IB syllabus > redox(sl) 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

9.4 - Voltaic cells


9.4.1: Explain how a redox reaction is used to produce electricity in a voltaic cell. This should include a diagram to show how two half-cells can be connected by a salt bridge to form a whole cell. Suitable examples of half-cells are Mg, Zn, Fe and Cu in solutions of their ions.


Electrochemicall cells

As we have seen certain species lose electrons (reducing agents) and other species gain electrons (oxidising agents) when reacting. If these species are not mixed together but connected electrically around an external circuit then these electrons will flow around the external circuit producing an electroic current.

Each of the reacting species is then called a half cell and the whole set up is called an electrochemical cell. It is the basis behind the electrical battery.

An electrochemical cell

In this cell the zinc metal has a tendency to dissolve as ions leaving itselectrons on the electrode. The copper, which is a weaker reducing agent, is forced to accept the electrons and use then to turn the copper ions into copper at the copper electrode. As these electrons flow around the outer (external) circuit they constitute an electric current

The 'salt bridge' is usually a filter paper soaked in potassium nitrate solution (neither of these ions react with any other ions in the experiment). This 'salt bridge' then allows ions to move in both directions equalising any build up of electrical charge in the beakers.

The zinc forces the copper ions to accept electrons and the overall cell equation can be constructed by adding together the two 'half equations' above.

Zn(s) Zn2+(aq) + 2e
Cu2+(aq) + 2e Cu(s)
overall: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

This type of cell can be constructed using any pair of reducing and oxidising agents. The greater the difference in the reactivity of one type of species (i.e. the reducing species) the greater the cell potential (voltage)

Consequently a cell constructed from zinc | zinc sulphate in one half cell and silver ! silver nitrate solution in the other half cell will have a greater voltage that the cell above (there is a greater difference in reactivity between zinc and silver than between zinc and copper)


Electrochemical cell - general

General cell construction

 

Copper - Iron cell

In this cell the iron is the more powerful reducing agent and will preferentially lose electrons. These electrons force the copper redox equation to go in the direction of receiving electrons (reduction) - i.e. the copper Cu2+ ions pick up electrons and are deposited on the electrode as copper metal atoms

The iron | Fe2+ solution beaker is called a half-cell and the copper | copper ions solution is said to be the other half-cell.


9.4.2: State that oxidation occurs at the negative electrode (anode) and that reduction occurs at the positive electrode (cathode)


This is potentially one of the most confusing areas for students of electrochemistry. The electrochemical cell seems to have the electrodes labelled opposite to the electrolytic cell.

Electrochemical cell Electrolytic cell

Anode (negative electrode)

Cathode (positive electrode)

Anode (positive electrode)

Cathode (negative electrode)

This arises because of the nature of the processes occurring in the two different cells. The cathode, for example, is the electrode that makes electrons available for the circuit in each case. In the electrochemical cell the cathode is the positive electrode where electrons are being supplied to positive ions in a reduction process.

For example the Cu2+¦Cu electrode in a zinc-copper electrochemical cell is positive and absorbs electrons from the system.

Cu2+ + 2e --> Cu

However in an electrolytic cell the cathode is the electrode where the electrons are being supplied to react with the positive ions in the electrolyte.

For example, the electrolysis of copper(II) chloride. At the cathode:

Cu2+ + 2e --> Cu

So in both cells the cathode supplies electrons and reduction takes place.


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