(a) The conductivity of 0.001 mol L−1 solution of CH3COOH is 3.905×10−5S cm−1. Calculate its molar conductivity and degree of dissociation α. Given λ∘(H+)=349.6 S cm2 mol−1 and λ∘(CH3COO−)=40.9 S cm2 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 CH3COOH is 3.905×

1.	(a) The conductivity of 0.001 mol L−1 solution of CH3COOH is 3.905×10−5S cm−1. Calculate its molar conductivity and degree of dissociation α. Given λ∘(H+)=349.6 S cm2 mol−1 and λ∘(CH3COO−)=40.9 S cm2 mol−1. (b) Define electrochemical cell. What happens if external potential applied becomes greater than E∘cell of electrochemical cell?
Answer» (a) The conductivity of 0.001 $m o l L^{- 1}$ solution of $C H_{3} C O O H$ is $3.905 \times 10^{- 5} S c m^{- 1}$ . Calculate its molar conductivity and degree of dissociation $α$ . Given $λ^{\circ} (H^{+}) = 349.6 S c m^{2} m o l^{- 1}$ and $λ^{\circ} (C H_{3} C O O^{-}) = 40.9 S c m^{2} m o l^{- 1}$ . (b) Define electrochemical cell. What happens if external potential applied becomes greater than $E_{c e l l}^{\circ}$ of electrochemical cell?