Electrochemistry

In the electrolysis of aqueous copper(II) bromine CuBr2(aq, 1.00 g of copper is deposited at the cathode. How many grams of bromine are produced at the anode?

96,485 C will deposit 63.54/2g Cu: therefore, how many C will deposit 1 g Cu? That's

96,485 x 2/63.54 = ?

Then 96,485 will deposit 160/2 g Br2. How man g will ? from above deposit.
That's (160/2) x ? from above = ?g Br2.
You should confirm all of the calculations AND confirm the atomic/molar masses of Cu and Br2.

Since 1 g of copper is deposited, that is 1/63.55 = 0.016 mol of Cu approximately.

Since the equation is CuBr2 -> Cu2+ + 2 Br- the mol ratio is 2 to 1.
2:1
Br:Cu
x:0.16
x=0.0315 mols of Br.
since m = nM,
m= 0.0315 x 79.9 (Molar mass of Br)
m=2.5145 g of Br

To determine the grams of bromine produced at the anode, we need to use the principles of electrolysis and Faraday's laws.

1. Write the balanced half-reactions for the electrolysis process:
At the cathode: Cu2+ + 2e- → Cu
At the anode: 2Br- → Br2 + 2e-

2. Calculate the moles of copper deposited at the cathode:
Molar mass of Cu = 63.55 g/mol
Mass of copper deposited = 1.00 g
Moles of copper = mass / molar mass = 1.00 g / 63.55 g/mol = 0.0157 mol

3. Use Faraday's laws to determine the moles of electrons transferred:
From the balanced equation at the cathode, we see that 2 moles of electrons are required to deposit 1 mole of copper. Therefore, moles of electrons transferred = 2 * moles of copper = 2 * 0.0157 mol = 0.0314 mol

4. Use Faraday's constant and the number of moles of electrons transferred to determine the charge passed:
Faraday's constant (F) = 96,485 C/mol
Charge passed (Q) = moles of electrons transferred * Faraday's constant = 0.0314 mol * 96,485 C/mol = 3,103 C

5. Calculate the moles of bromine produced at the anode:
From the balanced equation at the anode, we see that 2 moles of electrons are required to produce 1 mole of bromine. Therefore, moles of bromine produced = charge passed / (2 * Faraday's constant) = 3,103 C / (2 * 96,485 C/mol) = 0.0160 mol

6. Calculate the mass of bromine produced:
Molar mass of Br2 = 159.81 g/mol
Mass of bromine produced = moles of bromine * molar mass = 0.0160 mol * 159.81 g/mol = 2.56 g

Therefore, 2.56 grams of bromine are produced at the anode.

To determine the number of grams of bromine produced at the anode, we need to first understand the electrolysis of copper(II) bromide (CuBr2).

In electrolysis, an electric current is passed through an electrolyte (in this case, copper(II) bromide) to produce a chemical change. The electrolyte is made up of positive ions (cations) and negative ions (anions) that move towards the cathode (positive electrode) and anode (negative electrode), respectively.

At the cathode:
- The copper(II) cations (Cu2+) gain electrons and are reduced to copper (Cu), which gets deposited on the cathode surface.

At the anode:
- The bromide (Br-) anions lose electrons and are oxidized to bromine (Br2) gas, which is released at the anode.

To calculate the mass of bromine produced at the anode, we'll use Faraday's laws of electrolysis. Faraday's laws state that the amount of substance formed at an electrode is directly proportional to the quantity of electricity passing through the electrolyte.

We need to find the quantity of electricity (in coulombs) in order to apply the laws. To do this, we can use the equation:

Q = I x t

where:
Q = quantity of electricity (in coulombs)
I = current (in amperes)
t = time (in seconds)

Now, let's assume that the electrolysis reaction took place for a specific amount of time (t). We don't have the time given in the problem, so we'll leave it as a variable for now.

We are given that 1.00 gram of copper is deposited at the cathode. To find the quantity of electricity, we need to determine the number of moles of electrons involved in the copper deposition reaction.

The molar mass of copper is 63.55 g/mol, so we can calculate the number of moles of copper using the equation:

n = mass / molar mass

where:
n = number of moles
mass = mass of copper deposited
molar mass = molar mass of copper

n = 1.00 g / 63.55 g/mol
n = 0.0157 mol

Since 1 mol of electrons corresponds to 1 mol of copper in the reaction, the number of moles of electrons involved is also 0.0157 mol.

According to Faraday's laws, 1 mol of electrons corresponds to a charge of 1 Faraday (F), which is 96,485 Coulombs.

So, if 0.0157 mol of electrons corresponds to 96,485 Coulombs, we can calculate the quantity of electricity (Q) using the equation:

Q = n x F

where:
Q = quantity of electricity (in Coulombs)
n = number of moles of electrons
F = Faraday constant (96,485 C/mol)

Q = 0.0157 mol x 96,485 C/mol
Q = 1,517 C

Now that we have the quantity of electricity, we can use Faraday's laws to calculate the mass of bromine produced at the anode.

According to Faraday's first law, the mass of a substance formed during electrolysis is directly proportional to the quantity of electricity:

m = M x z x Q

where:
m = mass of substance formed
M = molar mass of the substance
z = number of moles of electrons transferred (stoichiometric coefficient)
Q = quantity of electricity (in Coulombs)

For the formation of bromine from bromide ions, the stoichiometric coefficient (z) is 2, as two electrons are transferred for each bromine molecule produced.

The molar mass of bromine is 79.90 g/mol, so we can calculate the mass of bromine using the equation:

m = M x z x Q

m = 79.90 g/mol x 2 x 1,517 C
m = 242,438 g

Therefore, approximately 242.438 grams of bromine are produced at the anode during the electrolysis of aqueous copper(II) bromide.