Electricity is passed through an acidic — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Eq of Cu = Eq of O2 3 O 300 10 n 63.54      2.36 × 10 –3 = O n When current is further passed O 600 28 60 n 96500 1000      O n = 2.611 × 10 –3 Total O2 released = [10 –3 ×(2.36 + 2.611)]×22400 ml = 111.35 ml