If an aqueous solution containing M2 + ions electrolyzed by a current of 3.00 A for 60.0 min., resulting in 3.66 g of solid M produced at a cathode during this time, what is the likely identity of the element M

Question

Adam Sweeney

Chemistry

13 September, 08:16

If an aqueous solution containing M2 + ions electrolyzed by a current of 3.00 A for 60.0 min., resulting in 3.66 g of solid M produced at a cathode during this time, what is the likely identity of the element M

+2

Answers (1)

Know the Answer?

Deacon · Answer 1

According to Faraday's Law,

W = I t e / F

Where,

W = Amount deposited = 3.66 g

I = Current = 3 A

t = Time = 60 min = 3600 s

e = Chemical equivalence

F = Faraday's Constant = 96500

Solving for e,

e = F W / I t

Putting Values,

e = (96500*3.66) : (3 * 3600)

e = 32.7

As we know,

e = M / 2

solving for M,

M = e * 2

Putting value of e,

M = 32.7 * 2

M = 65.4

M = Zinc