Question

A dilute solution of ${\text{KCl}}$ was placed between two ${\text{Pt}}$ electrodes $10.0{\text{ cm}}$ apart, cross which a potential difference of $6.0{\text{ volt}}$ was applied. ${{\text{K}}^ + }$ ions move $x{\text{ cm}}$ in $2{\text{ hour}}$ at ${25^ \circ }{\text{C}}$, the x is _____. (nearest integer)
Ionic conductivity of ${{\text{K}}^ + }$ at infinite dilution is $73.52{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}$ at ${25^ \circ }{\text{C}}$.

Answer 1

Hint: To solve this first calculate the ionic mobility of ${{\text{K}}^ + }$ ions at infinite dilution. Then calculate the potential gradient applied. Then calculate the speed of ${{\text{K}}^ + }$ ions. Remember to convert the time i.e. $2{\text{ hour}}$ to seconds.

Formula Used: $U_ \pm ^\infty = \frac{{\lambda _ \pm ^\infty }}{F}$

Complete step-by-step answer:
We know that the potential gradient is the rate of change of potential with respect to displacement. Thus, the potential gradient is,
Potential gradient $ = \dfrac{{6.0{\text{ volt}}}}{{10.0{\text{ cm}}}} = 0.6{\text{ volt c}}{{\text{m}}^{ - 1}}$
Thus, the potential gradient is $0.6{\text{ volt c}}{{\text{m}}^{ - 1}}$.
The ionic mobility of ions at infinite dilution is related to the ionic conductivity by the equation as follows:
$U_ \pm ^\infty = \dfrac{{\lambda _ \pm ^\infty }}{F}$
Where $U_ \pm ^\infty $ is the ionic mobility of ions at infinite dilution,
 $\lambda _ \pm ^\infty $ is the ionic conductivity at infinite dilution,
 $F$ is Faraday's constant.
Thus, the ionic mobility of ${{\text{K}}^ + }$ ions at infinite dilution is,
$U_{{{\text{K}}^ + }}^\infty = \dfrac{{73.52{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}}{{96500{\text{ C mo}}{{\text{l}}^{ - 1}}}}$
$U_{{{\text{K}}^ + }}^\infty = 7.6186 \times {10^{ - 4}}{\text{ S c}}{{\text{m}}^2}{\text{ }}{{\text{C}}^{ - 1}}$
Thus, the ionic mobility of ${{\text{K}}^ + }$ ions at infinite dilution is $7.6186 \times {10^{ - 4}}{\text{ S c}}{{\text{m}}^2}{\text{ }}{{\text{C}}^{ - 1}}$.
Now, calculate the speed of ${{\text{K}}^ + }$ ions using the equation as follows:
${\text{Speed of }}{{\text{K}}^ + }{\text{ ions}} = {\text{Ionic mobility}} \times {\text{Potential gradient}}$
Thus,
${\text{Speed of }}{{\text{K}}^ + }{\text{ ions}} = 7.6186 \times {10^{ - 4}}{\text{ S c}}{{\text{m}}^2}{\text{ }}{{\text{C}}^{ - 1}} \times 0.6{\text{ volt c}}{{\text{m}}^{ - 1}}$
${\text{Speed of }}{{\text{K}}^ + }{\text{ ions}} = 4.57 \times {10^{ - 4}}{\text{ cm se}}{{\text{c}}^{ - 1}}$
But we are given that the ${{\text{K}}^ + }$ ions move $x{\text{ cm}}$ in $2{\text{ hour}}$. Thus,
\[{\text{Speed of }}{{\text{K}}^ + }{\text{ ions}} = 4.57 \times {10^{ - 4}}{\text{ cm se}}{{\text{c}}^{ - 1}} \times \left( {2 \times 60 \times 60} \right){\text{ sec}}\]
\[{\text{Speed of }}{{\text{K}}^ + }{\text{ ions}} = 3.2904{\text{ cm}}\]

Thus, the x is $3{\text{ cm}}$.

Note: The conducting power of all the ions that are produced by dissolving one mole of any electrolyte in the solution is known as the molar conductivity of the solution. As the temperature of the solution increases, the molar conductivity of the solution increases. This is because as the temperature increases the interaction and the mobility of the ions in the solution increases. The molar conductivity of both strong and weak electrolytes increases as the dilution increases. This is because dilution increases the degree of dissociation and the total number of ions that carry current increases.