Request a call back

what is efficiency
Asked by sujitjana971 | 15 Dec, 2022, 08:09: PM
Dear Student,

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.

Answered by | 16 Dec, 2022, 02:29: PM
NEET neet - Chemistry
Asked by sr8834055 | 20 Mar, 2024, 02:54: PM
NEET neet - Chemistry
Asked by dshreya247 | 03 Feb, 2024, 11:32: AM
NEET neet - Chemistry
Asked by sujitjana971 | 15 Dec, 2022, 08:09: PM
NEET neet - Chemistry
Asked by zoyakhan98264 | 16 Jul, 2022, 01:59: PM
NEET neet - Chemistry
Asked by gurugubellisaivishal2705 | 30 Jun, 2022, 12:32: PM