Question

In an electrolysis cell, the passage of $6.02\times {10}^{23}$ electrons can produce

• A. 11.2 liters of $H_2$ gas (measured at standard condition) from dilute $H_{2}S{O}_4$ solution
• B. 22.4 liters of $O_2$ gas (measured at standard conditions) from dilute $H_{2}S{O}_4$ solution
• C. 1 mole of $C{l}_2$ gas from HCl solution
• D. 1 mole of metallic silver from $AgN{O}_3$
• E. 1 mole of metallic copper from $CuS{O}_4$ solution

Answer 1

The correct option is A 11.2 liters of $H_2$ gas (measured at standard condition) from dilute $H_{2}S{O}_4$ solution

$2H^{+}+2e^{-}\to H_2$

As, 2 moles of electrons are required for liberation of 1 mole of $H_2$ gas at cathode, therefore, for 1 mole of electron 0.5 mole of $H_2$ gas will be produced whose volume will be 11.2 liters.