Relation of conductivity with electrode distance [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 5 years ago.








[Improve this question](/posts/84308/edit) 



Conductivity is defined as the conduction of ions when the distance between the electrodes is 1 cm and length is 1 cm.  

Say for instance we have a solution where the distance between the electrodes is 2 cm and its height is also 2 cm, what will be the resulting conductivity of the solution?




sigh... lifted out of [Definitions](https://en.wikipedia.org/wiki/Conductivity_(electrolytic)#Definitions) section of wikipedia article on [Conductivity (electrolytic)](https://en.wikipedia.org/wiki/Conductivity_(electrolytic))


Resistance, $R$, is proportional to the distance, $l$, between the electrodes and is inversely proportional to the cross-sectional area of the sample, $A$. Writing $\rho$ (rho) for the specific resistance (or resistivity),


$R={\frac {l}{A}}\rho$ 


In practice the conductivity cell is calibrated by using solutions of known specific resistance, $\rho^\*$, so the quantities $l$ and $A$ need not be known precisely. If the resistance of the calibration solution is $R^\*$, a cell-constant, $C$, is derived.


$ R^{\*}=C\times \rho ^{\*}$


The specific conductance (conductivity), $\kappa$ (kappa) is the reciprocal of the specific resistance.


$\kappa ={\frac {1}{\rho }}={\frac {C}{R}} $


So if $l\_1 = 1 \text{ cm}$, $A\_1 = 1 \text{ cm}^2$ but $l\_2 = 2\text{ cm}$, $A\_2 = 4 \text{ cm}^2$


$\kappa\_1 ={\frac {C}{R}} $   



$\kappa\_2 ={\frac {2C}{R}} = 2\kappa\_1 $   



 1