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Question

# (a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4 and ZnSO4 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 (Λm) vs square root of concentration (C)1/2 following curves are obtained for two electrolytes A and B :

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Solution

## (a)Fe2+(aq)+2e−→Fe(s) Now Q=it,i=2amp ∴t=Q/i since 2 F charge is required to deposit 56 g of Fe2+ so for 2.8 g we need 9650 C from above equation t=9650/2=4825s Using Faraday's second law of electrolysis W1(Weight of Fe deposited)W2(Weight of Zn deposited)=E1(Equivalent weight of Fe)E2(Equivalent weight of Zn) 2.8W2=56/265.3/2 W2=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 λ0m i.e. molar conductance at infinite dilution. In the case of weak electrolytes, λm increases steeply on dilution. Therefore, λ0m cannot be obtained by extrapolation.

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