See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 H 10       m moles of 2 H 10 100 1      During electrolysis, following reactions take place. At cathode    2 Cu aq. 2e Cu s     At anode     2 2 1 H O O g 2H aq. 2e 2       From above reactions, it is clear that for 1 m mole of H+ produced at anode, 0.5 m mole of Cu+2 is reduced at cathode. Initial m mole of Cu+2 = 100 mL × 0.05 = 5 m mole After electrolysis Cu+2 remaining in solution = 5 – 0.5 = 4.5 m moles 2 2 2 2 2Cu 4I Cu I I      So, according to above reaction