An electrolytic cell contains a solution of Ag2SO4 and have platinum electrodes. A current is passed until 1.6 gm of O2 has been liberated at anode. The amount of silver deposited at cathode would be

Question

An electrolytic cell contains a solution of Ag2SO4 and have platinum electrodes. A current is passed until 1.6 gm of O2 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

Tardigrade Admin · Accepted Answer

( textWeight of oxygen/ textEq.wt.of oxygen) = ( textWt. of Ag/ textEq. wt of Ag) (Faraday's second law ) (1.6/8) = ( textWt. ofAg/108) Weight of Ag deposited = 21.60 gm [Eq. wt. of oxygen = ( textatomic weight / textvalency) = (16/2) = 8]

Q. An electrolytic cell contains a solution of $A g_{2} S O_{4}$ 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

107.88 gm

l.6gm

0.8 gm

21.60 gm

$\frac{Weight of oxygen}{Eq.wt.of oxygen} = \frac{Wt. of A g}{Eq. wt of A g}$
(Faraday's second law )
$\frac{1.6}{8} = \frac{Wt. of A g}{108}$
Weight of $A g$ deposited $= 21.60 g m$
[Eq. wt. of oxygen $= \frac{atomic weight}{valency}$ $= \frac{16}{2} = 8$ ]

Q. An electrolytic cell contains a solution of Ag2​SO4​ and have platinum electrodes. A current is passed until 1.6 gm of O2​ has been liberated at anode. The amount of silver deposited at cathode would be