Al(3+) is reduced to Al(s) in a electrolytic cell. If a current of 2.00 amperes

Question

Al(3+) is reduced to Al(s) in a electrolytic cell. If a current of 2.00 amperes is used and 270 g of Al is produced, how much time is needed to obtain the result? The reaction occurs according to the following equatoion:

Al(3+) + 3e -> Al(s)

Assume 100% current efficiency.


The #mol of electrons used in the electrolysis which is mentioned in the previous question is determined using the relation between the mole ratio of (electrons / Al )which is shown by the equation to be

The expression of #mol(e) is as follows: #mol(e) =_______________________(a digit)_________________

(aletter indicating a math operation sign) #mol(Al), which is found using the following relation: #mol(Al) =

_____________________(a math operation) MW =_____________________ a 3 digit number) mol, obtained after pluggin the available numbers in the operation.

Explanation / Answer

m= 270 g

I= 2amp

z=3

M=26.98g/mol

F= Faraday's constant = 96485C mol^-1

A/C to faraday law of electrolysis

m= It/F xM/z

t = mxFxz /MI = 270x96485x3 / 2x 26.98 = 810x96485/53.96= 1448347.85 second= 402.32 hr

Q= (ne-)F

mole of electricity used = Q/F = It/F = 2 x1448347.5 / 96485 = 30 mol

mole of Al = 270 / 26.98 = 10.0074

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