(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^(-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 ?
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 ?
Text Solution
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The correct Answer is:
`Lambda_(0) = 390.5 S cm^(2) mol^(-1), alpha =0.1`
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