108 g of silver (molar mass 108 $$gmol^{-1}$$) is deposited at cathode from $$AgNO_3(aq)$$ solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from water by the same quantity of electricity is ___________.

According to Faraday’s first law of electro-lysis,

$$m \;=\;\frac{Q\,M}{n\,F}$$

where  $$m$$ is the mass deposited, $$Q$$ is the charge passed, $$M$$ is the molar mass of the metal ion, $$n$$ is the number of electrons required per ion, and $$F$$ is Faraday’s constant $$(96500\ \text{C mol}^{-1})$$.

We are told that $$108\ \text{g}$$ of silver is deposited. The molar mass of silver is also $$108\ \text{g mol}^{-1}$$, so

$$n_{\text{Ag}}=\frac{108\ \text{g}}{108\ \text{g mol}^{-1}}=1\ \text{mol}.$$

The cathodic reaction for silver deposition is

$$\text{Ag}^+ + e^- \;\longrightarrow\; \text{Ag}. $$

Each mole of Ag+ needs $$1$$ mole of electrons $$(n=1)$$. Therefore, the charge that must have passed is

$$Q = n_{\text{e}^-}\,F = 1\ \text{mol}\times 96500\ \text{C mol}^{-1}=96500\ \text{C}.$$

Exactly the same charge is now sent through water to produce oxygen gas. The anode reaction in water electro-lysis is

$$2\,\text{H}_2\text{O} \;\longrightarrow\; \text{O}_2 + 4\,\text{H}^+ + 4\,e^-.$$

This equation shows that $$4$$ moles of electrons give $$1$$ mole of $$ \text{O}_2$$. Hence, the moles of oxygen obtained with the available charge are

$$n_{\text{O}_2}= \frac{n_{\text{e}^-}}{4}= \frac{1\ \text{mol}}{4}=0.25\ \text{mol}.$$

At $$273\ \text{K}$$ and $$1\ \text{bar}$$ the molar volume of an ideal gas is $$22.7\ \text{L mol}^{-1}$$. Therefore, the volume of oxygen liberated equals

$$V = n_{\text{O}_2}\times 22.7\ \text{L mol}^{-1} = 0.25\ \text{mol}\times 22.7\ \text{L mol}^{-1} = 5.675\ \text{L}.$$

Rounding to three significant figures gives $$5.66\ \text{L}$$.

So, the answer is $$5.66\ \text{L}.$$