Question

An electrolytic cell contains a solution of $Ag_2SO_4$ and have platinum electrodes. A current is passed until 1.6 gm of $O_2$ has been liberated at anode. The amount of silver deposited at cathode would be

• A. 107.88 gm
• B. l.6gm
• C. 0.8 gm
• D. 21.60 gm

Answer 1

Answer: (D)

21.60 gm

Answer 2

$\frac{\text{Weight of oxygen}}{\text{Ewt.of oxygen}} = \frac{\text{Wt. of} Ag}{\text{E wt of} Ag}$ (Faraday's second law ) $\frac{1.6}{8} = \frac{\text{Wt. of}Ag}{108}$ Weight of $Ag$ deposited $= 21.60 \,gm$ [E wt. of oxygen $= \frac{\text{atomic weight }}{\text{valency}}$ $= \frac{16}{2} = 8$]