JEE Main 2024 — Electrochemistry Question with Solution

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 $$22.4L/mol$$ (or $$22400mL/mol$$).

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

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

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:

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

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:

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

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 $$108\mathrm{g}$$.