If the cost of electricity required to deposit 1 g of magnesium is N5.00, how much would it cost to deposit 10g of aluminium?

Badmus7639

6 years ago

In the question it was said that the cost of 1g of Al is #5
So going back to our electrolysis 3e- is discharged by Al
3*96500=27g
X=1g I.e x=3*96500/27=10722.22
So during electrolysis 2e- is discharged by Mg
2*96500=24
Y=1g I.e y=2*96500/24=8041.67
1g of mg=8041.67*1=#5
10g of Al=10722.22*10=#Z
#Z=10722.22*10*5=536111/8041.67
Z=#66.67
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Ufot1

7 months ago

This is how I solve this:
Find the no of Faraday that cost N5
So, 24g Mg = 2F
then 1g Mg = 0.0833F = N5
Find no of Faraday in 10g Al
So, 27g Al = 3F
therefore, 10g Al = 1.11F
Lastly, Find the cost of 1.11F
If 0.0833F = N5
then, 1.11F = N66.6

wealthy7154

1 year ago

ohhh,what is this no explanation
I don't like it

ImohDan

7 months ago

Please where does the 27 in Al3+ come from

chydirim@gmail.com

10 years ago

please expalain how this is gotten