The conductivity of 0.001 `"mol" L^(-1)` solution of `CH_(3)COOH " is " 3.905 xx 10^(-5) "S" cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`.
`"Given" lambda^(@) (H^(+)) = 349.6 "S" cm^(2) "mol"^(-1) " and " lambda^(0) (CH_(3)COO^(-)) = 40.9 "S" cm^(2) "mol"^(-1)`
(b) Define electrochemical cell. What happens if external potential applied becomes greater than `E_(cell)^(@)` of electrochemical cell?

Explanation

`wedge_(m) = (K("conductivity") xx 1000)/(M)`
`= (3.905 xx 10^(-5) xx 1000)/(0.001)`
`=39.055 " cm"^(2)"mol"^(-1)`
`=(wedge^(c)m)/(wedge^(@)m) = (39.05)/(wedge^(@)m) " " [wedge_(CH_(3)COOH)^(@) = wedge_(CH_(3)COO^(-))^(@) + wedge _(H)^(@)+`
`= 349.6 + 6 +0.9`
`=390.5 "S"cm^(2)"mol"^(-1)]`
` = (39.5 xx 10)/(390.5 xx 100) = 0.1`
(b) Electrochemical Cell : It is a device to convert chemical energy into electrical energy. It is based upon the redox reaction which is spontaneous. The electrode on which oxidation takes place is called the anode and the electrode, where reduction takes place is called cathode. If, external potential applied becomes greater than `E_(cell)^(@)` of electrochemical cell, then the cell behave as electrolytic cell.