Chemistry · Electrolytic and metallic conduction, conductance in electrolytic solutions, molar conductivities and their variation with concentration, Kohlrausch's law and its applications
If is electrolysed completely with the current of efficiency then :
If \( 9 g m H_{2} O \) is electrolysed completely with the current of \( 50 \% \) efficiency then :
- A. 96500 charge is required
- B. \( 2 \times 96500 C \) charge is required
- C. \( 5.6 L \) of \( O_{2} \) at STP will be formed
- D. \( 11.2 L \) of \( O_{2} \) at STP will be formed
Step-by-step solution
The electrolysis of water follows: 2H₂O → 2H₂ + O₂. For 9 g H₂O (0.5 mol), the theoretical charge required is 1 mol e⁻ = 96500 C. With 50% current efficiency, the actual charge needed is doubled: 2 × 96500 C. The volume of O₂ produced (0.25 mol) at STP is 5.6 L, which is also correct but the primary effect of efficiency is on charge required.
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