In the experiment above, a current was passed for 10 minutes, and 0.63g of copper was found to be deposited on the cathode of CuSO4 cells. The weight of silver deposited on the cathode of AgNO3 cell during the same period would be? [Cu = 63, Ag = 108]
- 0.54g
- 1.08g
- 1.62g
-
2.16g
- 3.24g
The correct answer is: D
Explanation
Same period, Same current
\(Cu^{2+} + 2e^- → Cu; 2Ag^+ + 2e^-\) → Ag
From the equation above, 2 mol Ag ≡ 1 mol Cu
But 0.63g Cu = \(\frac{0.63}{63}\) = 0.01mole Cu
Hence 0.002 mole Ag is deposited
mass of Ag deposited = 0.002 x 108 = 2.16g.