Ncert Exercises & Solutions

3.16 Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO_3, and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Hint: Faraday’s Second Laws of Electrolysis

Step 1:

According to the reaction:

{Ag}^{+}_{(aq)} + e^{-} \to {Ag}_{(s)}

108 g

i.e., 108 g of Ag is deposited by 96487 C.

Therefore, 1.45 g of Ag is deposited by=

\frac{96487 \times 1.45}{108} C

= 1295.43 C

Step 2:

Given,

Current=1.5 A

∴

Time=

\frac{1295.43}{1.5} s

= 863.6 s

= 864 s

= 14.40 min

Step 3:

Again,

{Cu}^{2 +}_{(aq)} + 2 e^{-} \to {Cu}_{(s)}

63.5 g

i.e., 2 x 96487 C of charge deposit=63.5 g of Cu

Therefore, 1295.43 C of charge will deposit=

\frac{65.4 \times 1295.43}{2 \times 96487} g

= 0.439 g of Zn

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