ElectrochemistryMedium
Question
An electrolytic cell contains a solution of Ag2SO4 and platinum electrodes. A current is passed until 1.6 g of O2 has been liberated at anode. The amount of silver deposited at cathode would be
Options
A.108.0 g
B.1.6 g
C.10.8 g
D.21.6 g
Solution
$n_{eq}Ag = n_{eq}O_{2} \Rightarrow \frac{w}{108} \times 1 = \frac{1.6}{32} \times 4 \Rightarrow w = 21.6\text{ gm}$
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