First, we need to determine the amount of $$\mathrm{O}_2$$ (oxygen gas) in moles that is displaced by the quantity of electricity mentioned. To do this, we'll use the molar volume of a gas at Standard Temperature and Pressure (S.T.P.), which is approximately $$22.4 \mathrm{~L/mol}$$ (or $$22400 \mathrm{~mL/mol}$$).

The volume of $$\mathrm{O}_2$$ is given as $$5600 \mathrm{~mL}$$. Now, we convert this volume to moles:

$$ \text{moles of } \mathrm{O}_2 = \frac{\text{volume of } \mathrm{O}_2}{\text{molar volume}} = \frac{5600 \mathrm{~mL}}{22400 \mathrm{~mL/mol}} = 0.25 \mathrm{~mol} $$

Next, we'll use Faraday's laws of electrolysis to relate the moles of $$\mathrm{O}_2$$ to the moles of silver being displaced. In electrolysis, silver is deposited at the cathode according to the following half-reaction:

$$ \mathrm{Ag}^+ + e^- \rightarrow \mathrm{Ag} $$

This tells us that for each mole of $$\mathrm{Ag}^+$$ ions, only 1 mole of electrons is required to reduce it to silver metal $$\mathrm{Ag}$$. However, the liberation of oxygen gas involves the following half-reaction:

$$ 2\mathrm{H_2O} (l) \rightarrow \mathrm{O}_2 (g) + 4H^+ (aq) + 4e^- $$

From this reaction, we can see that 1 mole of $$\mathrm{O}_2$$ gas requires 4 moles of electrons to be produced. Therefore, the number of moles of electrons associated with the $$0.25 \mathrm{~mol}$$ of $$\mathrm{O}_2$$ will be:

$$ \text{moles of electrons for } \mathrm{O}_2 = 0.25 \mathrm{~mol} \times 4 = 1 \mathrm{~mol} $$

Now, since it takes 1 mole of electrons to reduce 1 mole of $$\mathrm{Ag}^+$$. The moles of electrons required is equal to the moles of $$\mathrm{Ag}$$ produced. Hence, we have 1 mole of $$\mathrm{Ag}$$ being deposited.

Finally, to calculate the mass of this silver, we use the molar mass of silver:

$$ \text{Mass of Ag} = (\text{moles of Ag}) \times (\text{Molar mass of Ag}) = 1 \mathrm{~mol} \times 108 \mathrm{~gmol}^{-1} = 108 \mathrm{~g} $$

So, the mass of silver displaced by the quantity of electricity that displaces 5600 mL of $$\mathrm{O}_2$$ at S.T.P. will be $$108 \mathrm{~g}$$.