\(R=50 \Omega\)
\(\kappa=1.45 S m^{-1}=1.4 \times 10^{-2} S \mathrm{cm}^{-1}\)
Now, \(R=\rho {\frac{\ell}{a}}=\frac{1}{\kappa} \times \frac{\ell}{a}\)
\(\Rightarrow \frac{\ell}{a}=R \times \kappa=50 \times 1.4 \times 10^{-2}\)
For \(0.5 M\) solution
\(R=280 \Omega ; \kappa=?\)
\(\frac{\ell}{a}=50 \times 1.4 \times 10^{-2}\)
\(\Rightarrow R=\rho \frac{\ell}{a}=\frac{1}{\kappa} \times \frac{\ell}{a}\)
\(\Rightarrow \quad \kappa=\frac{1}{280} \times 50 \times 1.4 \times 10^{-2}\)
\(=\frac{1}{280} \times 70 \times 10^{-2}=2.5 \times 10^{-3} S c m^{-1}\)
Now, \(\Lambda_{m}=\frac{\kappa \times 1000}{M}=\frac{2.5 \times 10^{-3} \times 1000}{0.5}\)
\(=5 S c m^{2} m o l^{-1}=5 \times 10^{-4} S m^{2} m o l^{-1}\)
(નજીકનાં પૂર્ણાંકમાં રાઉન્ડ ઑફ). $\left[\right.$ આપેલ $, E_{C u^{2+} / C u}^{o}=0.34\, V , E _{ NO _{3}^{-} / NO_2 }^{\circ}=0.96\, V$ $,E _{ NO _{3} / NO _{2}}^{\circ}=0.79 \,V$ $\left.\frac{ RT }{ F }(2.303)=0.059\right]$
$F{e^{ + 2 }} + 2{e^ - }\, \to \,Fe\,;\,\,\,\,{E^o} = - 0.440\,V$
$F{e^{ + 3 }} + 3{e^ - }\, \to \,Fe\,;\,\,\,\,{E^o} = - 0.036\,V$
તો $F{e^{ + 3 }} + {e^ - } \to \,F{e^{ + 2 }}$ માટે પ્રમાણિત ઇલેક્ટ્રોન પોટેન્શિયલ $({E^o})$ .............. $\mathrm{V}$ છે.
${Zn}({s})+{Cu}^{2+}(0.02 {M}) \rightarrow {Zn}^{2+}(0.04 {M})+{Cu}({s})$
${E}_{\text {cell }}=...... \,\times 10^{-2} \,{~V}$ { (નજીકના પૂર્ણાંકમાં) }
${\left[ {E}_{{Cu} / {Cu}^{2+}}^{0}=-0.34\, {~V}, {E}_{2 {n} / {Zn}^{2+}}^{0}=+0.76 \,{~V}\right.}$
$\left.\frac{2.303 {RT}}{{F}}=0.059\, {~V}\right]$