IB Chemistry HL

Revision Notes

19.1.1 Electrochemical Cells

Electrochemical Cells

  • We have seen previously that redox reactions involve simultaneous oxidation and reduction as electrons flow from the reducing agent to the oxidizing agent
  • Which way electrons flow depends on the reactivity of the species involved
  • Redox chemistry has very important applications in electrochemical cells, which come in two types:
    • Voltaic cells
    • Electrolytic cells

Voltaic cells

  • A voltaic cell generates a potential difference known as an electromotive force or EMF
  • The EMF is also called the cell potential and given the symbol E
  • The absolute value of a cell potential cannot be determined only the difference between one cell and another
  • This is analogous to arm-wrestling: you cannot determine the strength of an arm-wrestler unless you compare her to the other competitors
  • Voltaic (or Galvanic) cells generate electricity from spontaneous redox reactions
  • For example:

Zn (s)  + CuSO4 (aq)→ Cu (s)  + ZnSO4 (aq)

  • Instead of electrons being transferred directly from the zinc to the copper ions a cell is built which separates the two redox processes
  • Each part of the cell is called a half cell
  • If a rod of metal is dipped into a solution of its own ions, an equilibrium is set up
  • For example:

Zn (s)  ⇌  Zn2+ (aq) + 2e– 


When a metal is dipped into a solution contains its ions an equilibrium is established between the metal and it ions

  • This is a half cell and the strip of metal is an electrode
  • The position of the equilibrium determines the potential difference between the metal strip and the solution of metal
  • The Zn atoms on the rod can deposit two electrons on the rod and move into solution as Zn2+ ions:

                   Zn(s) ⇌ Zn2+(aq) + 2e– 

  • This process would result in an accumulation of negative charge on the zinc rod
  • Alternatively, the Zn2+ ions in solution could accept two electrons from the rod and move onto the rod to become Zn atoms:

                  Zn2+(aq) + 2e–  ⇌ Zn(s)

  • This process would result in an accumulation of positive charge on the zinc rod
  • In both cases, a potential difference is set up between the rod and the solution
  • This is known as an electrode potential
  • A similar electrode potential is set up if a copper rod is immersed in a solution containing copper ions (eg CuSO4), due to the following processes:

Cu2+(aq) + 2e–  ⇌ Cu(s)  – reduction (rod becomes positive)

Cu(s) ⇌ Cu2+(aq) + 2e–    – oxidation (rod becomes negative)

  • Note that a chemical reaction is not taking place – there is simply a potential difference between the rod and the solution

Creating an emf

  • If two different electrodes are connected, the potential difference between the two electrodes will cause a current to flow between them. Thus an electromotive force (EMF) is established and the system can generate electrical energy
  • A typical electrochemical cell can be made by combining a zinc electrode in a solution of zinc sulphate with a copper electrode in a solution of copper sulphate

The zinc-copper voltaic cell ( also known as the Daniell Cell)

  • The circuit must be completed by allowing ions to flow from one solution to the other
  • This is achieved by means of a salt bridge – often a piece of filter paper saturated with a solution of an inert electrolyte such as KNO3(aq)
  • The emf can be measured using a voltmeter
  • Voltmeters have a high resistance so that they do not divert much current from the main circuit
  • The two half cells are said to be in series as the same current is flowing through both cells
  • The combination of two electrodes in this way is known as a voltaic cell, and can be used to generate electricity

Cell Potential Calculations

  • Voltmeters measure potential on the right-hand side of the cell and subtract it from the potential on the left-hand side of the cell

EMF= ErightEleft

  • Sometimes this can be hard to remember, but it helps if you remember the phrase ‘knives & forks’



You hold your knife in your right hand and your fork in your left hand. EMF is right minus left

  • If the standard hydrogen electrode is placed on the left-hand side of the voltmeter by convention Eleft will be zero and the emf of the cell will be the electrode potential of the right-hand electrode
  • For example, if the standard zinc electrode is connected to the standard hydrogen electrode and the standard hydrogen electrode is placed on the left, the voltmeter measures -0.76V.

Zn2+(aq) + 2e ⇌ Zn(s)

  • The   Zn2+(aq) + 2e ⇌ Zn(s) half-cell thus has an electrode potential of -0.76V
  • If the Cu2+(aq) + 2e ⇌ Cu(s) electrode is connected to the standard hydrogen electrode and the standard hydrogen electrode is placed on the left, the voltmeter reads +0.34V
  • The Cu2+(aq) + 2e ⇌ Cu(s) half-cell thus has an electrode potential of +0.34V.

Standard electrode potential

  • The standard electrode potential of a half-reaction is the emf of a cell where the left-hand electrode is the standard hydrogen electrode and the right-hand electrode is the standard electrode in question
  • The equation EMF = ERHSELHS  can be applied to electrochemical cells in two ways:
    • Calculating an unknown standard electrode potential
    • Calculating a cell EMF
  • To be a standard electrode potential the measurements must be made at standard conditions, namely:
    • 1.0 mol dm-3 ions concentrations
    • 100 kPa pressure
    • 298 K

Calculating an unknown standard electrode potential

  • If the RHS and LHS electrode are specified, and the emf of the cell measured accordingly, then if the Eθ of one electrode is known then the other can be deduced.
    • For example, if the standard copper electrode (+0.34 V) is placed on the left, and the standard silver electrode is placed on the right, the emf of the cell is +0.46 V.
    • Calculate the standard electrode potential at the silver electrode.


+0.46 = EθAg – (+0.34 V)

EθAg = 0.46 + 0.34 = +0.80 V

Calculating a cell EMF

  • If both SEP’s are known, the EMF of the cell formed can be calculated if the right-hand electrode and left-hand electrode are specified
  • For example, if in a cell the RHS = silver electrode (+0.80V) and LHS is copper electrode (+0.34 V), then


EMF = +0.80 – 0.34 = +0.46 V

Conventional Representation of Cells

  • As it is cumbersome and time-consuming to draw out every electrochemical cell in full, a system of notation is used which describes the cell in full, but does not require it to be drawn.
  • An electrochemical cell can be represented in a shorthand way by a cell diagram





The conventional representation of voltaic cells

  • By convention, the half cell with the greatest negative potential is written on the left of the salt bridge, so Eθcell = Eθright Eθleft
    • In this case, Eθcell = +0.34 – -0.76 = +1.10 V.
  • The left cell is being oxidized while the right is being reduced
  • If there is more than one species in solution, and the species are on different sides of the half-equation, the different species are separated by a comma
  • This method of representing electrochemical cells is known as the conventional representation of a cell, and it is widely used
  • If both species in a half reaction are aqueous then an inert platinum electrode is needed which is recorded on the outside of the half cell diagram

Some Examples

  • For the iron(II) and iron(III) half cell reaction a platinum electrode is needed as an electron carrier
  • The half equation is

Fe3+(aq) + e ⇌ Fe2+(aq)

  • So the cell convention as a left hand electrode would be

Pt 丨Fe2+(aq), Fe3+(aq)

  • Notice the order must be Fe(II) then Fe(III) as the left side is an oxidation reaction, so Fe(II) is oxidised to Fe(III) by the loss of an electron
  • The platinum electrode is separated by the phase boundary (vertical solid line), but the iron(II) and iron(III) are separated by a comma since they are in the same phase
  • Non-metals will also require a platinum electrode
  • If chlorine is used as an electrode the reduction reaction is

Cl2(g) + 2e ⇌ 2Cl(aq)

  • The conventional representation of the half reaction would be

Cl2 (g), 2Cl(aq) Pt 

  • Notice that the half cell reaction is balanced; however, it would be also correct to write it as

Cl2 (g), Cl(aq) Pt 

  • This is because conventional cell diagrams are not quantitative- they are just representations of the materials and redox processes going on
    • Most chemists tend to show them balanced anyway
  • Combining these two half cells together gives

Pt 丨Fe2+(aq), Fe3+(aq)  ∥  Cl2 (g), 2Cl(aq) Pt

  • As you can see the overall cell diagram is not quantitative as the left side is a one electron transfer and the right side is a two electron transfer

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