Electrolysis/Electrochemistry

Summary

Introduction

Electrolysis of Molten Compounds

• • Molten/aqueous ionic compounds conduct electricity because ions free to move.

  • In solid state, these ions are held in fixed position within the crystal lattice.

  • Hence solid ionic compounds do not conduct electricity.

  • When molten binary compound is electrolysed, metal is formed on cathode while non-metal is formed on anode.

Example

Electrolysis of molten PbBr₂

To make molten lead(II) bromide, PbBr₂, we strongly heat the solid until it melts. To electrolyse it, pass current through the molten PbBr₂.

What happens:

Ions present: Pb²⁺ and Br^-

Reaction at Anode

Br- loses electrons at anode to become Br atoms. Br atoms created form bond together to make Br2 gas.

2Br^-(aq) --> Br₂(g)+ 2e^-

Reaction at Cathode

Pb²⁺ gains electrons at cathode to become Pb atoms becoming liquid lead (II).

Pb²⁺(aq) + 2e^- --> Pb(l)

Overall equation

PbBr₂(l) --> Pb(l) + Br₂(g)

Electrolysis of Aqueous Solution

• • Aqueous solutions contain additional H⁺ and OH^- ions of water, totaling 4 ions in the solution :

    • 2 from electrolyte, 2 from water.

    • Only 2 of these are discharged.

  • Electrolysis of aqueous solutions use the theory of selective discharge.

At cathode

• • In CONCENTRATED solutions of nickel/lead compound, nickel/lead will be discharged instead of hydrogen ions of water which is less reactive than nickel/lead.

  • In VERY DILUTE solutions, hydrogen, copper and silver ions are preferable to be discharged, according to its ease to be discharged.

  • Reactive ions (potassium, sodium, calcium, magnesium, aluminium) will NEVER BE DISCHARGED in either concentrated or dilute condition. Instead, hydrogen ions from water will be discharged at cathode.

At anode

• • In CONCENTRATED solutions, iodine/chlorine/bromine ions are preferable to be discharged, although it’s harder to discharged compared to hydroxide ions.

  • In VERY DILUTE solutions containing iodide/chloride/bromide ions, hydroxide ions of water will be discharged instead of iodide/chloride/bromide, according to ease of discharge.

  • Sulphate and nitrate are NEVER DISCHARGED in concentrated/dilute solutions.

Examples

A. Concentration Solutions

Electrolysis of Concentrated NaCl

What happens:

Ions Present: Na⁺, H⁺, OH^- and Cl^-

Reaction at Anode

• • Cl^- loses electrons at anode to become Cl atoms, although OH^- is easier to discharge.

  • Cl atoms created form covalent bond together to make Cl₂ gas.

  • 2Cl^- (aq) --> Cl₂ (g) + 2e^-

Reaction at Cathode

• • H⁺ gains electrons at cathode to become H atoms becoming hydrogen gas

  • 2H⁺ (aq) + 2e^- --> H₂ (l)

Overall Equation

2HCl(l) --> H₂(l) + Cl₂(g)

Note: any cation and anion left undischarged in solution forms new bonds between them.

E.g. in above, leftovers Na⁺ and OH^- combine to form NaOH.

B. Very Dilute Solutions

Electrolysis of Dilute H2SO4

What happens:

Ions Present: H⁺, OH^- and SO₄^2-

Reaction at Anode

• • OH- loses electrons at anode to become O₂ and H₂O.

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

Reaction at Cathode

• • H⁺ gains electrons at cathode to become H atoms becoming hydrogen gas.

  • 2H⁺(aq) + 2e^- --> H₂ (g)

Overall Equation

• • Both equations must be balanced first.

  • The cathode equation is short 2 electrons. Hence, we should first even them by multiplying cathode equation by 2.

    • (2H⁺(aq) + 2e^- --> H₂ (g)) x 2 = 4H ⁺(aq) + 4e^- --> 2H₂ (g)

  • Now we can combine the equations, forming:

    • 4H⁺ (aq) + 4OH⁺ (aq) --> 2H₂ (g) + O₂ (g) + 2H₂O (l)

  • 4H⁺ and 4OH⁺ ions, however, combine to form 4H₂O molecules.

    • Hence: 4H₂O (l) --> 2H₂ (g) + O₂ (g)+ 2H₂O (l)

  • H₂O molecules are formed on both sides.

    • Therefore, they cancel the coefficients: 2H₂O (l) --> 2H₂ (g) + O₂ (g)

  • Since only water is electrolysed, the sulfuric acid now only becomes concentrated.

Electrolysis using different types of electrodes

• • Inert Electrodes are electrodes which do not react with electrolyte or products during electrolysis.

    • Eg. platinum and graphite.

  • Active Electrodes are electrodes which react with products of electrolysis, affecting the course of electrolysis.

    • Eg. copper.

A. Electrolysis of CuSO₄ Using Inert Electrodes (e.g. carbon)

What happens:

Ions Present: Cu²⁺, H⁺, OH^- and SO₄^2-

Reaction at Anode

• • OH^- loses electrons at anode to become O₂ and H₂O.

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

Reaction at Cathode

• • Cu²⁺ gains electrons at cathode to become Cu atoms becoming liquid copper.

  • Hydrogen ions are not discharged because copper is easier to discharge.

  • Cu²⁺ (aq) + 2e^- --> Cu (s)

Overall Equation

• • Both equations must be balanced first.

  • The cathode equation is short 2 electrons. Hence, we should first even them by multiplying cathode equation by 2.

    • (Cu²⁺ (aq) + 2e^- --> Cu (s)) x 2 = 2Cu²⁺ (aq) + 4e^- --> 2Cu (s)

  • Now we can combine the equations, forming:

    • 2Cu(OH)₂ (aq) --> 2Cu (s) + O₂ (g) + 2H₂O (l)

  • Since copper ions in solution are used up, the blue colour fades.

  • Hydrogen and sulphate ions left forms sulphuric acid.

B. Electrolysis of CuSO₄ Using Active Electrodes (e.g. copper)

Ions Present: Cu²⁺, H⁺, OH^- and SO₄^2-

Reaction at Anode

• • Both SO₄^2- and OH^- gets attracted here but not discharged. Instead, the copper anode discharged by losing electrons to form Cu²⁺. So, the electrode size decreases.

  • Cu (s) --> Cu²⁺ (aq) + 2e^-

Reaction at Cathode

• • Cu²⁺ produced from anode gains electrons at cathode to become Cu atoms becoming copper. Hence, the copper is deposited here and the electrode grows.

  • Cu²⁺ (aq) + 2e^- --> Cu (s)

Overall Change

• • There is no change in solution contents as for every lost of Cu²⁺ ions at cathode is replaced by Cu²⁺ ions released by dissolving anode.

  • Only the cathode increases size by gaining copper and anode decreases size by losing copper.

  • We can use this method to create pure copper on cathode by using pure copper on cathode and impure copper on anode.

  • Impurities of anode falls under it.

Electroplating

• • Electroplating is coating an object with thin layer of metal by electrolysis. This makes the object protected and more attractive.

  • Object to be plated is made to be cathode and the plating metal is made as anode.

  • The electrolyte MUST contain plating metal cation.

Plating Iron object with Nickel

Reaction at Anode

• • Ni²⁺ discharged from anode into solution. So, the electrode size decreases.

  • Ni (s) --> Ni²⁺ (aq) + 2e^-

Reaction at Cathode

• • Ni²⁺ produced from anode gains electrons at cathode to become Ni atoms becoming nickel. Hence, the nickel is deposited here and the electrode grows.

  • Ni²⁺ (aq) + 2e^- --> Ni (s)

Overall Change

• • There is no change in solution contents while iron object receives nickel deposit.

  • Uses of Electroplating

Creation of Electric Cells by Electrolysis

• • A Simple cell or an Electric cell is a device that converts chemical energy into electrical energy, and it consists of 2 electrodes made of 2 metals of different reactivity.

• In a simple cell, the MORE REACTIVE metal/electrode is ALWAYS designated the NEGATIVE electrode

  • The anode (negative electrode) is made of more reactive metal. This is because they have more tendency of losing electrons.

  • The cathode (positive electrode) is made of less reactive metal.

  • The further apart the metals in the reactivity series, the higher the voltage created.

• The electrons in a simple cell will ALWAYS flow from the NEGATIVE electrode (made of the MORE reactive metal) to the POSITIVE electrode.

Eg. A simple electric cell using zinc and copper

Observation: bubbles of hydrogen gas appear at the copper rod.

Explanation: Zinc is more reactive than copper. Thus, it is more electropositive than copper, meaning that zinc loses electrons more easily than copper. As a result, oxidation occurs at the zinc rod (the anode) and zinc metal loses electrons to become zinc ions, that is, Zn (s) - 2e^- --> Zn²⁺(aq)

The electrons then flow from the zinc rod to the copper rod through the external circuit. At the copper rod, reduction occurs - the hydrogen ions in solution accept these electrons to form hydrogen gas;

2H⁺(aq) + 2e^- --> H₂ (g)

This explains why bubbles of gas are produced at the copper rod when the two rods are connected by a wire.

The magnitude of the voltage (potential difference) is related to the positions of the two metals in the reactivity series. The further apart the two metals, the larger will be the potential difference (voltage) produced.

Electrolytic cell vs Electrochemical cell

Factors affecting electrolysis

• • Concentration

  • Type of electrode

Concentration

• • If the concentration of a particular ion is high, then this can alter the preferential discharge

  • If dilute hydrochloric acid is electrolysed, hydrogen gas is given off at the cathode and oxygen gas at the anode. However, when concentrated hydrochloric acid is electrolysed, hydrogen gas is still given off at the cathode, but chlorine gas is given off at the anode.

  • This is because although the chloride ion is harder to discharge than the hydroxide ion, its high concentration makes it more likely to be discharged.

Type of electrode

Eg. electrolysis of aqueous copper(II) sulphate solution

Use Carbon Electrodes:

• • Carbon electrodes are inert and so do not affect the electrolysis

  • At the anode, we have a choice of sulphate or hydroxide ions, and hydroxide ions are easier to discharge so oxygen gas is given at the anode

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

  • At the cathode, we have a choice of copper or hydrogen ions. Copper ions are easier to discharge so we will see a pink deposit of copper metal on the carbon electrode

  • Cu²⁺ (aq) + 2e^- ---> Cu (s)

Use Copper Electrodes

• • Copper electrodes are active and so will affect electrolysis

  • At the anode, the copper electrode dissolves into solution

  • Cu (s) ---> Cu²⁺ (aq) + 2e^-

  • At the cathode, the copper ions are deposited as pink copper metal

  • Cu²⁺ (aq) + 2e^- ---> Cu (s)

MCQ Questions

1. Electricity can pass through molten lead(II) bromide because of the presence of

a. free electrons

b. moveable ions

c. moveable atoms

d. lead metal

2. When a dilute salt water is electrolysed, a colorless gas is given off at the anode. The gas is

a. hydrogen

b. steam

c. oxygen

d. chlorine

3. A solution of copper(II) sulphate is electrolysed, using carbon electrodes. The pinkish deposit which forms on one of the electrodes is

a. copper

b. copper(I) oxide

c. copper(II) oxide

d. copper(III) sulphide

4. A solution of copper(II) sulphate is electrolysed, using copper electrodes. Which of the following would happen?

a. the anode loses weight

b. the cathode loses weight

c. the solution darkens in color

d. the solution lightens in color

5. An electrolyte is always

a. an acid or alkali

b. an aqueous solution

c. a liquid

d. a molten solid

6. Anions are formed by

a. metals gaining electrons

b. metals losing electrons

c. non-metals gaining electrons

d. non-metals losing electrons

7. Which of these anions is never discharged at the positive electrode during electrolysis?

a. NO₃^-

b. OH^-

c. I^-

d. O^2-

8. In the electrolytic manufacture of aluminium, what is the anode made of?

a. copper

b. graphite

c. platinum

d. steel

9. In which electrolyte would a carbon cathode increase in mass during electrolysis?

a. aqueous copper(II) sulphate

b. concentrated hydrochloric acid

c. concentrated aqueous sodium chloride

d. dilute sulphuric acid

10. Chlorine is manufactured commercially by the electrolysis of aqueous sodium chloride (brine). Which other important products are made in the process?

a. hydrochloric acid and hydrogen

b. hydrogen and sodium

c. hydrogen and sodium hydroxide

d. sodium and sodium hydroxide

11. An electric current is passed through aqueous potassium sulphate, K₂SO₄.

Which electrode reactions will occur on closing the switch?

Anode reaction Cathode reaction

a. Copper dissolves preferentially. Copper is deposited.

b. Copper dissolves preferentially. Hydrogen is evolved.

c. Zinc dissolves preferentially. Hydrogen is evolved.

d. Zinc and copper both dissolve. Copper is deposited.

23. During the electrolysis of concentrated sodium chloride in a cell, chlorine, hydrogen, and sodium hydroxide are produced. What is the molar ratio of these products?

Chlorine Hydrogen Sodium hydroxide

a. 1 1 1

b. 2 1 2

c. 2 1 1

d. 2 2 1

24. Chlorine is manufactured commercially by the electrolysis of aqueous sodium chloride. Which other important products are made in this process?

a. hydrochloric acid and hydrogen

b. hydrogen and sodium

c. hydrogen and sodium hydroxide

d. sodium and sodium hydroxide

25. Why is cryolite used in the extraction of aluminium from aluminium oxide?

a. to dissolve aluminium oxide

b. to prevent the anodes from burning away

c. to prevent oxidation of the aluminium

d. to remove impurities from the aluminium oxide

Answers

1. b

2. c

3. a

4. a

5. c

6. c

7. a

8. b

9. a

10. c

11. a (H⁺ and K⁺ ions in the electrolyte migrate to the cathode. H⁺ are preferentially discharged to form hydrogen gas because it is lower down in the electrochemical series than K⁺ ions)

12. b (the negative chloride ions will migrate to the anode and become oxidised at the anode to form chlorine gas)

13. b (the ions attracted to the cathode are H⁺ and Na⁺ ions. H⁺ is preferentially discharged to form hydrogen gas)

14. b

15. c

16. b

17. a

18. c

19. b

20. d

21. d

22. c

23. b

24. c (chlorine gas is produced at the anode. The sodium ions are discharged at the mercury cathose to produce sodium which reacts with the mercury to form sodum amalgam. The sodium amalgam eventually passes into water to produce sodium hydroxide and hydrogen gas)

25. a (cryolite acts as an impurity to lower the melting point of aluminium oxide and also dissolves the aluminium oxide to form a molten electrolyte)

Structured Question Worked Solutions

1. Dilute sulphuric acid will conduct an electric current.

a. Give the formulae of all of the ions present in dilute sulphuric acid

b. Name the gaseous products which you would expect to be formed during the electrolysis of aqueous potassium sulphate using inert electrodes

at the anode:_____

at the cathode:______

c. Name a metal which is used to electroplate

i. bicycle handlebars

ii. teaspoon

d. Explain why a metal such as aluminium can conduct an electric current but a non-metal such as sulphur cannot conduct a current

Solution

a. H⁺, OH^-, SO₄^2-

b. cathode: Hydrogen

anode: Oxygen

ci. chromium

cii. silver

d. Aluminium consists of positively charged particles in a sea of electrons. The electrons are able to move freely and thus electricity can flow. In sulphur, the atomic arrangement is fixed, so there is no movement of electrons. When electricity is passed through sulphur, electricity will not be able to flow.

2a. When concentrated aqueous sodium chloride is electrolysed using graphite electrodes, hydrogen is collected at the cathode and chlorine at the anode.

When concentrated aqueous sodium chloride is electrolysed using iron electrodes, hydrogen is again collected at the cathode but much less chlorine is collected at the anode.

i. Give the equations for the electrode reactions by which hydrogen and chlorine are formed

ii. Explain why much less chlorine is collected when iron electrodes are used.

iii. Name the product, other than hydrogen and chlorine, which is manufactured by the electrolysis of concentrated aqueous sodium chloride. Give a major use of this product

b. Why is the electrolysis of concentrated hydrochloric acid not used for the manufacture of chlorine?

Answers

ai. 2H⁺ (aq) + 2e^- -----> H₂ (g)

2Cl^- (aq) -----> Cl₂ (g) + 2e^-

ii. When iron anode is used, some oxygen gas is produced at the same time. Some of the electrical energy is used to liberate oxygen. So less Cl₂ is produced.

iii. Sodium hydroxide. It is used to manufacture soap

b. because concentrated HCl is not a cheaply and readily available raw material. It is also a volatile acid. A lot of HCl gas will be emitted.

3.

a. Write ionic equations, with state symbols, for the reactions which take place at the anode in each cell.

b. Describe one change that you would see happen in both cells.

c. Describe one change that you would see happen in Cell 1 but not in Cell 2.

Solution

a. Anode reaction cell 1: 4OH^- (aq) --> O₂ (g) + 2H₂O (l) + 4e^-

Anode reaction cell 2: Cu (s) --> Cu²⁺ (aq) + 2e^-

b. The size of the cathode increases as copper metal is plated onto the cathode in both cells.

Cu²⁺ (aq) + 2e^- --> Cu (s)

c. The blue colour of the electrolyte in cell 1 fades when more and more Cu2+ ions are reduced to copper metal and plated onto the cathode as a pink deposit.

Cu²⁺ + 2e^- --> Cu

6a. Write an ionic equation for the reaction between zinc and aqueous copper(II) sulphate.

This reaction can be used to generate electricity in a cell.

i. Suggest the names of the two salts, A and B.

ii. Describe tests to confirm the identities of the three gases collected.

Solution

a.

b. reaction at anode: Ag (s) --> Ag⁺ (aq) + e^-

ci. Salt A: sodium sulphate

Salt B: sodium chloride

cii. Collect samples of each gas using test tubes. To test for oxygen, insert a glowing splint into the test tube of gas. The gas that relights the glowing splint is oxygen. To the remaining samples, place a lighted splint at the mouth of each test tube. The gas that extinguishes the lighted splint with a "pop" sound is hydrogen. To identity chlorine, place a piece of moist blue litmus at the mouth of test tube of gas. The litmus turns red and bleaches.

8. One important use of a gas Y is to sterilize swimming pool water. The gas is prepared in the laboratory by the electrolysis of a solution. A student tried to prepare gas Y by the electrolysis of a very dilute sodium chloride solution as shown below. Contrary to the student's expectation, a colorless gas, instead of gas X, was liberated at the anode.

a. What is the colorless gas liberated?

b. Suggest a chemical test for the colorless gas

c. The experiment was then modified to prepare gas Y.

i. Suggest how the experiment could be modified. Explain your answer.

ii. Suggest the solution left after the electrolysis

iii. Suggest one common use of the solution left.

Solution

8a. oxygen

8b. The gas relights a glowing splint.

8ci. Use concentrated sodium chloride solution instead of a very dilute sodium chloride solution. Concentration of chloride ions in the solution would be much greater than that of hydroxide ions. Therefore, chloride ions would be preferentially discharged to form chlorine gas.

8ciii. manufacture of bleach

9. A dilute copper(II) sulphate solution is electrolysed using carbon electrodes.

a. Describe and explain what would happen at the two carbon electrodes

b. Write half-equations for the reactions at carbon exlectrodes X and Y.

c. What will be the charge in the electrolyte as electrolysis proceeds for some time? Explain your answer.

d. Explain what could happen to the copper(II) sulphate solution if copper electrodes are used in the above experiment.

Solution

9a.

At carbon electrode X:

The sulphate ions and hydroxide ions migrate to electrode X. A hydroxide ion is a stronger reducing agent than a sulphate ion. So hydroxide ions are preferentially discharged.

At carbon electrode Y:

The copper(II) ions and hydrogen ions migrate to electrode Y. A copper(II) ion is a stronger oxidizing agent than a hydrogen ion. So copper(II) ions are preferentially discharged to form a deposit of copper on electrode Y.

9b.

electrode X:

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

electrode Y:

Cu²⁺ (aq) + 2e^- --> Cu (s)

9c. The solution becomes sulphuric acid because copper(II) ions and hydroxide ions are consumed in the electrolysis. Hydrogen ions and sulphate ions remain in the solution.

9d. The net effect is the transfer of copper from electrode X to electrode Y. The dilute copper(II) sulphate solution remains the same.

10. Tuning knobs on radios are often made of plastics plated with metal coatings. The plastic knobs are first coated with copper and then electroplated with nickel. The electroplating can be conducted using the following setup.

a. Explain the term 'electroplating'

b. Why is the plastic knob first coated with copper before electroplating?

c. Explain why nickel(II) sulphate solution can conduct electricity

d. Which is the anode, the nickel electrode or the copper-coated knob?

e. Write an ionic half-equation for the reaction at the copper-coated knob

f. Explain why it is better to use a nickel electrode than a carbon electrode in the above process

g. In a nickel-plating factory, the waste water is treated with sodium hydroxide solution to remove nickel(II) ions before discharge. Suggest 2 reasons why it is necessary to remove nickel(II) ions from the waste water before discharge

Solution

10a. It is the coating of an object with a thin layer of a metal by electrolysis

10b. to make the knob conduct electricity

10c. the solution contains mobile ions

10d. the nickel electrode

10e. Ni2+ (aq) + 2e- --> Ni (s)

10f. the concentration of nickel(II) ions in the electrolyte can be maintained

10g.

- to recover the nickel metal

- nickel(II) ions are harmful to marine lives. Humans may get poisoned by eating contaminated seafood

11. The following circuit is set up. Electrodes A and B are made of carbon while electrode C and D are made of copper.

a. What are the functions of the ammeter and rheostat respectively?

b. Explain why no current flows when potassium iodide is in solid state, but a current flows when water is added.

ci. What would be observed at electrode A and B respectively?

cii. Write ionic half-equations for the reactions at electrodes A and B

di. What would be observed at electrodes C and D respectively?

dii. Write ionic half-equations for the reactions at electrodes C and D

diii. Would you expect any color change in the dilute copper(II) sulphate solution during the process? Explain your answer.

Solution

11a. The ammeter is an instrument used to measure the electric current passing through the circuit. The rheostat is used to vary the resistance in the circuit and regulate the current.

11b. In solid state, the ions in potassium iodide are held together by strong attraction. They are not free to move. So solid potassium iodide does not conduct electricity. When water is added to the compound, the compound dissolves in the water and the ions become mobile and a current can then flow through the solution.

11ci. A brown color develops around electrode A. A colorless gas is given off from electrode B.

11cii.

electrode A:

2I^- (aq) --> I₂ (aq) + 2e^-

electrode B:

2H⁺ (aq) + 2e^- --> H₂ (g)

11di. Electrode C dissolves. A reddish brown deposit forms on electrode D.

11dii.

electrode C:

Cu (s) --> Cu²⁺ (aq) + 2e^-

electrode D:

Cu²⁺ (aq) +2e^- --> Cu (s)

11diii. The blue color of the dilute copper(II) sulphate solution does not change because the concentration of copper(II) ions in the solution remains the same.

12. A student used the following set-up for passing electricity through some solutions.

a. Write a half equation for the reaction that occurs in the acidified potassium permanganate solution. Explain whether the permanganate ion is oxidized or reduced.

b. What would be observed in the iron(II) sulphate solution after some time. Write a half equation for the reaction that would occur.

c. Identify the direction of electron flow in the external circuit.

d. Write an ionic equation for the reaction that occurs when acidified potassium permanganate solution and iron(II) sulphate solution are mixed together.

ei. Give the function of the salt bridge set up

eii. Explain whether a sodium sulphite solution can be used instead of a potassium nitrate solution in the salt bridge.

As an alternative to iron(II) sulphate and acidified potassium permanganate solution, potassium iodide solution and iron(III) sulphate solution may be used on the left hand side and the right hand side respectively.

fi. What would be observed in the potassium iodide solution after some time? Write a half equation for the reaction that would occur.

fii. What would be observed in the iron(III) sulphate solution after some time? Write a half equation for the reaction that would occur.

Solution

13a. MnO₄^- (aq) + 8H⁺ (aq) + 5e^- --> Mn²⁺ (aq) + 4H₂O (l)

MnO₄- is reduced because it receives electrons and the oxidation number of Mn changes from +7 to +2.

13b. The solution changes from green to yellow gradually because iron(II) ions are oxidised to iron(III) ions

Fe²⁺ (aq) --> Fe³⁺ (aq) + e^-

13c. From iron(II) sulphate solution to potassium permanganate solution

13d. 5Fe²⁺ (aq) + 8H⁺ (aq) + MnO₄^- (aq) --> Mn²⁺ (aq) + 5Fe³⁺ (aq) + 4H₂O (l)

13ei. to allow migration of ions between the two beakers or to complete the circuit

13eii. No. sodium sulphite reacts with potassium permanganate or the sulphite ions can be oxidised by permanganate ions

13fi. the solution turns brown or yellow

2I^- (aq) --> I₂ (aq) + 2e^-

13fii. the solution changes from yellow to green

Fe³⁺ (aq) + e- --> Fe²⁺ (aq)

13fiii. 2I^- (aq) + 2Fe³⁺ (aq) --> I₂ (aq) + 2Fe²⁺ (aq)

14. The diagram shows a dry cell

ai. which substance is the positive electrode?

aii. write an ionic half equation for the reaction at the positive electrode

bi. which substance is the negative electrode?

bii. write an ionic half equation for the reaction at the negative electrode

c. explain the function of manganese(IV) oxide

d. the voltage of the cell drops if current is drawn from the cell rapidly. Explain briefly.

e. explain why the zinc case of a used cell is thinner than that of a new cell

f. explain why disposal of 'flat' cells present a pollution problem

g. explain why zinc-carbon cells should be removed from electric appliances when not in use for a long period

Solution

14ai. carbon rod

14aii. 2NH₄⁺ (aq) + 2e^- --> 2NH₃ (aq) + H₂ (g)

14bi. zinc case

14bii. Zn (s) --> Zn²⁺ (aq) + 2e^-

14c. Hydrogen is produced and collected on the surface of the positive electrode. Since hydrogen is a poor conductor of electricity, the accumulation of hydrogen at the positive electrode may hinder further reactions and decrease the current of the cell. Manganese(IV) oxide, an oxidizing agent, is used to remove the hydrogen.

14d. If a current is drawn from the cell rapidly, the gaseous product cannot be removed fast enough. The voltage drops as a result.

14e. The zinc case undergoes oxidation to give zinc ions in the cell reaction.

14f. The materials inside the cells do not decompose even after a long time. These materials may combine with other compounds and form harmful substances which pollute the environment.

14g. There is a slow direct reaction between the zinc electrons and ammonium ions. After some time, the zinc case becomes too thin and the paste leaks out. This may cause damage to electric appliances.

15. Aqueous copper(II) sulphate can be electrolysed using either carbon electrodes or copper electrodes.

a. Give the formulae of the ions present in copper(II) sulphate.

b. When aqueous copper(II) sulphate is electrolysed using carbon electrodes, a colourless gas is formed at the anode (positive electrode).

i. What is the name of this gas?

ii. Describe a test for this gas.

c. Aqueous copper(II) sulphate is electrolysed using copper electrodes.

i. Explain why the anode decreases in mass.

ii. Explain why the cathode increases in mass.

iii. Write an equation for the reaction at the cathode.

Solution

15a. Cu²⁺ and SO₄^2-

15bi. oxygen

15bii. Place a glowing splint into a test tube of the gas. The glowing splint relights.

15ci. The anode dissolves to form Cu²⁺

15cii. Cu²⁺ migrates to the cathode and is reduced and deposited onto the copper cathode, so the cathode increases in mass.

15ciii. Cu²⁺ (aq) + 2e^- --> Cu (s)