13 — electrochemistry

Question 1

Electrolysis

Answer 1

Electrolysis is the process of using electricity to break down or decompose a compound (usually an ionic compound in the molten or aqueous state）

Question 2

Conditions for electrolysis to occur

Answer 2

1. A power supply or a power source
   - e- moves from the negative terminal (cathode) to positive terminal (anode)
2. Electrodes
   - contain delocalised mobile electrons to conduct electricity
   - anode is the positive electrode connected to the positive terminal of the power source
   - cathode is the negative electrode connected to the negative terminal of the power source
   - RED CAT AN OX
3. Electrolyte
   - conducts electricity due to mobile ions
   - aq or molten
   - if solid: ions held in fixed positions, immobile, cannot conduct electricity

Question 3

Inert vs reactive electrodes

Answer 3

IE:
- electrodes that do not undergo chemical changes n do not tk part in the electrolysis reaction
- eg graphite, platinum

RE:
-metal anodes undergo oxidation during electrolysis
- eg copper, silver

Question 4

Molten binary ionic compound

Answer 4

A molten binary ionic compound is typically a salt containing only one cation and one anion in the liquid state

Question 5

Electrolysis of molten NaCl

Answer 5

Sodium ions gain electrons and is reduced to form sodium atom. Grey globules (Liquid Metal) of sodium obtained at cathode.

Chloride ions loses electrons and is oxidised to form chlorine molecules. Yellow green chlorine gas obtained at anode.

(Bromide: dark reddish-brown gas)

Question 6

Graphite electrode

Answer 6

Advantages
- high melting point
- will not melt when used in the electrolysis of molten binary ionic compounds

Disadvantages
- graphite will react w oxygen gas under high temperatures to produce CO2
- graphite anodes might hv to be periodically replaced

Question 7

Platinum

Answer 7

Advantages
- does not tk part in the electrolysis reaction

Disadvantages
- lower melting point than graphite
- might melt when used in the electrolysis of molten binary ionic compounds
- mainly used in the electrolysis of aqueous electrolytes

Question 8

Equation for discharge of OH- ions

Answer 8

4 OH- (Aq) -> O2 (G) + 2H2O (l) + 4e-

Question 9

Equation for discharge of H+ ions

Answer 9

2H+ (aq) + 2e- ->H2 (g)

Question 10

Electrolysis of concentrated aq NaCl

Answer 10

The ratio of H2 to Cl2 produced is 1:1
The solution becomes alkaline as there is a net discharge of H+ ions. (Concentrated hence Cl- gets discharged at the anode) The remaining Na+ and OH- ions form NaOH, an alkaline solution. When a few drops of Universal Indicator is added to the electrolyte, the UI changes from green to violet/purple.

Question 11

Electrolysis of CuSO4 using inert electrodes

Answer 11

After Cu^2+ and OH- ions r discharged, H+ (aq) and SO4 ^2- (aq). Ions remains in the solution. Hence, the resulting electrolyte becomes increasingly acidic as the concentration of H+ is greater than the concentration of OH-.
Concentration of Cu^2+ ions decreases thus the blue colour of the electrolyte gradually fades and eventually turns colourless.

Question 12

Electroplating

Answer 12

Electroplating allows us to coat a thin layer of metal onto an object

Cathode is coated w a layer of copper metal and increases in mass
Anode dissolves to form Cu^2+ ions and decrease in mass

Question 13

State and explain if the concentration of the electrolyte changes in electroplating involving reactive electrodes.

Answer 13

The concentration of the electrolyte remains the same as 1 mol of Ag oxidises to form 1 mol of Ag+ ions at the anode and 1 mol of Ag+ ions reduce to form 1 mol of Ag at the cathode. Thus, there is no net change in concentration of Ag+ ions.

Question 14

Simple cells

Answer 14

A simple cell is a device that converts chemical energy into electrical energy

Question 15

Metals in a simple cell

Answer 15

More reactive metal:
- acts as anode
- oxidises forming cations that enter the electrolyte
- releases electrons that flow thru external circuit

Less reactive metal:
- acts as cathode
- causes cations from the electrolyte to gain e- and be reduced

The further apart the 2 metals r in the reactivity series, the greater the voltage produced

Question 16

Simple cell vs electrolytic cell

Answer 16

Source:
SC: electrical energy produced thru chemical reactions
EC:electrical energy supplied by an external source

Electron movement:
SC: electrons move from the anode to the cathode
EC: electrons move from the battery to the cathode, thru the electrolyte and into anode

Polarity
SC:
Anode — negative
Cathode: positive

EC:
Anode — positive
Cathode — negative

Question 17

Measuring the potential difference

Answer 17

Voltmeter is used
SI: V

Question 18

Hydrogen fuel cells

Answer 18

A fuel (hydrogen) is continuously added at the anode
An oxidiser (oxygen) is continuously added at the cathode

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

Anode:
H2 (g) + 2OH- (aq) -> 2H2O (l) + 2e-

Overall equation: 2H2(g) + O2(g) -> 2H2O (l)

Question 19

Advantage of using hydrogen as fuel

Answer 19

Hydrogen is a renewable fuel and can be obtained via the electrolysis of water or from the cracking of hydrocarbon

Hydrogen fuel cells produce only water as a by-product

It is moe efficient than fuel-burning electricity sources. A larger percentage of the chemical energy stored in the fuel ends up as useful electricity in a fuel cell

Question 20

Disadvantages of using hydrogen as a fuel

Answer 20

It is difficult to store and transport hydrogen safely as hydrogen is a highly flammable gas at room temperature and pressure. Hydrogen is ofte transported in high-pressure cylinders which can be dangerous to handle.

Large amount of energy is needed to produce hydrogen from electrolysis. However, this can be mitigated by using renewable sources like solar and wind energy to power electrolysis

Question 21

Describe a set-up that can electroplate a layer of tin onto an iron can. Include reference to electrodes and an electrolyte in your answer.

Answer 21

1. Connect the tin metal to the positive terminal of the cell to make it an anode.
2. Connect the iron can to the negative terminal of the cell to make it a cathode.
3. Immerse both electrodes in the electrolyte which is tin (II) nitrate solution.

Question 22

Explain why different combinations of metals produce different voltages.

Answer 22

When 2 diff metals w diff reactivities r connected in a simple cell, diff voltages r produced. The further apart the 2 metals r in the reactivity series, the greater the voltage produced.

Question 23

Electrolysis of aq NaCl forms H2 and O2 gas. Why is the theoretical ratio of H2 to O2 2:1?

Answer 23

The overall reaction is
4H2O (l) -> 2H2 (g) + O2 (g)

4 moles of H2O is broken down into 2 moles of H2 and 1 mole of O2.
Since mole ratio is equal to volume ratio for gases, volume of H2:O2 theoretically evolved is 2:1.

Question 24

Explain how and why the solubility of oxygen affects the ratio of hydrogen to oxygen that is collected in electrolysis of aq NaCl.

Answer 24

As O2 is more soluble in water than H2, less O2 will be collected as it dissolves in water. Thus, the ratio of H2 to O2 is 2:1.

Question 25

The difference from the expected ratio of H2:O2 collected in electrolysis of aq NaCl is greater initially but less noticeable after the electrolysis has been running for some time. Suggest why this happens.

Answer 25

As the electrolysis runs, lesser oxygen is dissolved in water as the solution becomes saturated w oxygen. Hence, more O2 is collected.

Question 26

What happens to the concentration of NaCl during the electrolysis? Explain your reasoning.

Answer 26

The concentration of NaCl increases over time. Water is electrolysed to form hydrogen and oxygen gases, and total volume of solution decreases. Concentration of NaCl increases since volume of NaCl remains constant.

Question 27

Give 1 similarity and 1 difference between the products of the electrolysis of dilute and concentrated aq NaCl.

Answer 27

Similarity:
In both setups, H2 gas will be collected at the cathode.
Difference:
In the electrolysis of dilute NaCl, O2 gas will be produced at the anode but chlorine gas will be produced at the anode in the electrolysis of conc NaCl.

Question 28

Platinum metal electrodes r used. Why is platinum a suitable material for use as an electrode?

Answer 28

Platinum is an inert electrical conductor and will not participate in the reaction.

Question 29

Suggest why the mass of the electrode does not change after some time of running the electrolysis.

Answer 29

All copper (II) ions in the electrolyte have been deposited.

Question 30

Difference between electrolysis of aq CuSO4 using inert electrode and copper (reactive) electrodes.

Answer 30

Mass of the cathode (negative electrode) will continue to increase after 60 minutes w copper electrodes. At the same time, mass of the anode (+ve electrode) will decrease as copper anode is oxidised to form Cu2+ ions.

The concentration of Cu2+ ions is constant bcos the amt of Cu2+ ions (from anode) oxidised to the electrolyte is equal to the amt of Cu2+ ions deposited on the cathode. Hence the mass of negative electrode increases for a longer period.

Question 31

After some time, the student observes that no more silver is deposited on the object in experiment 1 (anode: carbon, electrolyte: aq AgNO3) but more silver is still deposited on the object in experiment 2 (anode:silver, electrolyte aq AgNO3). Suggest a reason for this observation and describe how he could use aq NaCl to find out if his reasoning is correct.

Answer 31

In experiment 1, no silver ion is generated at the anode to replace the silver ion deposited on the cathode.

In experiment 2, silver ion is generated at the anode to replace the silver ion deposited on the cathode. NaCl can be added to test for the presence of silver ion as white precipitate of AgCl will be observed.

Question 32

If an iron object is placed in a beaker of aq AgNO3, a silver coating forms on the iron. If a gold object is placed in aq AgNO3, no reaction happens. Explain why.

Answer 32

Iron is higher than silver in the reactivity series and thus displaces silver ions from its solution.
Gold is lower than silver in the reactivity series and cannot displace silver ions, therefore no reaction occurs.

Question 33

Explain why the colour of the copper (II) sulfate solution remains the same.

Answer 33

As 1 mole of Cu oxidises to form 1 mole of Cu2+ ions at the anode, 1 mole of Cu2+ ions reduces to form 1 mole of Cu at the cathode. Thus, there is no net change in the concentration of Cu2+ ions -> no colour change observed.

Question 34

Explain why universal indicator in the concentrated sodium chloride has changed colour

Answer 34

H+ gets selectively discharged at the cathode. Thus, there is a higher conc of OH- ions and the resulting NaOH electrolyte causes the UI to change colour from green to purple.

Question 35

A hydrogen fuel cell uses aq potassium hydroxide. Suggest one reason why aq potassium hydroxide is used in the fuel cell rather than pure water.

Answer 35

Pure water is a simple molecule which cannot dissociate to produce ions. Aq potassium hydroxide can produce ions like hydroxide ions and take part in half reactions in the fuel cell.

Question 36

Experiment 1: dilute aq NaOH electrolysed. Experiment 2: dilute aq NaCl electrolysed.
Explain the observations for both experiments. Include explanation of the difference in vol of gas collected at the +ve and -ve electrodes, why ratio of vol of gases evolved is the same in both exp (2:1) and give half-equations for the reaction at each electrode. [6]

Answer 36

Both experiments:
Anode:
4OH- (aq)-> 2H2O(l) + O2 (g) + 4e-

Cathode (-ve electrode):
2H+ (aq) + 2e- -> H2(g)

The ratio of the gases collected at the +ve electrode: -ve electrode = 1:2.

The overall reaction is
4H2O (l) -> 2H2 (g) + O2 (g)

The ratio of the volume of gases collected is the same in both experiments since 4 moles of H2O is broken down into 2 moles of H2 at the cathode and 1 mole of O2 at the anode in both experiments.
Since mole ratio is equal to volume ratio for gases, volume of H2:O2 theoretically evolved is 2:1.

Question 37

How do the products of electrolysis of conc aq NaCl compare w dilute aq NaCl?

Answer 37

Chlorine gas, instead of oxygen gas, will be produced at the positive electrode while hydrogen gas will still be produced at the negative electrode.

Question 38

Predict the colour of the universal indicator in

Exp 1: electrolysis of dilute aq NaCl
Exp 2: electrolysis of conc aq NaCl

Answer 38

Exp 1: green
Since H+ and OH- ions are discharged at the cathode and anode, an increasing conc of sodium and chloride ions will be left in the electrolyte, resulting in a neutral solution.

Exp 2: violet
Since H+ and Cl- ions are discharged at the cathode and anode, an increasing concentration of sodium and OH- ions will be left in the electrolyte resulting in an alkaline solution.

Question 39

Predict and explain the expected observations at the end of experiment 1 and 2.
E1: displacement of Cu2+ ions from CuSO4 (aq) by zinc
E2: electrolysis of aq CuSO4 w carbon electrodes

Include:
Describe the expected observations in each experiment
Explain why each change occurs
Give half-equations for each change.

Answer 39

E1:
Observations:
Zinc rod decreases in size. Pink deposits are produced. Blue solution turns colourless.
Cu2+ (aq) + 2e- -> Cu (s)
Zn (s) -> Zn2+ (aq) + 2e-
Zinc is more reactive than copper and loses electrons readily, displacing copper (II) ions from the solution to form copper metal and zinc sulfate solution.

E2:
Pink deposits formed around negative electrode. Bubbles r produced at the positive electrode. Blue solution turns colourless.

Copper(II) ions r selectively discharged at the -ve elctrode and undergo reduction to form copper. OH- ions get oxidised at the +ve electrode to form O2 gas.
Cu2+ (aq) + 2e- -> Cu (s)
4OH- (aq) -> 2H2O (l) + O2 (g) + 4e-

Question 40

The volume and mass of CuSO4 solution is the same initially. The zinc rod and electrodes are removed. Predict how the mass of the solution is expected to change during exp 1 and 2. Explain.
E1: displacement of Cu2+ ions from CuSO4 (aq) by zinc
E2: electrolysis of aq CuSO4 w carbon electrodes

Answer 40

E1:
Mass of solution increases as zinc has a higher atomic mass than copper.
E2:
Mass of solution decreases as copper and oxygen have been produced from the electrolysis of the solution resulting in a lower mass.

Question 41

Observation of electrolysis of aq CuSO4 using copper electrodes (E2).

Answer 41

The colour of the electrolyte in E2 remains unchanged.
Cu2+ ions r selectively discharged at the -ve electrode (cathode) and undergo reduction to form copper.
The positive copper electrode undergoes oxidation to form copper.
This results in no net change in conc of Cu2+ ions in the electrolyte. Hence, colour of electrolyte remains unchanged.

Question 42

Electrolysis of molten NaCl w inert electrodes

Answer 42

Cathode:
Na+ discharged
Na+ (l) + e- -> Na (l)

Na+ ions gains electrons and is reduced to form sodium atom. Grey globules of sodium obtained at cathode.

Anode:
Cl- discharged
2Cl- (l) -> Cl2 (g) + 2e-

Cl- ions loses e- and is oxidised to form chlorine molecules. Yellow green chlorine gas obtained at anode.

Overall equation:
2NaCl (l) -> 2Na (l) + Cl2 (g)

Question 43

Electrolysis of dilute aq NaCl w inert electrode

Answer 43

Cathode:
H+ (aq) discharged

Since H2 is below NA in the reactivity series, H+ ions r SD to form colourless H2 (g).

Anode:
OH- (aq) discharged
4OH- (aq) -> 2H2O (l) + O2 (g) + 4e-
Since OH- ions r lower than Cl- in th electrochemical series, OH- ions r SD to form colourless O2 gas.

OE: 2H2O (l) -> 2H2 (g) + O2(g)

Ratio of H2 to O2 gas produced is 2:1.
The solution remains neutral as H+ and OH- ions r being discharged. The water lvl drops as the electrolysis continues. Conc of NaCl solution increases. If enuf water is lost, solution might become concentrated enuf for Cl- ions to be SD.

Question 44

Electrolysis of concentrated aq NaCl

Answer 44

Cathode:
H+ (aq) discharged
2H+ (aq) + 2e- -> H2 (g)

Anode:
Cl- (aq) discharged
Since conc of Cl- ions > OH- ions, Cl- ions r SD to form yellow green chlorine gas.

OE:
2H+ (aq) + 2Cl- (aq) -> H2 (g) + Cl2 (g)

Th ratio of H2 to Cl2 produced is 1:1
Solution becomes alkaline as there is a net discharge of H+ ions. Remaining Na+ and OH- ions form NaOH, an alkaline solution.

UI changes from green to violet.

Question 45

Electrolysis of copper (II) sulfate using copper electrodes

Answer 45

Cathode: object to be electroplated
Cu2+ discharged
Cu2+ (aq) + 2e- -> Cu(s)
Cu2+ SD to form red-brown copper. Copper cathode increases in mass.

Anode:
—
Cu (s) -> Cu2+ (aq) + 2e-
Copper anode is oxidised, dissolved to form Cu2+ ions and decreases in mass.

Copper anode is constantly oxidising and dissolving into electrolyte, replenishing the Cu2+ ions which r reduced at the cathode. Conc of Cu2+ ions remains unchanged, colour intensity of the blue electrolyte remains constant.

Question 46

Electrolysis — metal purification

Answer 46

Raw, impure copper is the anode while the pure copper is the cathode. Electrodes r placed in an electrolyte that contains Cu2+ ions eg aq CuSO4.

If the anode is replaced w an inert electrode, some pure copper can still be deposited at the cathode. This copper is produced from the Cu2+ ions in the electrolyte. The colour intensity of the electrolyte gradually fades from blue to colourless. Deposition of copper at the cathode ceases once Cu2+ ions r used up

Question 47

Context: a simple cell. State and explain any difference in the voltmeter readings if the experiment is repeated using filter paper soaked with the organic solvent methylbenzene. (So electrolyte becomes organic solvent)

Answer 47

Voltmeter reading will be 0V. Methylbenzene does not have mobile ions or mobile electrons to conduct electricity.

Question 48

Simple cell, metal strip of unknown metal and copper plate as electrodes. Explain why the metal strips and copper plate must be cleaned w sandpaper first.

Answer 48

To remove any oxides on the surface of both metals.