How many grams of copper and oxygen from Cu 2+ solution are liberated by the same current that liberates 0.504 g hydrogen in 2 hours?

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freemihai's profile pic

freemihai | College Teacher | (Level 2) Adjunct Educator

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By definition, the same current liberates masses of different substances that are proportional to the equivalent masses.

Therefore, we must determine the equivalent masse of copper, oxygen and hydrogen, so we have 31.8 g of copper, 8 g of oxygen and 1.008 g of hydrogen.

We must find first the number of eq in 0.504 g of hydrogen:

n = 0.504g/1.008(g/eq) = 0.5 eq

Now that we know the number of eq of hydrogen, we can easily calculate the mass of copper and the mass of oxygen liberated by the same current. So, we just multiplicate each equivalent mass by 0.5 eq:

Mass of copper = 31.8*0.5 = 15.9 g

Mass of oxygen = 8*0.5 = 4 g

You answer is: 15 g liberated copper and 4 g liberated oxygen.

valentin68's profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

The Faraday law of electrilysis states that for a charge Q that passed through a solution the mass of the substance liberated is

m = (Q/F)*(M/z)

where F =96485 C/mol is the Faraday constant

M is the molecular mass of substance and

z is the valence of the substance ions in solution

Thus for liberated hydrogen (z=1, M=1 g/mol) the total charge Q is

Q = m*F*z/M =0.504*96485*1/1 =46628.44 C

For Cu(2+) (z =2, M=63.5 g/mol) the liberated mass is

m=(Q/F)*(M/z) = (46628.44/96485)*(63.5/2) =15.343 g

For oxygen O(2-) (z=2, M=16 g/mol) the liberated mass is

m = (Q/F)*(M/z) =(46628/96485)*(16/2) =3.867 g

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

`1.008 gm` hydrogen is liberated by the passage of `96500 C ` electric charge.

`0.504 gm` hydrogen is liberated by the passage of `(96500*0.504)/1.008 C= 48250 C` electric charge.

For `Cu^(2+)` solution:

`2*96500 C ` electric charge deposists `63.5 gm` copper from `Cu^(2+)` solution

Hence, `48250 C` electric charge deposists `(63.5*48250)/(2*96500) =15.875` `gm` copper from `Cu^(2+)` solution

For oxygen:

`2*96500 C` electric charge liberates 16 gm copper from `Cu^(2+) ` solution

Hence, `48250 C` electric charge liberates `(16*48250)/(2*96500) ` =`4 gm` oxygen from `Cu^(2+)` solution

Therefore, 15.875 gm copper and 4 gm oxygen from `Cu^(2+) ` solution are liberated by the same current that liberates 0.504 g hydrogen in 2 hours.

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