A current of 1 .93 ampere is passed through 200 mL of 0 .5 M Zinc sulphate (aq .) solution for 50 min with a current efficiency of 80% . If volume of solution remain constant, then [ Z n 2+ ] after deposition of Z n 2+ is:

Step-by-step solution

No. of Faraday = 1.93 × 50 × 60 96500 = 0.06 ∴ Moles of Z n 2+ deposited = 80 100 × 0.06 2 = 0.024 ∴   Zn 2+   =   0 .5   ×   0 .2   −   0 .024 0 .2 =   0 .38   M