(a) The conductivity of 0.001molL^{-1} acetic acid is 4.95\times 10^{-5}Scm^{-1}. Calculate the dissociation constant if \wedge _{m}^{o} for acetic acid is 390.5Scm^2mol^{-1}.

(b) Write Nernst equation for the reaction at 25^{\circ}C:

   2Al(s)+3Cu^{2+}(aq)\rightarrow \rightarrow 2Al^{3+}(aq)+3Cu(s)

(c) What are secondary batteries? Give an example. 

 

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

(a) Given C=0.001molL^{-1}=10^{-3}molL^{-1}

K=4.95\times 10^{-5}Scm^{-1}

\wedge _{m}^{c}=\frac{k\times 1000}{C}=\frac{4.95\times 10^{-5}\times 1000}{10^{-3}}=49.5Scm^2mol^{-1}

\alpha =\frac{\wedge _{m}^{c}}{\wedge \tfrac{m}{o}}=\frac{49.5}{390.5}=0.1267=12.67 _{}^{o}\textrm{/}_o

CH_3COOH\rightleftharpoons CH_3COO^-+H^+

    C                                 O                      O

c(l-\alpha)                         c\alpha                    c\alpha

K_\alpha=\frac{\left [ CH_3COO^- \right ][H^+]}{[CH_3COOH]}=\frac{C\alpha^2}{1-\alpha}

        =\frac{1\times 10^{-3}\times (0.127)^2}{1-0.127}=1.844\times 10^{-5}molL^{-1}

(b) 2Al + 3Cu^{2+}\rightarrow2Al^{3+}+3Cu

       (s)         (aq)             (aq)            (s)

 

Nernst equation :

E_{cell}=E^ocell-\frac{0.0591}{n}\log \frac{\left [ Al^{3+} \right ]^2}{\left [ Cu^{2+} \right ]^3}

=E^ocell-\frac{0.0591}{6}\log \frac{\left [ Al^{3+} \right ]^2}{\left [ Cu^{2+} \right ]^3}

(c) Secondary Batteries : Secondary batteries are those which can be recharged by passing an electric current through them and hence can be used over and again. 

e.g. Lead storage battery and Nickel cadmium storage cell. 

 

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