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Question: A dilute solution of \({\text{KCl}}\) was placed between two \({\text{Pt}}\) electrodes \(10.0{\text...

A dilute solution of KCl{\text{KCl}} was placed between two Pt{\text{Pt}} electrodes 10.0 cm10.0{\text{ cm}} apart, cross which a potential difference of 6.0 volt6.0{\text{ volt}} was applied. K+{{\text{K}}^ + } ions move x cmx{\text{ cm}} in 2 hour2{\text{ hour}} at 25C{25^ \circ }{\text{C}}, the x is _____. (nearest integer)
Ionic conductivity of K+{{\text{K}}^ + } at infinite dilution is 73.52 S cm2 mol173.52{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} at 25C{25^ \circ }{\text{C}}.

Explanation

Solution

To solve this first calculate the ionic mobility of K+{{\text{K}}^ + } ions at infinite dilution. Then calculate the potential gradient applied. Then calculate the speed of K+{{\text{K}}^ + } ions. Remember to convert the time i.e. 2 hour2{\text{ hour}} to seconds.

Formula Used: U±=λ±FU_ \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 =6.0 volt10.0 cm=0.6 volt cm1 = \dfrac{{6.0{\text{ volt}}}}{{10.0{\text{ cm}}}} = 0.6{\text{ volt c}}{{\text{m}}^{ - 1}}
Thus, the potential gradient is 0.6 volt cm10.6{\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±=λ±FU_ \pm ^\infty = \dfrac{{\lambda _ \pm ^\infty }}{F}
Where U±U_ \pm ^\infty is the ionic mobility of ions at infinite dilution,
λ±\lambda _ \pm ^\infty is the ionic conductivity at infinite dilution,
FF is Faraday's constant.
Thus, the ionic mobility of K+{{\text{K}}^ + } ions at infinite dilution is,
UK+=73.52 S cm2 mol196500 C mol1U_{{{\text{K}}^ + }}^\infty = \dfrac{{73.52{\text{ S c}}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}}{{96500{\text{ C mo}}{{\text{l}}^{ - 1}}}}
UK+=7.6186×104 S cm2 C1U_{{{\text{K}}^ + }}^\infty = 7.6186 \times {10^{ - 4}}{\text{ S c}}{{\text{m}}^2}{\text{ }}{{\text{C}}^{ - 1}}
Thus, the ionic mobility of K+{{\text{K}}^ + } ions at infinite dilution is 7.6186×104 S cm2 C17.6186 \times {10^{ - 4}}{\text{ S c}}{{\text{m}}^2}{\text{ }}{{\text{C}}^{ - 1}}.
Now, calculate the speed of K+{{\text{K}}^ + } ions using the equation as follows:
Speed of K+ ions=Ionic mobility×Potential gradient{\text{Speed of }}{{\text{K}}^ + }{\text{ ions}} = {\text{Ionic mobility}} \times {\text{Potential gradient}}
Thus,
Speed of K+ ions=7.6186×104 S cm2 C1×0.6 volt cm1{\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}}
Speed of K+ ions=4.57×104 cm sec1{\text{Speed of }}{{\text{K}}^ + }{\text{ ions}} = 4.57 \times {10^{ - 4}}{\text{ cm se}}{{\text{c}}^{ - 1}}
But we are given that the K+{{\text{K}}^ + } ions move x cmx{\text{ cm}} in 2 hour2{\text{ hour}}. Thus,
Speed of K+ ions=4.57×104 cm sec1×(2×60×60) sec{\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}}
Speed of K+ ions=3.2904 cm{\text{Speed of }}{{\text{K}}^ + }{\text{ ions}} = 3.2904{\text{ cm}}

Thus, the x is 3 cm3{\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.