Question

(a) The conductivity of 0.001 $mol{L}^{-1}$ solution of $C{H}_{3}COOH$ is $3.905\times {10}^{-5}Sc{m}^{-1}$.
Calculate its molar conductivity and degree of dissociation $\alpha$.
Given ${\lambda }^{\circ }\left(H^{+}\right)=349.6Sc{m}^{2}mo{l}^{-1}$ and ${\lambda }^{\circ }\left(C{H}_{3}CO{O}^{-}\right)=40.9Sc{m}^{2}mo{l}^{-1}$.

(b) Define electrochemical cell. What happens if external potential applied becomes greater than $E_{cell}^{\circ }$ of electrochemical cell?

Answer 1

(a) Conductivity of 0.01 $mol{L}^{-1}$ solution of $C{H}_{3}COOH$
$=3.905\times {10}^{-5}Sc{m}^{-1}$
$=3.905\times {10}^{-3}S{m}^{-1}$
Concentration $=0.001mol{L}^{-1}$
$=1000\times 0.001mol{m}^{-3}$
$=1mol{m}^{-3}$
Molar conductivity, ${\wedge }_m=\frac{K}{C}$
${\wedge }_m=\frac{3.905\times {10}^{-3}S{m}^{-1}}{1mol{m}^{-3}}$
${\wedge }_m=3.905\times {10}^{-3}S{m}^{2}mo{l}^{-1}$
Limiting molar conductivity of $C{H}_{3}COOH,$ ${\wedge }_m^{\circ }$
$={\lambda }^{\circ }\left(C{H}_{3}CO{O}^{-}\right)+{\lambda }^{\circ }\left(H^{+}\right)$
$=349.6Sc{m}^{2}mo{l}^{-1}+40.9Sc{m}^{2}mo{l}^{-1}$
$=390.5Sc{m}^{2}mo{l}^{-1}$
$=3.905\times {10}^{-2}S{m}^{2}mo{l}^{-1}$
Degree of dissociation $\left(\alpha \right)=\frac{\wedge m}{{\wedge }_m^{\circ }}$
$\alpha =\frac{3.905\times {10}^{-3}S{m}^{2}mo{l}^{-1}}{3.905\times {10}^{-2}S{m}^{2}mo{l}^{-1}}$
$\alpha =0.1$

(b) The arrangement which converts chemical energy of a redox reaction into electrical energy is called electrochemical cell. If external potential applied becomes greater than $E_{cell}^{\circ }$ of electrochemical cell, then the reaction in the cell starts in opposite direction.