{\text{-}}}}\]

The transport number for the chlorine in $\text{HCl}$,$\text{NaCl}$ ,$\text{KCl}$ and $\text{CsCl}$ are as follows:

For $\text{HCl}$,

\[{{t}_{-}}_{_{(HCl)}}=\dfrac{\text{Current carried by anion in HCl }}{\text{Total current}}\text{=}\dfrac{{{\text{u}}_{-}}_{_{(C{{l}^{-}})}}}{{{\text{u}}_{_{^{\text{+}}\text{(}{{\text{H}}^{\text{+}}}\text{)}}}}\text{ + }{{\text{u}}_{-}}_{_{(C{{l}^{-}})}}}\]

For$\text{NaCl}$,

\[{{t}_{-}}_{_{(NaCl)}}=\dfrac{\text{Current carried by anion in NaCl }}{\text{Total current}}\text{=}\dfrac{{{\text{u}}_{-}}_{_{(C{{l}^{-}})}}}{{{\text{u}}_{_{^{\text{+}}\text{(N}{{\text{a}}^{\text{+}}}\text{)}}}}\text{ + }{{\text{u}}_{-}}_{_{(C{{l}^{-}})}}}\]

For$\text{KCl}$,

\[{{t}_{-}}_{_{(KCl)}}=\dfrac{\text{Current carried by anion in KCl }}{\text{Total current}}\text{=}\dfrac{{{\text{u}}_{-}}_{_{(C{{l}^{-}})}}}{{{\text{u}}_{_{^{\text{+}}\text{(}{{\text{K}}^{\text{+}}}\text{)}}}}\text{ + }{{\text{u}}_{-}}_{_{(C{{l}^{-}})}}}\]

For$\text{CsCl}$,

\[{{t}_{-}}_{_{(CsCl)}}=\dfrac{\text{Current carried by anion in KCl }}{\text{Total current}}\text{=}\dfrac{{{\text{u}}_{-}}_{_{(C{{l}^{-}})}}}{{{\text{u}}_{_{^{\text{+}}\text{(C}{{\text{s}}^{\text{+}}}\text{)}}}}\text{ + }{{\text{u}}_{-}}_{_{(C{{l}^{-}})}}}\]

From the transport number for chlorine in the compound, we observe that the transport number for chlorine depends on the ionic mobility’s (${{\text{u}}_{\text{+}}}$) of the cation.

The ionic mobility’s or the velocities of the ions is inversely related to the size of the ion.Larger the size of the ion slower it moves or it moves with less velocity. The transport number for such ions is less. The transport number is inversely proportional to for the ionic mobility’s of ion. Thus the transport number of the ion increases with the decrease in size.

Thus here the trend of velocity is as:

${{\text{u}}_{{{\text{H}}^{\oplus }}}}>{{\text{u}}_{\text{N}{{\text{a}}^{\oplus }}}}> {{\text{u}}_{{{\text{K}}^{\oplus }}}}> {{\text{u}}_{\text{C}{{\text{s}}^{\oplus }}}}$

From the transport formula for chlorine, the ionic mobility of ion is inversely related to the transport number of cation. Thus the transport number for the chloride ion in $\text{CsCl} $ is maximum than the $\text{KCl}$ followed by the $\text{NaCl}$ and minimum or least in $\text{HCl}$.

\[{{\text{t}}_{Cl_{(HCl)}^{-}}}< \text{ }{{\text{t}}_{\text{Cl}_{\text{(NaCl)}}^{\text{-}}}}< \text{ }{{\text{t}}_{\text{Cl}_{\text{(KCl)}}^{\text{-}}}}< \text{ }{{\text{t}}_{\text{Cl}_{\text{(CsCl)}}^{\text{-}}}}\]

Thus the chloride $\text{C}{{\text{l}}^{\text{-}}}$ having the least transport number is $\text{HCl}$

Hence, (A) is the correct option.

Note: Transport number is also affected by the temperature, concentration of electrolyte, nature of co-ion, and on the hydration of ions in solution. Remember that the total transport of cation and anion of electrolyte is equal to one ${{\text{t}}_{+}}\text{+}{{\text{t}}_{\text{-}}}\text{=1}$