(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO₄and ZnSO₄ until 2.8 g 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 = 56 g mol^–1 Zn = 65.3 g mol^–1, 1F = 96500 C mol^–1)

(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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Detailed Solution

(a) I =2 Ampere
56 g of Fe requires =2× 96500 C
$∴ 2.8 g will require = 2.8 \times \frac{2 \times 96500}{56} = 9650 C$
$\begin{array}{l} Q = It \\ t = \frac{Q}{I} = \frac{9650}{2} = 4825 s \end{array}$
Using Faraday's second law of electrolysis
$\frac{w_{1}}{w_{2}} = \frac{E_{1}}{E_{2}}$
Where w₁ = mass of Fe and w₂= mass of Zn
$\begin{array}{l} \frac{2 .8}{w_{2}} = \frac{56}{2} \times \frac{2}{65 .3} \\ w_{2} = 3 .265 g \end{array}$
(b) (i) A is a strong electrolyte and B is a weak electrolyte.

(ii) For B, 𝛬_m value at zero concentration cannot be obtained by extrapolation as it is a weak electrolyte. For A, the intercept will give the 𝛬_m value at zero concentration.

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