Q. 25

# E° cell for the given redox reaction is 2.71Mg(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 VOR(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 ofFe 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)

2.8g of Fe requires = (2) (96500)(2.8)/(56)

= 9650 Coulombs

Q= It = (9650)(2) = 4825 sec

(ii)

(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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