Two electrolysis cells are set up in series so that the same current flows through each. In one compartment 5.40 g of Ag (molar mass = 108 g/mol) is deposited from a Ag+ solution. What mass of Cu (molar mass = 64 g/mol) is deposited simultaneously if the other cell contains Cu2+?
I don't know how to do this. I am confused that both are positive? and it doesn't have a redox reaction? Help on how to do?
How many coulombs were needed to deposit the 5.4 g Ag. That's 96,485 coulombs x (5.4/108) = approx 4824. How many grams Cu will be deposited with 96,485? That is 64/2 = 32 g. You didn't have 96,485; you had 4824. So
32 x (4,824/96,495) = ?
Redox reactions are
Ag^+ + e ==> Ag(s)
Cu^2+ + 2e ==> Cu(s)
The oxidation reaction(s) depend(s) upon the anions in the water solutions.
To calculate the mass of Cu deposited in the other cell, we need to consider the stoichiometry of the electrochemical reaction. In this case, there is a redox reaction taking place at each electrode.
First, let's start with the Ag+ solution. The overall balanced redox reaction for the deposition of Ag metal is:
2Ag+ + 2e- -> 2Ag
From this reaction, we can see that for every 2 moles of electrons (2e-), 2 moles of Ag+ ions are reduced to 2 moles of Ag metal. Since the molar mass of Ag is 108 g/mol, we can say that 2 moles of Ag is equal to 216 g.
In the Ag electrode of the electrolysis cell, 5.40 g of Ag is deposited. We can calculate the number of moles of Ag deposited using the formula:
moles = mass / molar mass
moles = 5.40 g / 108 g/mol = 0.05 mol
Since 2 moles of Ag are produced for every 2 moles of electrons, it means 1 mole of Ag is deposited for every 1 mole of electrons. Therefore, the number of moles of electrons involved in the Ag deposition is also 0.05 mol.
Now, let's move on to the Cu2+ solution in the other cell. The balanced redox reaction for the deposition of Cu metal is:
Cu2+ + 2e- -> Cu
From this reaction, we can see that for every 2 moles of electrons (2e-), 1 mole of Cu2+ ions are reduced to 1 mole of Cu metal. Since the molar mass of Cu is 64 g/mol, we can say that 1 mole of Cu is equal to 64 g.
Since the number of moles of electrons involved in the Ag deposition is 0.05 mol, it also means that 0.05 mol of Cu2+ ions are reduced to 0.05 mol of Cu metal in the other cell.
Now, we can calculate the mass of Cu deposited using the formula:
mass = moles x molar mass
mass = 0.05 mol x 64 g/mol = 3.20 g
Therefore, simultaneously with the deposition of 5.40 g of Ag, 3.20 g of Cu is deposited in the other cell.