A solution of $\text{NaCl}$ was placed between two $\text{'pt'}$ electrodes 10.0 cm apart, across which a potential difference of 5.0 V was applied. How far ${\text{Na}}^{\text{+}}$ions move in 10 hours at 298K in cm is … 

(Given${\lambda }^{\text{o}}{\text{Na}}^{\text{+}}={\text{\hspace{0.17em}50.0\hspace{0.17em}S\hspace{0.17em}cm}}^{\text{2}}{\text{\hspace{0.17em}eq}}^{\text{-1}}\text{\hspace{0.17em}\hspace{0.17em}at\hspace{0.17em}\hspace{0.17em}298K)}$

Ionic mobility ${\mu }_{\text{Na+}}^o\text{=}\frac{{\text{λ}}^{\text{0}}{\text{Na}}^{+}}{\text{F}}\text{=}\frac{\text{50}}{\text{96500}}{\text{=\hspace{0.17em}5\hspace{0.17em}×\hspace{0.17em}10}}^{\text{-4}}{\text{cm}}^{\text{2}}{s}^{\text{-1}}{v}^{\text{-1}}$

Political gradient applied  $\frac{\text{5.0\hspace{0.17em}V}}{\text{10.0\hspace{0.17em}cm}}{\text{=0.5V\hspace{0.17em}cm}}^{\text{-1}}$ 

Speed of ${\text{Na}}^{\text{+}}$= Ionic mobility $\times$ political gradient

                           =${\text{5.0×10}}^{\text{-4}}{\text{cm}}^{\text{2}}{\text{s}}^{\text{-1}}{\text{v}}^{\text{-1}}{\text{×0.5V\hspace{0.17em}cm}}^{\text{-1}}$

                           = ${\text{2.5×10}}^{\text{-4}}\text{cm\hspace{0.17em}}{S}^{\text{-1}}$

Distance travelled by ${\text{Na}}^{\text{+}}$  ion in 10 hours

$\Rightarrow$ Speed of ${\text{Na}}^{\text{+}}$ion $\times$ time (in sec)

$\Rightarrow$ $2.5\times {10}^{-4}\times 10\times 60\times 60$ 

$\Rightarrow$9 cm