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  • Give answer! - Redox Reaction and Electrochemistry - JEE Main

If the resistivity of a subsatnce is 4 \Omega m and the conductance of the cell is 0.25, then what is the numerical value of the cell constant?

  • Option 1)

    16

  • Option 2)

    0.0625

  • Option 3)

    1

  • Option 4)

    none

 

Answers (1)

best_answer

As we have learnt,

 

Cell Constant -

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

G^{*} =\frac{l}{A}=\frac{R}{\rho }=\kappa R

 

- wherein

The quantity \frac{l}{A} is called cell constant denoted by the symbol \varphi^{*}. It depends on the distance between the electrodes and their area of cross section.

 

 

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

\frac{1}{G} = \rho \frac{l}{A} \Rightarrow \frac{1}{0.25} = 4\times \left (\frac{l}{A}\right)\Rightarrow \frac{l}{A} = 1 m^{-1}


Option 1)

16

Option 2)

0.0625

Option 3)

1

Option 4)

none

Posted by

satyajeet

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