NEET Class neet Answered

Mass of copper deposited in grams = M/nF × it × e,

Where,

M is the atomic mass of the element.

For Cu, Mis 63.5 g.

n is the number of electrons for the anodic or cathodic reaction.

Cathodic reduction for Cu is Cu²⁺ + 2e⁻ → Cu(s)

Hence, n = 2 electrons

F is the Faraday constant = 96500

i is the current passed in Amperes = 20 A

t is the duration of current passed in seconds = 965 s

e is the electrolytic efficiency = 50% = 0.5

Substituting respective values,

Therefore mass of Cu deposited (g) = 3.175 g

Hence, the amount of Cu deposited at the cathode will be 3.175 g.