Question

(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes $FeS{O}_4$ and $ZnS{O}_4$ until 2.8g of Fe deposited at the cathode of cell X .How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y . ( Molar mass :Fe=56g/mol ,Zn=65.3g/mol ,1 F=96500C/mol)

(b) In the plot of molar conductivity $\left({\Lambda }_m\right)$ vs square root of concentration $(C{)}^{1/2}$ following
curves are obtained for two electrolytes A and B :

Answer 1

(a)$F{e}_{\left(aq\right)}^{2+}+2e^{-}\to F{e}_{\left(s\right)}$
Now $Q=it,i=2amp$
$\therefore t=Q/i$
since 2 F charge is required to deposit 56 g of $F{e}^{2+}$ so for 2.8 g we need 9650 C
from above equation
$t=9650/2=4825s$
Using Faraday's second law of electrolysis
$\frac{W_1\left(WeightofFedeposited\right)}{W_2\left(WeightofZndeposited\right)}=\frac{E_1\left(EquivalentweightofFe\right)}{E_2\left(EquivalentweightofZn\right)}$
$\frac{2.8}{W_2}=\frac{56/2}{65.3/2}$
$W_2=3.265g$
(b) We can conclude from the graph that A is a strong electrolyte and B a is week electrolyte. On extrapolating the curve towards zero concentration for strong electrolytes, we get the value of ${\lambda }^{0}m$ i.e. molar conductance at infinite dilution. In the case of weak electrolytes, $\lambda m$ increases steeply on dilution. Therefore, ${\lambda }^{0}m$ cannot be obtained by extrapolation.