More on Electrolysis

Note: This post is mainly for Single Science although it could be good background information for Double Award anyway. :)

SS 1.53 describe simple experiments for the electrolysis, using inert electrodes, of aqueous solutions of sodium chloride, copper (II) sulfate and dilute sulfuric acid and predict the products

So, electrolysis can be used to decompose molten compounds as described in an earlier post on Electrolysis: 

http://askmichellechemistry.blogspot.com/2012/04/electrolysis.html

However, for Single Science, you also need to know about electrolysis of compounds in aqueous solutions. Predicting the reactions and working out the products for aqueous solutions are less straightforward than for molten compounds. 

An aqueous solution of a compound is a mixture of two electrolytes, it's a compound dissolved in water really (so water is the solvent). For example, an aqueous solution of copper (II) sulphate contains two electrolytes: copper (II) sulphate and water. It therefore contains copper (II) sulphate ions (Cu²⁺) and sulphate ions (SO₄^2-), and also small amounts of hydrogen ions (H+) and hydroxide ions (OH^-) from the dissociation of water molecules.

H₂O (l) à H⁺(aq) + OH^-(aq)

These ions compete with the ions from copper (II) sulphate for discharge at the electrodes. 

In general, when an aqueous solution of an ionic compound is electrolysed, a metal or hydrogen is produced at the cathode. At the anode, a non-metal, for example oxygen or a halogen, is given off. 

Let's see if this is the case in the electrolysis of dilute sodium chloride solution. 

Electrolysis of Dilute Sodium Chloride Solution

Note: There is a difference in the products between the electrolysis of dilute sodium chloride solution and concentrated sodium chloride solution. I will elaborate later.

An aqueous solution of sodium chloride contains four different types of ions. They are

• Ions from sodium chloride – Na⁺ (aq) and Cl^- (aq)
• Ions from water – H⁺(aq) and OH^- (aq)

When dilute sodium chloride solution is electrolysed using inert electrodes, the Na⁺ and H⁺ ions are attracted to the cathode. The Cl^- and OH^- ions are attracted to the anode. 

At the cathode: 

The H⁺ and Na⁺ ions are attracted to the platinum cathode. H⁺ ions gains electrons from the cathode to form hydrogen gas. (The hydrogen ions accept electrons more readily than the sodium ions. As a result, H⁺ ions are discharged as hydrogen gas, which bubbles off. I will explain why H⁺ ions are preferentially discharged later.)

2H⁺(aq) + 2e^- à H₂(g)

Na⁺ ions remain in solution. 

At the anode:

OH^- and Cl^- are attracted to the platinum anode. OH^- ions give up electrons to the anode to form water and oxygen gas. 

4OH^-(aq) à 2H₂O(l) + O₂(g) + 4e^-

Cl^- ions remain in solution. 

Summary: 

The overall reaction is: 

  2H₂O(l)   à 2H₂(g) + O₂(g)

Since water is being removed (by decomposition into hydrogen and oxygen), the concentration of sodium chloride solution increases gradually. The overall reaction shows that the electrolysis of dilute sodium chloride solution is equivalent to the electrolysis of water.

Another important thing to note is that twice as much hydrogen is produced as oxygen. This is because for every 4 electrons that flows around the circuit, you would get one molecule of oxygen. But four electrons would produce 2 molecules of hydrogen. Hence in a diagram, you would see the volume of hydrogen produced is twice that of oxygen. Refer to the equations above and note the number of electrons involved to help you understand. 

This diagram is just to illustrate how twice as much hydrogen gas is produced. 

Electrolysis of Concentrated Sodium Chloride Solution

The only difference is that at the anode, Cl^-  ions are more numerous than OH^- ions. Consequently, Cl^- ions are discharged as chlorine gas, which bubbles off. 

2Cl^- (aq) à Cl₂(g) + 2e^-

The OH^- ions remain in solution.   

One volume of hydrogen gas is given off at the cathode and one volume of chlorine gas is produced at the anode. The resulting solution becomes alkaline because there are more OH^- than H⁺ ions left in the solution.    

Comparison:

Compare the electrolysis of molten sodium chloride, dilute sodium chloride solution and concentrated sodium chloride solution:

Molten sodium chloride:

• Cathode: Na⁺ions discharged
• Anode: Cl^-discharged

Dilute NaCl solution: 

• Cathode: H⁺ions discharged
• Anode: OH^-discharged

Concentrated NaCl solution:

• Cathode: H⁺ discharged
• Anode: Cl^-discharged

So you can see that Na⁺ and Cl^- ions are not always discharged even though in all 3 of the above, the electrolytes contained these ions. For example in the electrolysis of dilute NaCl solution, H⁺ are discharged in preference to Na⁺ ions. OH^- ions are discharged in preference to Cl^- ions. Before I talk about the electrolysis of copper(II) sulfate and dilute sulfuric acid, I will discuss why one type of cation (or anion) in the electrolyte is more readily discharged than another type. If you already know this, just scroll right down. :)

Note: Most of the following is taken from a G.C.E. 'O' Level textbook, but I find it useful. :) 

Reactivity Series and Selective Discharge of Ions

In electrolysis, when more than one type of cation or anion is present in a solution, only one cation and one anion are preferentially discharged. This is known as the selective discharge of ions. 

How do you predict which ions are discharged in the electrolysis of a compound in aqueous solution?

If inert electrodes are used during electrolysis, the ions discharged and hence the products formed depend on three factors:

1. The position of the metal (producing the cation) in the reactivity series. 
2. The relative ease of discharge of an anion. 
3. The concentration of the anion in the electrolyte. 

The ease of discharge of cations and anions during electrolysis is shown below.

Cations	NB: Ease of discharge increases as you go down the table	Anions
Potassium ion, K+		Chloride ion, Cl-
Sodium ion, Na+		Bromide ion, Br-
Calcium ion, Ca2+		Iodide ion, I-
Magnesium ion, Mg2+		Hydroxide ion, OH-
Zinc ion, Zn2+		Note: sulphate ions (SO42-) and nitrate ions (NO3-) will not be discharged during electrolysis.
Iron ion, Fe2+
Lead ion, Pb2+
Hydrogen ion, H+
Copper ion, Cu2+
Silver ion, Ag+

Selective discharge of cations during electrolysis

The cations of an element lower in the reactivity series are discharged at the cathode in preference to cations above it in the solution. This is because cations of a less reactive element accept electrons more readily. For example, if a solution containing Na⁺ and H⁺ ions is electrolysed,  H⁺ ions are discharged in preference to Na⁺ ions. The more reactive the metal, the more stable its compound. They have lost a lot of energy and have lost electrons to form stable cations, so cations lower down the reactivity series are more readily discharged.

Selective discharge of anions during electrolysis
Sulphate (SO₄^2-) and nitrate (NO₃^-) ions remain in the solution and are not discharged during electrolysis. If a solution containing SO₄^2-, NO₃^- and hydroxide (OH^-) ions is electrolysed, the OH^- ions will be discharged in preference to SO₄^2- and NO₃^- ions. The OH^- ions give up electrons most readily during electrolysis to form water and oxygen.

4OH^- (aq) à 2H₂O (l) + O₂ (g) + 4e^-  

Effect of concentration on selective discharge of anions

An increase in the concentration of an anion tends to promote its discharge. For example, in the electrolysis of concentrated sodium chloride solution, two types of ions are attracted to the anode: Cl^- and OH^- ions. According to their relative ease of discharge, OH^- ions should be discharged preferentially. However, in concentrated sodium chloride solution,  Cl^- ions are far more numerous than OH^- ions and so are discharged at the anode instead. 

2Cl^- (aq) à Cl₂ (g) + 2e^-

What are the general rules for predicting selective discharge?

The following rules can be applied when predicting the products of electrolysis of any aqueous solution (using inert electrodes):

Rule 1	Identify the cations and anions in the electrolysis. Remember that an aqueous solution also contains H+ and OH- ions from the dissociation of water molecules.
Rule 2	At the anode, the product of electrolysis is always oxygen unless the electrolyte contains a high concentration of the anions, Cl-, Br­- or I- ions.
Rule 3	At the cathode, reactive metals such as sodium and potassium are never produced during electrolysis of the aqueous solution. If the cations come from a metal above hydrogen in the reactivity series, then hydrogen will be liberated (liberate=release). If the cations come from a metal below hydrogen, then the metal itself will be deposited.
Rule 4	Identify the cations and anions that remain in the solution after electrolysis. They form the product remaining in solution. Summarise the reactions. For example, in the electrolysis of dilute sodium chloride solution, Na+ and Cl- ions remain in solution after H+ and OH- ions have been discharged. Hence the solution of sodium chloride becomes more concentrated after electrolysis.

Electrolysis of Copper (II) sulphate solution
Copper (II) sulphate solution can be electrolysed using inert platinum electrodes. (Sometimes inert carbon electrodes in the form of graphite are used.)

During electrolysis, the cathode is coated with a layer of reddish-brown solid copper. The blue colour of the solution fades gradually as more copper is deposited. The resulting electrolyte also becomes increasingly acidic.

An aqueous solution of copper (II) sulphate contains four types of ions:

• Ions from copper (II) sulphate: Cu²⁺ and SO₄^2-
• Ions from water: H⁺ and OH^-

At the anode:
OH^- ions and SO₄^2- ions are attracted to the anode. OH^- ions give up electrons more readily than SO₄^2- ions. Consequently, OH^- ions are preferentially discharged to give oxygen gas.

4OH^- (aq) à 2H₂O (l) + O₂ (g) + 4e^-

The SO₄² ions remain in solution. 

At the cathode:

H⁺ ions and Cu²⁺ ions are attracted to the cathode. Copper is lower than hydrogen in the reactivity series. Cu²⁺ ions accept electrons more readily than H⁺ ions. As a result, Cu²⁺ ions are preferentially discharged as copper metal (atoms). 

Cu²⁺ (aq) + 2e^- à Cu (s)

The H⁺ ions remain in solution. 

Summary: 

When aqueous copper (II) sulphate is electrolysed using platinum electrodes, copper metal is deposited at the cathode and oxygen gas is given off at the anode. The overall reaction is:

2CuSO₄ (aq) + 2H₂O (l) à 2Cu (s) + O₂ (g) + 2H₂SO₄(aq)

Electrolysis of dilute sulfuric acid solution
Inert carbon or platinum electrodes are used.
At the cathode:
In this case, the only positive ions arrive at the cathode are the hydrogen ions from the acid and the water. (Adding acid to water forces it to split up/hydrolyse.) These are discharged to give hydrogen gas.

2H⁺ (aq) + 2e^- à H₂(g)

At the anode:

At the anode, SO₄^2- ions  and OH^- ions (from the water) accumulate. OH^- ions are discharged to give O₂ gas. 

4OH^- (aq) à 2H₂O (l) + O₂ (g) + 4e^-

The amount of hydrogen produced is twice that of oxygen. Just like in the electrolysis of dilute sodium chloride solution.  For every 4 e^- that flows around the circuit, you would get one molecule of O₂ . But four electrons would produce 2 molecules of H₂. 

Electrolysis

Electrolysis

1.47 understand an electric current as a flow of electrons or ions

Current is basically a flow of charged particles and in a metal wire, it is a flow of electrons. These have a negative 1 charge. Ions are charged particles too as they have lost or gained electrons. The movements of these ions are responsible for the conduction of electricity. 

1.48 understand why covalent compounds do not conduct electricity

Covalent compounds do not have spare free electrons that can move and carry the charge; and neither do they contain ions-it’s a covalent compound not an ionic compound.

1.49 understand why ionic compounds conduct electricity only when molten or in solution

When they are a solid the ions are not free to move and carry the charge. When they are molten-as in it is in melted form-the ions are free to move. Remember in liquids the particles are able to slide over each other and move whereas in solids the particles can only vibrate around a fixed position. When the ionic compound is dissolved in a solvent to form a solution the ions are also made free to move.

1.50 describe simple experiments to distinguish between electrolytes and non-electrolytes

http://www.dummies.com/how-to/content/how-to-distinguish-electrolytes-from-nonelectrolyt.html

1.51 recall that electrolysis involves the formation of new substances when ionic compounds conduct electricity

Passing an electric current through a compound which is either molten or in solution causes chemical changes, the chemical reactions produce new products—new substances.

1.52 describe simple experiments for the electrolysis, using inert electrodes, of molten salts such as lead (II) bromide

Right hand electrode: Cathode-attracts cations-is the negative electrode, as it is attracting positive ions.

Left hand electrode: Anode-attracts anions-is the positive electrode, as it is attracting negative ions.

[Opposite charges attract]

Nothing happens until the lead (II) bromide melts.

Lead (II) bromide is an ionic compound. The solid consists of a giant structure of lead (II) ions and bromide ions packed regularly in a crystal lattice. It doesn’t have any mobile electrons, and the ions are locked tightly in the lattice and aren’t free to move. The solid lead (II) bromide doesn’t conduct electricity.

As soon as the solid melts, the ions do become free to move around, and it is this movement that enables the electrons to flow in the external circuit.

Electrodes are made out of carbon-which is inert (unreactive).

As soon as you connect the power source, it pumps any mobile electrons away from the left-hand electrode towards the right-hand one. The excess of electrons in the right-hand electrode makes it negatively charged-called the cathode. The left-hand electrode is positively charged because it is short of electrons. There is a limit to how many electrons can squeeze into the negative electrode (Cathode) because of the repulsion by the electrons already there.

The positive lead (II) ions are attracted to the cathode, which is the negative electrode. When they get there, they gain 2 electrons each from the electrode and forms neutral lead atoms. These fall to the bottom of the container as molten lead.

Pb²⁺ (l) + 2e^-à Pb (l)

This leaves spaces in the cathode that more electrons can move into. The power source pumps new electrons along the wire to fill those spaces.

Bromide ions are attracted to the positive anode. When they get there, the extra electron which makes the bromide ion negatively charged moves onto the anode because this electrode is short of electrons.

The loss of the extra electron turns each bromide ion into a bromine atom. These join in pairs (bond covalently) to form bromine molecules. Overall:

2Br^-(l) à Br₂(g) + 2e^-

The new electrons on the anode are pumped away by the power source to help fill the spaces being created at the cathode.

The ions are discharged at the electrodes. Discharging an ion simply means that it loses its charge-either giving up electron(s) to the electrode or receiving electron(s) from it.

Redox reaction

Look back at the ionic equations, see that the lead (II) ions gain electrons at the cathode. Gain of electrons is reduction. The lead (II) ions are reduced to lead atoms.

The bromide ions lose electrons at the anode. Loss of electrons is oxidation. The bromide ions are oxidized to bromine molecules.

1.54 write ionic half-equations representing the reactions at the electrodes during electrolysis

For the electrolysis of lead (II) bromide, PbBr₂

Cathode:

Pb²⁺ (l) + 2e^-à Pb (l)

Anode:

2Br^-(l) à Br₂(g) + 2e^-