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9.2 Production of materials: 4. Electrochemical methods

Syllabus reference (October 2002 version)
4. Oxidation-reduction reactions are increasingly important as a source of energy	Students learn to: explain the displacement of metals from solution in terms of transfer of electrons identify the relationship between displacement of metal ions in solution by other metals to the relative activity of metals account for changes in the oxidation state of species in terms of their loss or gain of electrons describe and explain galvanic cells in terms of oxidation/reduction reactions outline the construction of galvanic cells and trace the direction of electron flow define the terms anode, cathode, electrode and electrolyte to describe galvanic cells	Students: perform a first-hand investigation to identify the conditions under which a galvanic cell is produced   perform a first-hand investigation and gather first-hand information to measure the difference in potential of different combinations of metals in an electrolyte solution   gather and present information on the structure and chemistry of a dry cell or lead-acid cell and evaluate it in comparison to: button cell fuel cell vanadium redox cell lithium cell liquid junction photovoltaic device (eg the Gratzel cell) in terms of: chemistry cost and practicality impact on society environmental impact   solve problems and analyse information to calculate the potential Eø requirement of named electrochemical processes using tables of standard potentials and half equations

Extract from Chemistry Stage 6 Syllabus (Amended October 2002). © Board of Studies, NSW.
[Edit: 7Jul09]

Prior learning: Preliminary modules 8.3.2.

Background: Reactions of metals usually require the transfer of electrons.Metals can be arranged in an activity series from most active to least active:

It is recommended that you become familiar with the names for the metals signified by the symbols listed. A Periodic Table that gives the names as well as symbols will be provided in the HSC examination.

explain the displacement of metals from solution in terms of transfer of electrons

• More active metals will displace less active metal ions from solution in an oxidation-reduction reaction.
  
  
• When an active metal is placed in a solution containing ions of a less active metal, the active metal displaces the less active metal from solution. This occurs because a more active metal atom loses one or more electrons and becomes a positive ion. The electrons lost are transferred to the ions of the less active metal, resulting in them becoming metal atoms.

  For example, if an iron nail is placed in a solution of blue copper (II) salt, some of the iron nails dissolves.

  

  At the same time, the blue colour of Cu²⁺ ions disappears and a dark copper coating appears on the nail surface.
  

  The overall reaction is:
  

  

  The electrons lost by iron atoms undergoing oxidation are used to reduce copper (II) ions to copper atoms. Oxidation–reduction reactions (also called redox reactions) involve transfer of electrons.

  

identify the relationship between displacement of metal ions in solution by other metals to the relative activity of metals

• If a metal is higher in the activity series, the metal atoms will react when put in a solution of ions of a metal that is lower in the activity series. The less active metal ions are displaced from solution as they form atoms.
• In reacting, the more active metal atom (M) changes to a metal ion (Mⁿ⁺) by losing one or more electrons to form a cation.

  

• Metal reactions can be related to the activity series. For example:
  • the metals from K to Pb react with dilute acids releasing hydrogen gas
  • the metals from K to Mg react with liquid water
  • the metals from Al to Ni require water to be in the form of steam before reacting.

account for changes in the oxidation state of species in terms of their loss or gain of electrons

The oxidation state (also called oxidation number) of an element is zero. The oxidation state of a metal cation (Mⁿ⁺) is (+)n.

• When a metal atom undergoes a loss of electrons (oxidation), there is an increase in the oxidation number of the metal from 0 to n.
• When a metal reacts with dilute acid and releases hydrogen, the metal undergoes oxidation (loss of electrons) while the hydrogen ions in the acid undergo reduction (gain of electrons).
  Example:  

  Magnesium changes from oxidation state 0 to 2. This is an increase, thus this is oxidation.

  Hydrogen changes oxidation state from +1 (in H⁺) to 0 (in the element H₂). This is a decrease, thus this is reduction.

  Memory aid

  A mnemonic you can use to remember oxidation and reduction is OILRIG.
  OIL Oxidation Is Loss (of electrons)
  RIG Reduction Is Gain (of electrons).

  
  
  
• It is important to recognise that when a substance acts as a reductant, causing reduction, it is oxidised.
  
  
• 
  When a substance acts as an oxidant, causing oxidation, it is reduced.

  

solve problems and analyse information to calculate the potential E^ø of named electrochemical processes using tables of standard potentials and half equations

• To solve these types of problems you will need to analyse information from the Table of standard potentials. It gives measurements of potentials,  E^ø values, for standard conditions, (temperature 25^oC, 10⁵ Pa pressure and solution concentrations of 1 mol L^-1). From the information, the potential or voltage produced by an oxidation-reduction reaction can be predicted.
  
  
• The E^ø value given in the table is the value for the equation as it is written from left to right. If you want the value for the reverse process simply change the sign of the E^ø value.

  In the table of standard potentials, a metal will displace the ions of any metal below it. Thus iron will displace Cu²⁺, but not Mg²⁺, from solution.

  oxidant +	electron/s		reductant	Eø (volts)
  Li+	+ e–		Li	–3.04
  K+	+ e–		K	–2.94
  Ca2+	+ 2e–		Ca	–2.87
  Na+	+ e–		Na	–2.71
  Mg2+	+ 2e–		Mg	–2.36
  Al3+	+ 3e–		Al	–1.68
  H2O	+ e–		1/2H2(g) + OH–	–0.83
  Zn2+	+ 2e–		Zn	–0.76
  Fe2+	+ 2e–		Fe	–0.44
  Cd2+	+ 2e–		Cd	–0.40
  Ni2+	+ 2e–		Ni	–0.24
  Pb2+	+ 2e–		Pb	–0.13
  H+	+ e–		1/2H2(g)	0.00
  Cu2+	+ 2e–		Cu	0.34
  1/2O2(g)+H2O	+ 2e–		2OH–	0.40
  Cu+	+ e–		Cu	0.52
  1/2I2(aq)	+ e–		I–	0.62
  Fe3+	+ e–		Fe2+	0.77
  Ag+	+ e–		Ag	0.80
  1/2Br2(aq)	+ e–		Br–	1.10
  1/2O2(g)+2H+	+ 2e–		H2O	1.23
  1/2Cl2(aq)	+ e–		Cl–	1.40
  1/2F2(g)	+ e–		F–	2.89

  The strongest agent of reduction is Li at the top right hand corner of the table. Lithium metal is the strongest reducing agent, or reductant, shown.

  The strongest agent of oxidation is F₂, located at the bottom left hand corner of the table. Fluorine is the strongest oxidising agent, or oxidant, shown.

  The equations in this table are half equations.

  The equations show reduction (gain of electrons) when read from left to right. They show oxidation (loss of electrons) when read from right to left. In English, we read from left to right, so the table is usually called a Table of reduction potentials rather that a Table of oxidation potentials.

• Calculating a potential E^ø:
  • Because the reduction of zinc () has the potential E^ø equal to -0.76 V, then the oxidation of zinc
    () has the potential E^ø value equal to +0.76 V.

    From the table, the half equation,  .

    A galvanic cell involving zinc and copper can be represented using the following IUPAC (International Union of Pure and Applied Chemistry) convention.

    Cu | Cu²⁺ || Zn²⁺ | Zn

    The single vertical line represents the boundary between phases. The double vertical line represents a salt bridge or porous pot through which ions can move.

    In this galvanic cell, the changes are:
    

    Thus, this galvanic cell is predicted to produce 1.10 volts.

perform a first-hand investigation to identify the conditions under which a galvanic cell is produced

perform a first-hand investigation and gather first-hand information to measure the difference in potential of different combinations of metals in an electrolyte solution

The activities described in the above two syllabus points can be done using the same experimental investigation and equipment.

• Perform the first investigation using a procedure, such as the one below, and identify and record the conditions under which a galvanic cell is produced. Carefully and safely dispose of any waste materials produced during the investigation.
  
  
  Suggested procedure
   

  1. Construct a galvanic cell using two different metals. Each metal should sit in a solution of its own ions (use a water-soluble salt of the metal—all nitrates and all acetates are water-soluble).
      
  2. Connect the two metal electrodes using a voltmeter and two wires. If the meter needle goes away from the scale, reverse the connections so the needle goes on scale and you can measure the difference in potential (i.e. the voltage).
     
     What happens when you remove the salt bridge from one of the solutions? Can you explain why this makes a difference?
      
  3. Connect two identical half-cells.
     
     Can you measure a difference in potential? Can you explain why there is no difference?
  
  
  
• Modify the procedure to perform the second investigation. Investigate a number of different combinations of metal electrodes and gather and record measurements of the potentials for each combination. The reliability of the investigation can be increased by the use of repeat trials for the various combinations you have chosen.

outline the construction of galvanic cells and trace the direction of electron flow

describe and explain galvanic cells in terms of oxidation/reduction reactions

define the terms anode, cathode, electrode and electrolyte to describe galvanic cells

The notes that follow provide information related to the above three syllabus points.

• Oxidation–reduction reactions normally take place by direct transfer of electrons between the reductant and the oxidant. For example, if zinc metal is placed in a solution of blue copper(II) ions, the blue colour fades as the zinc goes into solution (as colourless ions) and the copper metal comes out of solution as atoms.
  
  
• 
  
  
• A galvanic cell is a device constructed so that a reductant and oxidant are physically separated, but connected by an external circuit made of a conductor (to carry electrons) and a salt bridge (to carry charged ions in solution). A galvanic cell is thus composed of two half-cells, a reductant half-cell and an oxidant half-cell. This arrangement ensures that electrons cannot go directly from the reductant to the oxidant, but they will move through the external circuit.
  
  
  The electrons move through the conductor from the reductant half-cell to the oxidant half-cell.
  
  The energy of the moving electrons is electrical energy that can be used to turn an electric motor, produce heat or light energy in a light globe.
• A galvanic cell makes electrical energy available from chemical energy.
   
• The larger the galvanic cell, the more chemical energy is stored and the more electrical energy can be obtained from that cell.
  
  An historical application
  
  Many of the first telephone exchanges were run using electrical energy from galvanic cells consisting of a copper rod in a copper(II) salt solution sitting in a porous pot surrounded by zinc salt solution in a zinc bucket. When the zinc bucket and copper rod were connected, electrons flowed from the strongest reductant, zinc, to the strongest oxidant, copper(II) ions, through the connecting wire and copper rod.
• In a school laboratory, a galvanic cell can be made in the following way.
  
  
  
• The parts where electrons flow out of or into half-cells are electrical conductors called electrodes. Some galvanic cells use inert platinum or graphite electrodes.

  Any solution containing ions is called an electrolyte. Electricity flows through electrolytes by the movement of charged ions, not electrons.

  Electricity flows in electrodes (metals or graphite), or through connecting wires, by the movement of electrons. Electrons do not move through water or water solutions containing ions.

  The anode is the electrode where oxidation occurs. In the galvanic cell example, this is the zinc electrode. The following reaction occurs here.

  

  The cathode is the electrode where reduction occurs. In the galvanic cell example, this is the surface of the copper electrode where electrons are available for the following reaction to occur (resulting in a coating of copper).

  

  The salt bridge could be filter paper soaked in a conducting solution such as potassium nitrate solution. Potassium ions (K⁺) and (NO₃^-) ions do not form insoluble precipitates with other ions. The salt bridge allows the movement of ions between the half-cells. This prevents the build up of positive charge in the zinc half-cell as negative electrons leave and the build up of negative charge in the copper half-cell as negative electrons arrive. Positive and negative ions moving through the salt bridge keep a balance of negative and positive charge in each half-cell.

gather and present information on the structure and chemistry of a dry cell or lead-acid cell and evaluate it in comparison to:

• button cell
• fuel cell
• vanadium redox cell
• lithium cell
• liquid junction photovoltaic device (eg the Gratzel cell)

in terms of:

• chemistry
• cost and practicality
• impact on society
• environmental impact

• Gather from your class texts, a copy of a labeled diagram that shows the structure of a dry cell or a lead-acid cell. If possible, also collect a diagram of a button cell, a lithium cell, a vanadium-redox cell and a Gratzel cell. Extract information from the diagrams and from other information provided in the texts.
  
  
• Present the information as a tabulated summary. The table that follows indicates the type of information to extract and present.

Cell feature	Type of cell
	dry	lead-acid	button	fuel	vanadium redox	lithium	Gratzel
anode
cathode
electrolyte
energy density (kWh/kg)	0.090	0.030	0.125	-	-	0.150	-
cost and practicality
impact on society
environmental impact

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