Electrolysis of Copper sulfate (CuSO₄) Solution

Copper sulfate is soluble in water and gives a blue colour aqueous solution though anhydrous copper sulfate solid is white. In an aqueous solution, Copper sulfate completely dissociates to copper 2+ (Cu²⁺) cation and sulfate (SO₄^2-) anion. Therefore, CuSO₄ is a strong electrolyte. We can perform electrolysis of copper sulfate in different methodologies to perform copper plating and will discuss those stuff in detail in this tutorial.

In an electrolysis process, you should have the capability of deciding which chemical will be oxidized and which one will be reduced. Follow below steps to decide products of electrolysis and to observe the behavior of electrolysis.

1. Identify anode and cathode
2. List out compounds, anions and cations in the aqueous solution
3. Mark distribution of compounds, anions and cations around anode,cathode and aqueous solution
4. Decide which compounds, anions and cations will be oxidized and reduced according to the electrochemical series and decide final products.
5. Observe physical changes such as colour changes, gas formations, precipitating

Electrolysis of copper sulfate solution with inert and active anode and for copper plating

We will study several ways of electrolysis of CuSO₄ solution by changing the type of electrode.

1. Electrolysis of copper sulfate solution with Graphite / Platinum anode (inert electrode) and Iron cathode
2. Electrolysis of copper sulfate solution with Copper anode (active) and Iron cathode

Electrolysis of copper sulfate solution with graphite / platinum anode and Iron cathode

In this experiment, we are going to deposit metallic copper layer in the surface of a iron piece. We use graphite or platinum electrode because they do not react easily with water and is not oxidized or reduced too.

1. Identify anode and cathode
   • Anode (connected to the positive terminal of DC power supply): Graphite or platinum electrode
   • Cathode (connected to the negative terminal of DC power supply): Iron
   
   
2. List out compounds, anions and cations in the aqueous solution:
   • Compounds: Water
   • Anions: SO₄^2-, OH^-
   • Cations: Cu²⁺, H⁺
   
   
3. Mark distribution of compounds, anions and cations around anode,cathode and aqueous solution
   • Towards anode: SO₄^2-, OH^-
   • Towards cathode: Cu²⁺, H⁺
   
   
4. Decide which compounds, anions and cations will be oxidized and reduced according to the electrochemical series and decide the final products.

Anode reaction / Oxidation reaction

There are two anions (SO₄^2-, OH^-) around anode. Oxidation of sulfate ion is almost impossible. Due to this reason, electrochemical series is not used to identify which anion will be oxidized. Therefore, hydroxyl ions (OH^-) ion is oxidized to oxygen gas molecules and respective half reaction is written as below.

4OH^-_(aq) → O_2(g) + 2H₂O_(aq) + 4e

Cathode reaction / Reduction reaction

There are two cations (Cu²⁺, H⁺) around cathode. To decide the, which cation will be reduced, electrochemical series with standard potential is used as below. Let's consider standard potential values of Cu²⁺ and H⁺.

Because standard potential of Cu²⁺ cation reduction (to Cu) is more positive than standard potential of H⁺ reduction (to H₂), Cu²⁺ cations are reduced at cathode. When Cu²⁺ concentration become very low, H⁺ ions are reduced to H₂. Therefore, there should be enough concentration of Cu²⁺ ions in the solution.

Cu²⁺_(aq) + 2e → Cu_(s)

Physical changes such as colour changes, gas formations, precipitating

1. Blue colour of aqueous CuSO₄ solution is reduced with time because Cu²⁺ cation concentration is reduced.
2. You can see formation of gas bubbles (O₂) close to the anode.
3. pH of the solution is decreased.

Electrolysis of copper sulfate solution with Copper anode (active) and Iron cathode

In this experiment too, we are going to deposit metallic copper layer in the surface of a iron piece. But, We use a Copper electrode electrode here. So, there is a change in the overall reactions and we will see how products are given.

1. Identify anode and cathode
   • Anode (connected to the positive terminal of DC power supply): Copper electrode
   • Cathode (connected to the negative terminal of DC power supply): Iron metal piece
   
   
2. List out compounds, anions and cations in the aqueous solution:
   • Compounds: Water
   • Anions: SO₄^2-, OH^-
   • Cations: Cu²⁺, H⁺
   
   
3. Mark distribution of compounds, anions and cations around anode,cathode and aqueous solution
   • Towards anode: SO₄^2-, OH^-
   • Towards cathode: Cu²⁺, H⁺
   
   
4. Decide which compounds, anions and cations will be oxidized and reduced according to the electrochemical series and decide the final products.

Anode reaction / Oxidation reaction

• There are two anions (SO₄^2-, OH^-) around anode. Also, Copper metal piece behave as an active anode. So there are three available stuffs to being oxidized at the anode.
• Oxidation of sulfate ion is almost impossible.
• Because there is an active anode, we have to decide which reaction will be occurred at the anode. So, electrochemical series is used to identify which stuff will be oxidized.
• There are two possible reactions;
  
  1. Hydroxyl ions (OH^-) ions are oxidized to oxygen gas molecules.
  2. Copper is oxidized to Cu²⁺ cations.
• To decide which reaction will be occurred, standard potential values of each half reaction are considered.

When, positive value of standard potential is increased for a oxidizing half reaction is increased, possibility of that half reaction becomes low. That suggest that, possibility of Copper's oxidization is much higher than OH^- 's oxidation to oxygen gas.

Anode reaction: Cu_(s) → Cu²⁺_(aq) + 2e

So, copper atoms in the copper piece are oxidized and Cu²⁺ ions are released to the aqueous solution. Released electrons due to the oxidation process are flown towards cathode through the DC power supply. As well as, due to the released of Cu²⁺ cations to the solution, Cu²⁺ cation concentration will be kept at a constant value.

Cathode reaction / Reduction reaction

There are two cations (Cu²⁺, H⁺) around the cathode. To decide, which cation will be reduced, electrochemical series with standard potential is used as below. Let's consider standard potential values of Cu²⁺ and H⁺.

Because standard potential value of Cu²⁺ cation's reduction (to Cu) is more positive than standard potential of H⁺ cation's reduction (to H₂), Cu²⁺ cations are reduced at cathode. When Cu²⁺ concentration become very low, H⁺ ions are reduced to H₂ molecules. Therefore, there should be enough concentration of Cu²⁺ ions in the solution to complete the copper plating process.

Cathode reaction: Cu²⁺_(aq) + 2e → Cu_(s)

Physical changes such as colour changes, gas formations, precipitating

1. Blue colour of aqueous CuSO₄ solution is not changed with time because Cu²⁺ cations are continuously generated at the anode.
2. Copper metal piece is dissolved and Copper is plated on the iron metal piece.

Example problems for calculating current requirement in CuSO₄ electrolysis

A DC electricity is supplied to the CuSO₄ aqueous solution for the copper plating. For a perfect electrolysis process, enough and higher current should be supplied continuously. According to the following factors, we can calculate requirement of current flow.

Let's try few questions to understand this.

1.27 g of Copper should be plated on a Iron spoon by using a Copper bar as an anode. If electrolysis process should be completed within 10 minutes, Calculate the current.

Step 1: Calculate amount of copper plating on the Iron spoon

• Amount of copper plating = 1.27 g / 63.5 g mol^-1
• Amount of copper plating = 0.02 mol

Step 2: Decide how much electrons are exchange when Cu²⁺ are reduced to Cu

Cu²⁺ + 2e → Cu

Required quantity of electrons is twice as plated Cu amount.

• Required quantity of electrons = 2 * 0.02 mol
• Amount of copper plating = 0.04 mol

Step 3: Calculate total charge due to the flow of electrons

Charge of 1 mol of electrons = 96500 C

• Total charge due to the flow of electrons = 96500 C mol^-1 * 0.04 mol
• Total charge due to the flow of electrons= 3,860C

Step 3: Calculate current by using the equation of Q = It; Q = Charge , I = current , t = time of electrolysis

• Required current = 3860 C / (10*60 s)
• Total charge due to the flow of electrons= 6.43 A

A copper sulphate solution (100 cm³) is electrolyzed using graphite electrodes. Initially, CuSO₄ concentration was 1.0 mol dm^-3. If constant 1A current is supplied, what is the concentration of Cu²⁺ ion after 1 minute?

Step 1: Calculate supplied total charge to the CuSO₄ solution by Q = It

• Supplied charge = 1 * (1*60)
• Supplied charge = 60 C

Step 2: Calculate supplied amount of electrons

• Supplied amount of electrons = 60C / 96500 C mol^-1
• Supplied amount of electrons = 0.000621 mol

Step 3: Calculate deposited amount of Copper according to the balanced half equation

• Deposited amount of Copper = 0.000621 mol / 2
• Deposited amount of Copper = 0.000310 mol

Step 4: Calculate initial amount of Cu²⁺ in the solution [n = CV]

• Initial amount of Cu²⁺ in the solution = 1 mol dm^-3 * 100 * 10^-3 dm³
• Initial amount of Cu²⁺ in the solution = 0.1 mol

Step 5: Calculate remaining amount of Cu²⁺ in the solution after 1 minute

• Remaining amount of Cu²⁺ in the solution after 1 minute = Initial amount of Cu²⁺ - Deposited amount of Copper
• Initial amount of Cu²⁺ in the solution = 0.1 mol - 0.000310 mol
• Initial amount of Cu²⁺ in the solution = 0.09969 mol

Step 6: Calculate concentration of Cu²⁺ in the solution after 1 minute [C=n/V]

• Concentration of Cu²⁺ in the solution after 1 minute = 0.09969 mol / (100 * 10^-3 dm³)
• Concentration of Cu²⁺ in the solution after 1 minute = 0.9969 mol dm^-3

Questions

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