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

Inert vs reactive electrodes

Answer 2

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 3

Molten binary ionic compound

Answer 3

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

Question 4

Electrolysis of molten NaCl with inert electrodes

Answer 4

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)

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 5

Graphite electrode

Answer 5

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 6

Platinum

Answer 6

Advantages
- is an INERT ELECTRICAL CONDUCTOR and remains chemically unchanged hence 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 7

Equation for discharge of OH- ions

Answer 7

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

Question 8

Equation for discharge of H+ ions

Answer 8

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

Question 9

Electrolysis of concentrated aq NaCl

Answer 9

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.

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 10

Electrolysis of CuSO4 using inert electrodes

Answer 10

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 11

Electroplating

Answer 11

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 12

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

Answer 12

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 13

Simple cells

Answer 13

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

Question 14

Metals in a simple cell

Answer 14

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 15

Simple cell vs electrolytic cell

Answer 15

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 16

Measuring the potential difference

Answer 16

Voltmeter is used
SI: V

Question 17

Hydrogen fuel cells

Answer 17

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 18

Advantage of using hydrogen as fuel

Answer 18

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 19

Disadvantages of using hydrogen as a fuel

Answer 19

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 20

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 20

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 aqueous tin (II) nitrate solution.

Question 21

Explain why different combinations of metals produce different voltages.

Answer 21

When there is a difference in reactivity between 2 metals, a voltage is produced. The further apart the 2 metals r in the reactivity series, the greater the difference in reactivity between the metals, the greater the voltage produced.

Question 22

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

Answer 22

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

2 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 23

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

Answer 23

As O2 is more soluble in water than H2, less O2 will be collected, causing the volume ratio to bebgreater than 2:1

Question 24

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 24

As the electrolysis runs, lesser oxygen is dissolved in water as the solution becomes saturated w oxygen. Hence, more O2 is collected, bringing the ratio closer to 2:1

Question 25

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

Answer 25

The concentration of NaCl increases over time. Water is removed via the decomposition of water to form hydrogen and oxygen gases, and total volume of solution decreases. Concentration of NaCl increases since volume of Na+ ions and Cl- ions remains constant.

Question 26

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

Answer 26

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

Question 27

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

Answer 27

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

Question 28

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

Answer 28

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

Question 29

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

Answer 29

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 30

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 30

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 31

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 31

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 32

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

Answer 32

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 33

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

Answer 33

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 34

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 34

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 35

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

Answer 35

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 36

Predict the colour of the universal indicator in

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

Answer 36

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 37

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 37

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 38

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 38

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 39

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

Answer 39

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.

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 40

Electrolysis of dilute aq NaCl w inert electrode

Answer 40

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 41

Electrolysis — metal purification

Answer 41

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 42

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 42

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

Question 43

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 43

To remove any oxides on the surface of both metals.

Question 44

State one reason why aqueous KOH is used in the fuel cell rather than pure water. [1]

Answer 44

KOH contains mobile ions which enables it to conduct electricity but pure water is a simple molecule which cannot dissociate to produce ions thus does not have mobile ions or electrons and is unable to conduct electricity.

Question 45

Explain how you used the voltage data to determine the reactivities of the metal. [1]

Answer 45

The greater the difference in reactivity of the metals, the higher the voltage generated. [1]

Question 46

Describe briefly the function of the porous pot in the battery cell. [1]

Answer 46

Allows the movement of ions to maintain electrical conductivity. [1]

Question 47

When will the battery cell cease to function [1]

Answer 47

When the zinc rod (anode) is completely oxidised to form Zn2+ ions/
When all the Cu2+ ions have been completely reduced at the cathode

Question 48

A student set up a Daniell battery cell. He replaced the Zn rod with an aluminium strip instead. However, there was no reading registered in the voltmeter. Suggest an explanation. [2]

Answer 48

Al strip reacts with the H2SO4 to form an insoluble layer of aluminium oxide on the aluminium strip.
Aluminium oxide does not conduct electricity in solid state as the ions are held in fixed positions hence no mobile ions present to conduct electricity.

Question 49

Predict and explain what would happen is electrolyte in cell is replaced with ethanol [1]

Answer 49

No current flowing through as there are no mobile ions in ethanol [1]

Question 50

Describe the differences between what happens when electricity passes through the platinum electrodes compared to when electricity passes thru solution [2]

Answer 50

Electricity is able to pass through the solution due to presence of mobile ions, where silver ions and hydroxide ions are selectively discharged to form silver and oxygen gas respectively. Hence, a chemical change has occured. [1]

Electricity passes thru the electrodes due to the presence of mobile electrons [1] and the platinum electrodes remains chemically unchanged. [1]

Question 51

Explain why hydrogen gas is not formed at the start of electrolysis [2]

Answer 51

Silver is less reactive than hydrogen in the reactivity series hence silver ions are selectively discharged than hydrogen ions [1] at cathode.
Once all the silver ions are used up, hydrogen ions will be reduced to form hydrogen gas. [1]

Question 52

Use examples from the table to explain the difference between an inert electrode and an electrode that is not inert. [2]

Answer 52

Anode that is not inert will oxidase and become smaller in size to form Cu2+ ions as shown in experiment 4 where Cu anode oxidases to form Cu2+ ions. [1]
An inert electrode will not react as shown in experiment 1 or 2 where the carbon anode remains the same in size and anion gets preferentially discharged and oxidised to form oxygen/chlorine at the anode. [1]

Question 53

Predict the colour of the universal indicator in both experiments and explain. [3]

Answer 53

2: green
3: purple
In experiment 2, H+ and OH ions are preferentially discharged at the cathode and anode respectively. Concentration of H+ and OH- ions remains unchanged thus solution remains neutral.

In experiment 3, H+ and Cl- ions are preferentially discharged at cathode and anode respectively. Thus, concentration of H+ ions will be lower than concentration of OH- ions as concentration of OH- ions remains unchanged in the solution, there is a net discharge of H+ ions. Thus solution becomes alkaline.

Question 54

Predict how the mass of the solution is expected to change in both experiments. EYA. [3]

Answer 54

1: mass of solution increases as Zn displaces Cu from its SALT SOLUTION, forming ZnSO4 and Zn HAS A HIGHER Ar THAN CU
2: Mass of solution decreases as cu2+ is constantly preferentially discharged to form Cu (s) at the cathode.
+ equations

Note: when asked to describe changes:
1. Change in colour: turns from X to y/remains X/blue solution decolourises
2. Size of electrodes: increase/decrease
3. Mass of electrodes: increase/decrease
4. Product formed (reddish brown solid, gas is evolved (test of gas) etc)
5. pH of solution

Question 55

Electrolyte

Answer 55

A substance that conducts electricity in the molten or aqueous state and is not limited to ionic compounds. Mobile ions present in the electrolyte allow it to conduct electricity.