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Question

The copper anode of a cell containing silver nitrate solution weighs 60.0 g. After passing current for some time, it is found that 3.24 g of silver if deposited on the platinum cathode. What is the final weight of the anode? (Ag = 108, Cu = 64)

Options

A.0.96 g
B.60 g
C.59.04 g
D.60.96 g

Solution

neq Ag deposited = neq Cu dissolved

or, $\frac{3.24}{108} \times 1 = \frac{w}{64} \times 2$⇒ w = 0.96 gm

∴ Final mass of Cu anode = 60 – 0.96 = 59.04 gm

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