(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes $F e S O_{4}$ and $Z n S O_{4}$ 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 $(Λ_{m})$ vs square root of concentration $(C)^{1 / 2}$ following
curves are obtained for two electrolytes A and B :

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Solution

(a) $F e_{(a q)}^{2 +} + 2 e^{-} \to F e_{(s)}$
Now $Q = i t, i = 2 a m p$
$∴ t = Q / i$
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
$t = 9650 / 2 = 4825 s$
Using Faraday's second law of electrolysis
$\frac{W_{1} (W e i g h t o f F e d e p o s i t e d)}{W_{2} (W e i g h t o f Z n d e p o s i t e d)} = \frac{E_{1} (E q u i v a l e n t w e i g h t o f F e)}{E_{2} (E q u i v a l e n t w e i g h t o f Z n)}$
$\frac{2.8}{W_{2}} = \frac{56 / 2}{65.3 / 2}$
$W_{2} = 3.265 g$
(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 $λ^{0} m$ i.e. molar conductance at infinite dilution. In the case of weak electrolytes, $λ m$ increases steeply on dilution. Therefore, $λ^{0} m$ cannot be obtained by extrapolation.

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F e S O_{4}

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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

(Λ_{m})

vs square root of concentration

(C)^{1 / 2}

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