Question

Calculate how long it will take to deposit 1.0 g of chromium when a current of 1.25 ampere flows through a solution of chromium (III) sulphate. (Atomic weight of Cr = 52, 1 Faraday = 96,500 coulombs)

Answer 1

Given: Atomic weight of Cr = 52

1 Faraday = 96,500 coulombs

I = 1.25 ampere

m = 1.0 g

\[\ce{Cr^{3+} + 3e^- → Cr}\]

52 g of chromium requires current = 3 × 96,500 coulombs

1 g of chromium requires current = `(3 xx 96500 xx 1)/52`

= `289500/52`

= 5567.3

Q = I × t

t = `Q/I`

= `5567.3/1.25`

= 4453.8 seconds