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(a) The conductivity of 0.001 mol L^-1 solution of $CH_{3}COOH$ is $3.905 \times 10^{-5}Scm^{-1}$ . Calculate its molar conductivity and degree of dissiciation $(\propto )$ .

Given $\lambda ^{\circ}(H^{+})=349.6Scm^{2}mol^{-1}$ and $\lambda ^{\circ}(CH_{3}COO^{-})=40.9Scm^{2}mol^{-1}$

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

Answers (1)

(a)

$\lambda ^{\circ}(H^{+})= 349.6Scm^{2}mol^{-1}$

$\lambda ^{\circ}(CH_{3}COO^{-})=40.9Scm^{2}mol^{-1}$

$K\left (CH_{3}COOH \right ) =$ $3.905\times 10^{-5}Scm^{-1}$

$\propto =?$

$\Lambda _{m}=?$

$c = 0.001molL^{-1}=0.001mo;cm^{-3}$

$\Lambda _{m}=\frac{K\times 1000}{c}$

$\Lambda _{m}=\frac{3.905\times 10^{-5}\times 10^{3}}{0.001}$

$\Lambda _{m}=3.905 \times 10^{-5+3-3}=3.905\times 10$

$\Lambda _{m}=39.05Scm^{2}mol^{-1}$

$\propto = \frac{\Lambda _{m}}{\Lambda ^{\circ}_{m}}= \frac{39.05}{349.6+40.9}=0.1$

(b)

Electrochemical cell is a device which converts chemical energy of an indirect redox reaction into electrical energy. It is formed by two redox couples.

If the external potential applied becomes greater than $E^{\circ}_{cell}$ of the electrochemical cell the reaction gets reversed along with the flow of currents.

Posted by

Sumit Saini

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