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CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

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Question : 29 of 37
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When a steady current of 2A2A was passed through two electrolytic cells AA and BB containing electrolytes ZnSO4\mathrm{ZnSO}_4 and CuSO4\mathrm{CuSO}_4 connected in series, 2 g2\,\mathrm{g} of Cu\mathrm{Cu} were deposited at the cathode of cell BB. How long did the current flow? What mass of Zn\mathrm{Zn} was deposited at cathode of cell A?
[Atomic mass: Cu=63.5 g mol−1, Zn=65 g mol−1\mathrm{Cu}=63.5\,\mathrm{g}\,\mathrm{mol}^{-1},\,\mathrm{Zn}=65\,\mathrm{g}\,\mathrm{mol}^{-1} ; 1 F=96500 C mol−11\,\mathrm{F}=96500\,\mathrm{C}\,\mathrm{mol}^{-1}
Solution:  
  Zn2+(aq)+  2e−2 mol→  Zn(s)1 mol\; \mathrm{Zn}^{2+}(\mathrm{aq}) + \; \underset{2\,\mathrm{mol}}{2\mathrm{e}^{-}} \rightarrow \; \underset{1\,\mathrm{mol}}{\mathrm{Zn}(\mathrm{s})}
  Cu2++  2e−2 mol→  Cu(s)1 mol\; \mathrm{Cu}^{2+} + \; \underset{2\,\mathrm{mol}}{2\mathrm{e}^{-}} \rightarrow \; \underset{1\,\mathrm{mol}}{\mathrm{Cu}(s)}
( 2 gm given)
The charge QQ on a mole of electrons, Q=n FQ = n\,\mathrm{F}
Calculation of time for the flow of current:
  n=1 mol\; n = 1\,\mathrm{mol}
  Q=1×96500 C mol−1=96500 C\; Q = 1 \times 96500\,\mathrm{C}\,\mathrm{mol}^{-1} = 96500\,\mathrm{C}
Molar mass of Cu=63.5 g mol−1\mathrm{Cu} = 63.5\,\mathrm{g}\,\mathrm{mol}^{-1}
  ∵  63.5 g of Cu is deposited by electric charge\; \because \; 63.5\,\mathrm{g} \,\text{of}\, \mathrm{Cu} \,\text{is deposited by electric charge}
  =96500 C\; = 96500\,\mathrm{C}
  ∴  2 g of Cu is deposited by electric charge\; \therefore \; 2\,\mathrm{g} \,\text{of}\, \mathrm{Cu} \,\text{is deposited by electric charge}
  =9650063.5×2=3039.37 C\; = \frac{96500}{63.5} \times 2 = 3039.37\,\mathrm{C}
Let 2 A2\,\mathrm{A} of current be passed for time tt, quantity of electricity used =2 A×t=3039.37 C= 2\,\mathrm{A} \times t = 3039.37\,\mathrm{C}
  or, t=3039.37 C2=1519.68 s\;\text{or, } t = \frac{3039.37\,\mathrm{C}}{2} = 1519.68\,\mathrm{s}
  =25 min .33 s\; = 25\,\mathrm{min}\,.33\,\mathrm{s}
Calculation of mass of Zn\mathrm{Zn} deposited:
  W1W2=E1E2= Mass of Zn Mass of Cu\; \frac{W_1}{W_2} = \frac{E_1}{E_2} = \frac{\text{ Mass of } \mathrm{Zn}}{\text{ Mass of } \mathrm{Cu}}
  Molar mass of Zn/ Charge on CuMolar mass of Cu/ Charge on Cu\; \frac{\text{Molar mass of }\mathrm{Zn} / \text{ Charge on }\mathrm{Cu}}{\text{Molar mass of }\mathrm{Cu} / \text{ Charge on }\mathrm{Cu}}
Amount of Zn\mathrm{Zn} deposited:
=2×6526352=2.0472 g= 2 \times \frac{\frac{65}{2}}{\frac{635}{2}} = 2.0472\,\mathrm{g}
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