E° cell for the given redox reaction is 2.71
Mg(s) + Cu2+ (0.01 M) ––– Mg2+ (0.001 M) + Cu(s)
Calculate Ecell for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and
(ii) greater than 2.71 V
OR
(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 2.8 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 (^m) vs square root of concentration (c1/2), following curves are obtained for two electrolytes A and B :

Answer the following :
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of ^m to concentration approaching zero for electrolytes A and B?
Mg(s) + Cu2+ (0.01 M) ––– Mg2+ (0.001 M) + Cu(s)

(i) less than 2.71V, current will flow from copper to magnesium
(ii) more than 2.71V, curret will from magnesium to copper
OR
a) (i)
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2.8g of Fe requires = (2) (96500)(2.8)/(56)
= 9650 Coulombs
Q= It = (9650)(2) = 4825 sec
(ii)
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(2) (96500) C of electricity deposit Zn = 65.3g
Therefore, 96500 C of electricity deposit Zn = (65.3) (96500)(9650) = 3.265g
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