(a) The conductivity of 0.001 M solution of CH_3COOH 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^(@)(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 M solution of CH_3COOH is 3.905 xx 10^(-

1.	(a) The conductivity of 0.001 M solution of CH_3COOH 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^(@)(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» ANSWER :`Lambda_(0) = 390.5 S cm^(2) mol^(-1), ALPHA =0.1`