Given $k =5 \times 10^{-5} S cm ^{-1}$
$C =0.0025\,M$
$\wedge_{ m } =\frac{5 \times 10^{-5} \times 10^3}{0.0025}=\frac{5 \times 10^{-2}}{2.5 \times 10^{-3}}$
$=20\,S\,cm ^2\,mol ^{-1}$
$\alpha =\frac{20}{400}=\frac{1}{20}$
$\alpha=\frac{20}{400}=\frac{1}{20}$
$K_a=\frac{C \alpha^2}{1-\alpha} =\frac{0.0025 \times \frac{1}{20} \times \frac{1}{20}}{\frac{19}{20}}$
$=\frac{0.0025}{19 \times 20}=6.6 \times 10^{-6}$
$=66 \times 10^{-7}$

$Pt|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|0.1{\mkern 1mu} M{\mkern 1mu} HCl||{\mkern 1mu} {\mkern 1mu} 0.1{\mkern 1mu} M\,C{H_3}COOH|{H_2}_{\left( {1{\mkern 1mu} atm} \right)}|Pt$
$Pt/ M/M^{3+}(0.001 \,mol\, L^{ -1})/Ag^+(0.01\, mol\, L^{-1})/Ag$
$298\, K$ પર સેલનો $emf$ $0.421\, volt$ હોવાનું જાણવા મળ્યું છે. $298\, K$ પર અર્ધ પ્રક્રિયા $M^{3+} + 3e \to M$ નો પ્રમાણિત પોટેન્શિયલ ......... $\mathrm{volt}$ હશે .
(આપેલ છે: $298\, K$ પર $E_{A{g^ + }/Ag}^o \,=\, 0.80\, Volt$ )