CHAPTER 8 (Module 4)

ELECTROCHEMISTRY – cont

8.4 FUEL CELLS

1. A fuel cell is a cell in which a chemical reaction between a fuel and oxygen is used to create a voltage. The fuel and oxygen flow into the cell continuously and the products flow out of the cell. Therefore the cell does not need to be recharged. The most widely used fuel cell is the hydrogen-oxygen fuel cell:

2. A fuel cell, like a regular electrochemical cell, consists of two half-cells connected by a semi-permeable membrane. An aqueous solution of sodium hydroxide is used as the electrolyte.

Oxygen is pumped into one of the half-cells:O2(g) + H2O(l) + 4e- = 4OH-(aq) E0 = +0.40 V

Hydrogen is pumped into the other half-cell:H2O(l) + 2e- == H2(g) + 2OH-(aq) E0 = -0.83 V

3. The oxygen half-cell is more positive and therefore undergoes reduction. The hydrogen half-cell is more negative and undergoes oxidation:

O2(g) + H2O(l) + 4e- 4OH-(aq) reduction

H2(g) + 2OH-(aq) H2O(l) + 2e- oxidation

O2(g) + 2H2(g) 2H2O(l) overall cell reaction, emf = 1.23 V

4. Hydroxide ions are generated in the oxygen half-cell and travel through the membrane into the hydrogen half-cell, where they are used up. Water is the product of the reaction and it is allowed to run off.

5. There are a number of advantages of fuel cells as a way of producing energy:

a) The hydrogen-oxygen fuel cell produces water as the only product. It therefore does not produce any of the greenhouse or polluting gases associated with combustion engines. The process of generating hydrogenfor use in fuel cells does produce a small quantity of carbon dioxide, but much less than would be generated by a combustion engine.

b) Fuel cells are more efficient than combustion engines. Typically fuel cells are approximately 50% efficient but combustion engines are approximately 20% efficient.

6. However there are also a number of limitations of fuel cells as a way of producing energy:

a) Hydrogen is a flammable gas with a low boiling point. It is therefore both difficult and dangerous to store and transport. It can be stored as a liquid under pressure or as a solid adsorbed to the surface of a solid, but both of these techniques are expensive.

b) As a result, obtaining hydrogen as a fuel is difficult and this means that people will not buy hydrogen-powered vehicles.

c) Fuel cells use toxic chemicals in their manufacture.d) Fuel cells have a limited lifetime

Rechargeable and non-rechargeable cells

1. Electrochemical cells are the basis for all batteries. Batteries contain two separate half-cells. The solutions are connected by a salt bridge or semi-permeable membrane which allows ions to flow through without allowing complete mixing of the solutions. The electrodes are connected to the terminals of the battery, and when the battery is connected to an electrical device a current can flow.

2. If the reactions taking place in the half-cells are reversible, the battery is rechargeable. By connecting the battery to another power supply with a larger emf, electrons and ions are forced around the circuit in the opposite direction. This reverses the spontaneous chemical reaction and hence recharges the battery.

3. If the reactions taking place in the half-cells are irreversible the battery is non-rechargeable.

Non-rechargeableZinc-Carbon AlkalineMercuryAir-Zinc

RechargeableLead-AcidNiCadMetal-hydrideLithiumFuel CellsResearch Areas

Zinc-Carbon Cell

(LeClanche Dry Cell)

Carbon rod(cathode)Cathode mix:

MnO2, graphite

Electrolyte: ZnCl2,NH4Cl, flour, starchZinc can (anode)

Insulating tubeSteel jacketExpansion Chamber

Alkaline Cell

Steel jacketCathode mix:MnO2, graphiteAbsorbent /

separator

Zn and KOH anodepaste

Brass CurrentCollector

KOH electrolyte

anode

Zinc-Carbon Cell

E °½= +0.76 V

E °cell=

Zn2+(aq) + 2 e– Zn(s)Zn(s) + 2 MnO2(s) + 2 NH4

+(aq)Zn2+(aq) + 2 NH3(aq) + Mn2O3(s) + H2O(l)overall

2 NH3(aq) + Mn2O3 (s) + H2O(l) cathodeE °½=

2 MnO2(s) + 2 NH4+(aq) + 2 e–

Nernst Equation

E Ecell cell=on

ZnNHNH- ×+

+00592 232

42

. log[ ] [ ][ ]Mercury

CellZinc amalgam

and KOH anodeKOH inabsorbentelectrolyte

materialCathode mix:HgO, graphite

Steel orstainlesssteel case

anode

Alkaline Cell

E °½=

E °cell=Zn(s) + 2 MnO2(s)ZnO(s) + Mn2O3(s)overall

cathodeE °½=

Mn2O3(s) + 2 OH–(aq) 2 MnO2(s) + H2O(l) + 2 e–

Zn(s) + 2 OH–(aq)ZnO(s) + H2O(l) + 2 e–

E Ecell cell n= - ×o 00592 11

. log()()

anode

Mercury Cell

Zn(Hg) + HgO(s)ZnO(s) + Hg(l)overall

cathode Hg(l) + 2 OH–(aq) HgO(s) + H2O(l) + 2 e–

Zn(Hg) + 2 OH–(aq)ZnO(s) + H2O(l) + 2 e–

A metal dissolved in mercury is called an“amalgam”

anode

Lead Acid Cell

E °½= +1.685 V

E °cell= +2.041 V

overall

cathodeE °½= +0.356 V

PbO2(s) + SO42–(aq) + 4 H+(aq) + 2 e–PbSO4(s)+ 2 H2O(l)

Pb(s) + SO42–(aq)PbSO4(s)+ 2 e–

2 PbSO4(s)+ 2 H2O(l)Pb(s) + PbO2(s) + 2 SO42–(aq) + 4 H+(aq)

8.5 ELECTROLYSIS

1. The process of decomposing a compound using electricity is called electrolysis.

2H2O(l) → 2H2(g) + O2(g)

Lead-Acid Battery

Plasticbarriersseparatingeach cell

SulfuricAcid

Cathodeplates:lead grillsfilled withPbO2

Anodeplates:lead grillsfilled withspongy lead

2. Electrolysis is a most important industrial process with a wide application. However, the largest application is that of the manufacture of chlorine and sodium hydroxide from concentrated aqueous sodium chloride in the Chlor-alkali industry. The cell consists of two conducting rods called electrodes, dipped into a compound in a molten state or in solution, that is able to conduct electricity called the electrolyte.

3. Electrolysis only works with direct current (the charge flows in only one direction). Direct current causes one electrode to take on a positive charge called the anode. The second electrode takes on a negative charge called the cathode. Note: Anions (negative ions) are attracted to the anode. Cations (positive ions) are attracted to the cathode.

4. An example of electrolysis: If place the molten sodium chloride into the electrolysis cell and allow current to flow, the following reactions occur at the electrodes:

At the cathode: Reduction (Na+(l) + e- → Na(l))

At the anode: Oxidation (2Cl-(l) → Cl2(g) + 2e-)

These two half equations give the overall reaction: (2Na+(l) + 2Cl-

(l) → Na(l) + Cl2(g))

Predicting products in electrolysis

5. In the example above, of molten sodium chloride, there was one cation Na+ and one anion Cl-. Therefore, the products were easily calculated. However, if a compound is in an aqueous solution, then we have two more ions to deal with, OH- and H+.

So how do we decide with ions will appear at the electrodes and which remain in solution?

6. In the case of the cathode, we must decide which of the two cations, H+ or Na+ is reduced most easily back to their atoms. Hydrogen ions are most easily reduced so the following cathode reaction occurs:

2H+(aq) + 2e- → H2(g)

7. In the case of the anode, we must decide which of the two anions, OH- or Cl- will be most easily oxidised back to atoms. Chloride ions are most easily oxidised, hence the following anode reaction:

2Cl-(aq) → Cl2 + 2e-

This leaves Na+ (aq) and OH-(aq) in solution i.e. sodium hydroxide. Please refer to the electrochemical

series, redox series or reactivity series on the 3rd module for this information.

CARBON

CARBON

Na⁺

H⁺

OH⁻

Cl⁻

Na⁺H⁺

Cl⁻OH⁻

Cathode: 2H⁺ +2e⁻H2

Anode: 4OH⁻ O2 + 2H2O + 4e⁻

Calculating mass of products in electrolysis

1. In 1832, Michael Faraday deduced that: The quantity of electricity passed is proportional to the amount of substance discharged at the electrode.

Quantity of electricity (charge) = current x time

The units used are, coulombs for charge, amps for current and seconds for time.Note 1: One mole of electrons has a charge of 96500C. This quantity of charge is called the Faraday constant, F. Thus, the Faraday constant is related to Avagadro's constant, L, and the charge on the electron, e.

F = L x e

Note 2: The number of moles of electrons required to discharge 1 mole of ions is equal to the charge on the ion. For example, ions with a double charge, such as Cu2+ - it will take two moles of electrons to deposit one mole of copper.

8.6 APPLICATION OF ELECTROCHEMISTRY

Aluminium Production (The Hall process for aluminium production)

(Fuel Cell)

Corrosion (Rusting of Iron)

1. Rust is in fact the compound hydrated iron (III) oxide with the general formula, Fe2O3.xH2O. Rusting occurs when both water and oxygen are present. It is an electrochemical process involving two redox half-equations.

The corrosion starts with the oxidation of iron to Fe2+: Fe(s) → Fe2+(aq) + 2e- E = -0.44V

2. The iron acts as a negative terminal. The released electrons travel through the iron until they make contact with oxygen. This forms the positive electrode.

O2(aq) + 2H2O(l) + 4e- → 4OH-(aq) E = +1.23V.

3. The following step is that the Fe2+ and OH- ions combine to form Fe(OH)2(s). This is readily oxidised to iron (III) hydroxide, which then partially dehydrates to give rust, Fe2O3.xH2O.

2Fe + O2 (g) + 2H2O 2Fe(OH)2 (s) Eocell= +0.84 v

4. There are three requirements for the corrosion of iron are Iron, Water and Oxygen. Corrosion is spontaneous or an electrochemical cell.

carbon-lined steel vesselacts as cathode

CO2 bubbles

Al (l)Al2O3 (l) Drawoff Al (l)

-

+

Cathode: Al+3 + 3e- Al (l)Anode: 2 O-2 + C (s) CO2 (g) + 4e-

frompowersource

Al+3

O-2O-

2

Al+3

O-2

graphite anodes

e-

e- Electrolysis of molten Al2O3 mixed with cryolite – lowers melting point.

Cathode: Al+3 + 3e- Al (l)Anode: 2 O-2 + C (s) CO2 (g) + 4e-

4 Al+3 + 6 O-2 + 3 C (s) 4 Al (l) + 3 CO2 (g)The graphite anode is consumed in the process.

Preventing rust

1. Coating the Iron: By adding a layer of paint, oil or grease to the iron, you prevent oxygen and water from coming into contact with the iron.

2. Sacrificial Protection: A most effective way of preventing rust is to coat the iron with zinc, this is called galvanising. It works due to zinc's greater reactivity, i.e. zinc has a greater tendency to form ions, hence any Fe2+ present are reduced to Fe atoms.

3. Alloying: Iron can be alloyed with nickel, chromium or carbon. The presence of other elements helps prevent Fe2+ forming and electrons been released, hence preventing the rusting process starting.

4. Electroplating

Fe FeFe

Fe2

+

Fe2+OH-

OH-

Fe(OH)

2(s)

Fe(OH)

2(s)

Rust- low solubility

e-

e-

+ 2e-Cathode

Fe(s) → Fe2+ + 2e- Anode

Water Drop

Iron Surface

→ 2OH- O2 + H2O 1/2

Cations to cathode and anions to anode

The anode reaction is the oxidation of FeThe cathode reaction is the reduction of O2 and H2O- outer circle of drop Electrons flow from the anode to the cathode

Fe Nail in WaterAttach a piece of Zn or Mg

All lower than Fe.Zn AnodeFe

Zn → Zn2+ + 2e-

Sacrificial anode--

-

---

Electrons flow from anode to cathode to protect the Fe

Cathode or Reduction--

Exercise 8c

1. Concentrated nitric acid is a powerful oxidising agent. Concentrated nitric acid oxidises sulphur to sulphuric acid. Nitrogen dioxide and another product are also formed.

(i) Suggest a balanced equation for this reaction.

(ii) Deduce the change in oxidation number of nitrogen in this reaction.from ............... to ...............

2. NO2 reacts with oxygen and water to form nitric acid, HNO3. In the atmosphere, this contributes to acid rain. Construct a balanced equation for this formation of nitric acid and use oxidation numbers to show that this is a redox reaction.

3. The standard electrode potential of the 2

1

Cl2/ Cl– half-cell may be measured using

the following apparatus.

C

H (g)2

B

A

D

Cl (aq)–

salt bridge

(a) Suggest suitable labels for A, B, C and D.(b) The half cell reactions involved are shown below.

2

1

Cl2 + e– Cl– Eο = +1.36 V

H+ + e– 2

1

H2 Eο = 0.00V(i) Use an arrow to show the direction of flow of electrons in the diagram of

the apparatus. Explain your answer.

(ii) The values of Eο are measured under standard conditions. What are the standard conditions?

(c) The half cell reaction for ClO3–/ 2

1

Cl2 is shown below.

ClO3– + 6H+ + 5e– 2

1

Cl2 + 3H2OEο = +1.47 V

What does this tell you about the oxidising ability of ClO3– compared with

4. Chlorine gas may be prepared in the laboratory by reacting hydrochloric acid with potassium manganate(VII). The following standard electrode potentials relate to this reaction.

2

1

Cl2 + e– Cl– Eο = +1.36 V

MnO4– + 8H+ + 5e– Mn2+ + 4H2O Eο = +1.52 V

(a) Use the half-equations above to:(i) construct an ionic equation for the reaction between hydrochloric acid

and potassium manganate(VII);(ii) determine the oxidation numbers of chlorine and manganese before and

after the reaction has taken place;(iii) state what is oxidised and what is reduced in this reaction.

(b) If potassium manganate(VII) and very dilute hydrochloric acid are mixed, there is no visible reaction. Suggest why there is no visible reaction in this case.

5. Aluminium metal is extracted from molten bauxite (Al2O3.2H2O) using electrolysis. Cryolite (AlF3) added to the ore in order to lower the melting point required and thus the energy required by the process.(a) Write an half equation to show how aluminium metal is produced fromthe ore.(b) What mass of aluminium metal would be produced if a current of 30,000Ais applied to a cell for 1 hour.(c) In the molten mixture there is a mixture of anions which mostly consists of O2

- and F-. Write an equation to show which of these anions will be oxidised in the cell?

6. What is the maximum mass of copper that could be plated out by electrolyzing aqueous CuCl2 for 16.0 hours at a constant current of 3.00 amperes? (1 faraday = 96500 coulombs)

7. Zn(s) + Cu2+ Zn2+ + Cu(s)

An electrolytic cell based on the reaction above was constructed from zinc and copper half-cells. The observed voltage was found to be 1.00 volt instead of the standard cell potential, Eo, of 1.10 volts. Which of the following could correctly account for this observation?(A) The copper electrode was larger than the zinc electrode.(B) The Zn2+ electrolyte was Zn(NO3)2, while the Cu2+ electrolyte was CuSO4.(C) The Zn2+ solution was more concentrated than the Cu2+ solution.(D) The solutions in the half-cells had different volumes.(E) The salt bridge contained KCl as the electrolyte.

8. Fe2+ + 2e- Fe(s) Eo = -0.44 volt  Ni2+ + 2e-

Ni(s) Eo = -0.23 voltThe standard reduction potentials for two half reactions are given above. The Nernst

4. Chlorine gas may be prepared in the laboratory by reacting hydrochloric acid with potassium manganate(VII). The following standard electrode potentials relate to this reaction.

2

1

Cl2 + e– Cl– Eο = +1.36 V

MnO4– + 8H+ + 5e– Mn2+ + 4H2O Eο = +1.52 V

(a) Use the half-equations above to:(i) construct an ionic equation for the reaction between hydrochloric acid

and potassium manganate(VII);(ii) determine the oxidation numbers of chlorine and manganese before and

after the reaction has taken place;(iii) state what is oxidised and what is reduced in this reaction.

(b) If potassium manganate(VII) and very dilute hydrochloric acid are mixed, there is no visible reaction. Suggest why there is no visible reaction in this case.

5. Aluminium metal is extracted from molten bauxite (Al2O3.2H2O) using electrolysis. Cryolite (AlF3) added to the ore in order to lower the melting point required and thus the energy required by the process.(a) Write an half equation to show how aluminium metal is produced fromthe ore.(b) What mass of aluminium metal would be produced if a current of 30,000Ais applied to a cell for 1 hour.(c) In the molten mixture there is a mixture of anions which mostly consists of O2

- and F-. Write an equation to show which of these anions will be oxidised in the cell?

6. What is the maximum mass of copper that could be plated out by electrolyzing aqueous CuCl2 for 16.0 hours at a constant current of 3.00 amperes? (1 faraday = 96500 coulombs)

7. Zn(s) + Cu2+ Zn2+ + Cu(s)

An electrolytic cell based on the reaction above was constructed from zinc and copper half-cells. The observed voltage was found to be 1.00 volt instead of the standard cell potential, Eo, of 1.10 volts. Which of the following could correctly account for this observation?(A) The copper electrode was larger than the zinc electrode.(B) The Zn2+ electrolyte was Zn(NO3)2, while the Cu2+ electrolyte was CuSO4.(C) The Zn2+ solution was more concentrated than the Cu2+ solution.(D) The solutions in the half-cells had different volumes.(E) The salt bridge contained KCl as the electrolyte.

8. Fe2+ + 2e- Fe(s) Eo = -0.44 volt  Ni2+ + 2e-

Ni(s) Eo = -0.23 voltThe standard reduction potentials for two half reactions are given above. The Nernst

11. A student places a copper electrode in a 1 M solution of CuSO4 and in another beaker places a silver electrode in a 1 M solution of AgNO3. A salt bridge composed of Na2SO4 connects the two beakers. The voltage measured across the electrodes is found to be + 0.42 volt.

(a) Draw a diagram of this cell.(b) Describe what is happening at the cathode (Include any equations that may be useful.(c) Describe what is happening at the anode. (Include any equations that may be useful.(d) Write the balanced overall cell equation. (e) Write the standard cell notation.(f) The student adds 4 M ammonia to the copper sulfate solution, producing the complex ion Cu(NH3)+ (aq). The student re-measures the cell potential and discovers the voltage to be 0.88 volt. What is the Cu2+ (aq) concentration in the cell after the ammonia has been added?