1.0 M aqueous solutions of
AgNO3
, Cu(NO3
)2
and
Au(NO3
) 3
are electrolyzed
in the apparatus shown, so
the same amount of
electricity passes through
each solution. If 0.10 moles
of solid Cu are formed how
many moles of Ag and Au
are formed?
Well, it looks like these solutions are having a little electric party! Let's see what's happening.
We know that the same amount of electricity passes through each solution. So, if 0.10 moles of solid Cu are formed, we can assume that 0.10 moles of Cu are deposited on the electrode.
Now, let's look at the chemical reactions that are taking place at the electrodes.
For silver (Ag):
Ag+ + e- -> Ag
For copper (Cu):
Cu2+ + 2e- -> Cu
And for gold (Au):
Au3+ + 3e- -> Au
From these reactions, we can see that one mole of electrons is required to deposit one mole of each metal. So, if 0.10 moles of Cu are formed, it means that 0.10 moles of electrons have flowed through the solution.
Now, if 0.10 moles of electrons have flowed through the solution, and one mole of electrons is required to deposit one mole of each metal, it means that 0.10 moles of Ag and 0.10 moles of Au will be formed as well.
So, to answer your question, 0.10 moles of Ag and 0.10 moles of Au will be formed along with 0.10 moles of Cu. Now, let's hope they all have a good time at the metal party!
To determine the moles of Ag and Au formed, we first need to calculate the number of moles of electrons transferred in the reaction. This can be done using Faraday's law of electrolysis:
1 mole of electrons = 1 Faraday = 96,485 C
Given that the same amount of electricity passes through each solution, we can equate the charge passed through each solution:
Q(Ag) = Q(Cu) = Q(Au)
Now, let's calculate the moles of electrons transferred:
For Cu:
1 mole of Cu^2+ + 2 moles of electrons -> 1 mole of Cu
Since 0.10 moles of Cu are formed, it means 2 * 0.10 = 0.20 moles of electrons were transferred.
Now, let's calculate the moles of Ag and Au formed using the same number of electrons:
For Ag:
1 mole of Ag+ + 1 mole of electron -> 1 mole of Ag
0.20 moles of electrons would produce 0.20 moles of Ag.
For Au:
3 moles of Au^3+ + 3 moles of electrons -> 1 mole of Au
0.20 moles of electrons would produce 0.20/3 = 0.067 moles of Au.
Therefore, 0.20 moles of Ag and 0.067 moles of Au are formed.
To find out how many moles of Ag and Au are formed, we need to first understand the concept of electrolysis.
During electrolysis, the passage of electric current causes the breakdown of ionic compounds into their constituent elements. This process occurs at the electrodes, which are usually made of inert materials like graphite or platinum.
In the given apparatus, the aqueous solutions of AgNO3, Cu(NO3)2, and Au(NO3)3 are being electrolyzed. Each solution contains metal ions (Ag+, Cu2+, and Au3+) and nitrate ions (NO3-) that can undergo electrolytic reactions.
Based on Faraday's laws of electrolysis, the amount of substance formed at the electrode is directly proportional to the amount of electric charge passed through the solution.
Let's assume that it takes one mole of electrons to produce one mole of solid Cu. This means that 0.10 moles of Cu formed indicates that 0.10 moles of electrons have passed through the Cu(NO3)2 solution.
Since the same amount of electricity passes through each solution, the number of moles of electrons passed through the AgNO3 and Au(NO3)3 solutions would also be 0.10 moles. We can use this information to determine the amount of Ag and Au formed.
To find the number of moles of Ag and Au formed, we need to know their respective reduction half-reactions and their stoichiometry. Here are the half-reactions and their balanced equations:
Ag+ + e- -> Ag (Reduction half-reaction for Ag+)
Au3+ + 3e- -> Au (Reduction half-reaction for Au3+)
From the balanced equations, we can see that for every mole of electrons (e-) transferred, one mole of Ag and one mole of Au are formed. Therefore, the number of moles of Ag and Au formed is equal to the number of moles of electrons passed through the solutions, which is 0.10 moles.
Thus, 0.10 moles of Ag and 0.10 moles of Au are formed in this electrolysis process.
The short way of doing it is
0.2 C/2e = 0.1 mol Cu
0.2 C/1e = 0.2 mol Ag
0.2 C/3e = 0.0667 mol Au