(a)The conductivity of 0.001 mol L^(-1)solution of CH_3COOH is 3.905 xx 10^(-5) S cm^(-1). Calculate its molar conductivity and degree of dissociation (alpha). Givenlambda^(@) (H^(+)) = 349.6 S cm^(2) mol^(-1)and lambda^(@)(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 ?

(a)The conductivity of 0.001 mol L^(-1)solution of CH_3COOH is 3.905 x

1.	(a)The conductivity of 0.001 mol L^(-1)solution of CH_3COOH is 3.905 xx 10^(-5) S cm^(-1). Calculate its molar conductivity and degree of dissociation (alpha). Givenlambda^(@) (H^(+)) = 349.6 S cm^(2) mol^(-1)and lambda^(@)(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 ?
Answer» Solution :(a) Molar conductivity is given by the relation : `Lambda_(m) = k/C` Substituting the values in the above equation, we have: `Lambda_(m) = (3.905 xx10^(-5) S cm^(-1))/(0.001 mol L^(-1)) = (3.905 xx 10^(-5) S cm^(-1))/(0.001 mol (cm^(3))^(-1))` or `Lambda_(0) = lambda_(H^(+))^(@) + lambda_(CH_(3)COO^(-))^(@)` `=349.6 + 40.9 =390.5 S cm^(2) mol^(-1)` `alpha = Lambda_(m)/Lambda_(0) = (39.05)/(390.5) = 0.1` ENERGY. An electrochemical cell consists of two parts an anode and a cathode. For example a Zn — Cu cell. If external potential APPLIED becomes greater than £°ceU the ELECTRONS flow in the opposite direction i.e., from cathode to anode and current flows from anode to cathode.