Chemistry - Electrochemistry Question with Solution | TestHub

At the anode, water is oxidized to liberate O₂:

$$2H_{2}O\to O_2+4H^{+}+4e^{-}$$

4 moles of electrons are involved in the liberation of 1 mole of O₂ (32 g).

At the cathode, Ag⁺ is reduced to Ag:

$$A{g}^{+}+e^{-}\to Ag$$

1 mole of electron deposits 1 mole of Ag (107.88 g).

1.6 g of O₂ corresponds to $\frac{1.6}{32}$ = 0.05 moles of O₂.

Therefore, moles of electrons involved = 4 * 0.05 = 0.2 moles.

Amount of silver deposited = 0.2 * 107.88 = 21.576 g ≈ 21.60 g