9.6 Electrolysis Of Aqueous Solutions

Question 1

Why do the products of electrolysis of aqueous solution differs from those of a molten salt?

Answer 1

The difference arises because water can be oxidised reduced at the electrodes as well as individual ions. Water can be oxidised to oxygen at the anode or reduced to hydrogen at the cathode.

Question 2

What are the products of electrolysis of an aqueous solution in general?

Answer 2

Anode (positive) - Oxygen or halogen

Cathode (negative) - metal or hydrogen

Question 3

What are the main three factors that affect the products formed at the electrodes?

Answer 3

The standard electrode potentials of the species in the solution
The concentration of the electrolyte
The material from which the electrodes are made

Question 4

In general what happens is a metal has a very negative electrode potential?

Answer 4

Hydrogen will be discharged at the cathode.

Question 5

In general what happens is a metal has a positive electrode potential?

Answer 5

The metal metal will be discharged at the cathode.

Question 6

What is the effect of concentration of the electrolyte (mainly for solutions donating chloride ions)?

Answer 6

At very low chloride concentrations the product of electrolysis is mainly oxygen (oxidation of water) but with more concentrated solutions chlorine in the major product.

Question 7

For the electrolysis of copper sulfate how will the products depend on the electrode if it is made of graphite/Pt?

Answer 7

The anode is positive the cathode is negative, the product at the anode is O2 and the product at the cathode is Cu.
The half equation at the anode is:
2H2O (l) —–> O2 (g) + 4H+ (aq) + 4e-
The half equation at the cathode is:
Cu2+ (aq) + 2e- —–> Cu (s)
Observations during electrolysis at the cathode is coated in copper, bubbles of O2 is given off at the anode, blue colours of the solution fades as the concentration of copper ions decreases, the solution gets more acidic since H+ ions are produced.

Question 8

For the electrolysis of copper sulfate how will the products depend on the electrode if it is made of copper?

Answer 8

The anode is positive the cathode is negative, the anode dissolves and the product at the cathode is Cu.
The half equation at the anode is:
Cu (s) —–> Cu2+ (aq) + 2e-
The half equation at the cathode is:
Cu2+ (aq) + 2e- ——-> Cu (s)
Observations during electrolysis at the anode is that it gets smaller, the cathode is coated with copper and the solution remains the same colour since the concentrations of copper ions does not change.

Question 9

Why is the oxidation of copper at the copper anode more favourable than the oxidation of water?

Answer 9

Because the electrode potential is less negative (more positive).

Question 10

What is electroplating?

Answer 10

The process of coating an object with a thin layer of metal using electrolysis.

Question 11

What should the cathode be?

Answer 11

The object to be coated should be used as the cathode since metal ions are positive.

Question 12

What should the anode be?

Answer 12

The metal with which the object is to be plated with.

Question 13

What should the electrolyte be?

Answer 13

A solution contains the ions of the coating metal.

Question 14

Why does sulphuric acid need to be added to distilled water for its electrolysis?

Answer 14

Since distilled water is a very poor conductor of electricity and so a small amount of sulphuric acid is dully added to increase the concentration of ions present.

Question 15

What ratio of hydrogen to oxygen should be collected from the electrolysis of water?

Answer 15

Twice as much hydrogen as oxygen. When four electrons are lost at the anode one molecules of O2 is formed but when four electrons are gained at the cathode two molecules of H2 are formed therefore two molecules of H2 are formed for every one molecule of O2 formed.

Question 16

What are the half equation for the oxidation and reception of water?

Answer 16

Anode: 2H20 (l) —-> O2 (g) + 4H+ (aq) + 4e-
Cathode: 4H20 (l) + 4e- ——> 2H2 (g) + 4OH- (aq)

Question 17

How is the current related to the amount of product formed at the electrodes?

Answer 17

Proportional to the number of moles of product.

Question 18

How is the time of electrolysis related to the amount of product formed at the electrodes?

Answer 18

Proportional to the number of moles of product.

Question 19

How is the charge on the ion related to the amount of product formed at the electrodes?

Answer 19

Inversely proportional to the number of moles of product.

Question 20

What happens to the amount of product produced when the current to time is doubled?

Answer 20

Twice as much charge will pass and twice as many electrons will be transferred to or from ions therefore twice as many moles of substance will be produced.

Question 21

How does the charge of the ion affect the amount of product formed?

Answer 21

The higher the charge on the ion the more electrons will be required to produce 1 mol of atoms.
Na+ + e- —–> Na
Mg2+ + 2e- —–> Mg
2 moles of electrons produce 2 moles of Na but only 1 mole of Mg.

Question 22

How do you work out the reactive amounts of products formed during electrolysis?

Answer 22

Look at the number of electrons that must flow to produce 1 mole of substance by writing out the half equations. If two cells are connected in series the current is the same in each, the same number of lexctronsx will flow through each in a certain amount of time.