A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is ______ cm$$^3$$. (Nearest integer) [Given: Faraday constant F = 96500 C mol$$^{-1}$$ at STP, molar volume of an ideal gas is 22.7 L mol$$^{-1}$$]

We need to find the total volume of gases produced at STP during electrolysis of dilute H$$_2$$SO$$_4$$.

Current (I) = 0.10 A. Time (t) = 2 hours = 2 × 3600 = 7200 s. F = 96500 C mol$$^{-1}$$. Molar volume at STP = 22.7 L mol$$^{-1}$$ = 22700 cm$$^3$$ mol$$^{-1}$$.

$$Q = I \times t = 0.10 \times 7200 = 720$$ C

At cathode: $$2H^+ + 2e^- \rightarrow H_2$$. At anode: $$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$.

Moles of electrons = $$\frac{Q}{F} = \frac{720}{96500}$$ = 0.007461 mol

2 moles of electrons produce 1 mole of H$$_2$$. Moles of H$$_2$$ = $$\frac{0.007461}{2}$$ = 0.003731 mol

4 moles of electrons produce 1 mole of O$$_2$$. Moles of O$$_2$$ = $$\frac{0.007461}{4}$$ = 0.001865 mol

Total moles = 0.003731 + 0.001865 = 0.005596 mol

Volume = total moles × molar volume = 0.005596 × 22700 = 127.03 cm$$^3$$

$$\approx 127$$ cm$$^3$$

The total volume of gases produced at STP is 127 cm$$^3$$.