Answer to Question #145804 in General Chemistry for ken Tanaka

Question

A technician plates a faucet with 0.86 g of Cr Metal by electrolysis of aqueous Cr²(SO⁴)3. If 12.5 min is allowed for the plating, what current is needed?

Answer 1

Dissociation of Cr₂(SO₄)₃ is as follows

Cr₂(SO₄)₃ ---> 2Cr³⁺ + SO₄^2-

The plating by Cr3+ follow the reaction,

Cr³⁺ + 3e ---> Cr

This is a 3 electron transfer reaction,

So, z = 3

So, equivalent mass of Cr³⁺ ,

E = M/z g.eqv-1

= (52/3) g.eqv-1

= 17.334 g.eqv-1

We know the Faraday equation,

W =(I×t×E)/F

Or, I = (W×F)/(t×E)__________[a]

Where, W = mass of Cr metal accumulated

E = equivalent mass of Cr³⁺

I = total current pass through the electrolyte

t = time of pussin the current

F= Faraday constant

= 96500 C

Given,

t = 12.5 min

= 12.5×60 s

= 750 s

W = 0.86 g

Putting the values in equation [a] we get

I = (W×F)/(t×E)

Or,I = (0.86×96500)/(750×17.334)

Or,I = 6.384

Hence, 6.384 A current is needed.