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Doc Brown's A Level Chemistry Revision Notes

Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 7.2

7.2 Simple cell notation and construction, electrode potential chart

You should know that redox reactions take place in electrochemical cells where electrons are transferred from the reducing agent to the oxidising agent indirectly via an external circuit. How do you write electrochemical cell notation? how do you make a simple cell from two half–cells to make a simple battery? What is an electrode potential chart? How do you work out the voltage produced by a simple cell from the two half–cell electrode potentials involved?

Chemical Equilibrium Notes Part 7 Index


7.2 Constructing a simple cell or battery, and electrode potential chart

Fig.2 below shows the full explanatory diagram of how a zinc–copper Daniel cell works

  • A simple cell like the Daniel cell ('battery') is made by setting up a situation in which the two half–cell reactions are kept physically separated.

  • To complete the electrical circuit, the zinc and copper(II) ion solutions are connected by a salt bridge (a supported ion solution).

  • Wires connect strips of zinc and copper metal to a voltmeter or other device to complete the circuit.

  • The energy release from the exothermic reaction is not in the form of heat energy but in the form of electrical energy because the electrons from the two electron transfers of the two half–cell reactions are 'forced' around the circuit due to the potential difference between the two half–cells.

cell1 Fig.2

  • The diagram shows how to set up a simple electrochemical cell (galvanic/voltaic cell or battery), and relates the  direction of chemical changes (on electrodes) to the +ve and –ve terminals and the direction of electron flow. The left and right metal–metal ion solution 'beakers' constitute the two half–cells which together make up the full cell.

    • Cell notation (matching diagram, NOT IUPAC convention?)

    • Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)

      • Note the standard conventions in common use

        • the | notation indicates a phase boundary

          • the || notation 'divides' but connects the two half cells and

            • the oxidation state increases 'towards' it

    • Note a single vertical line | signifies an electrode interface or phase boundary and a double vertical line  ||  signifies a salt bridge of a suitable electrolyte such as potassium chloride or ammonium nitrate.

    • When two half–cells are connected together with the high resistance voltmeter to make a complete working cell, a salt bridge, a conducting electrolyte solution/gel, is needed to complete the electrical circuit and allows the slow drift of ions in both directions, NOT electrons – they only flow between the metal electrodes via the external circuit wire and high resistance voltmeter.

      • The best salt bridge contains a gel of potassium chloride solution, but even a bit of filter paper soaked in saturated potassium chloride solution will work for simple school/college experiments.

    • Cell notation: See Ecell calculation below and 7.3 The hydrogen electrode and standard conditions

    • If a half-cell contains a mixture of aqueous ions involved in a redox change, both ions must be shown, separated by a comma.

      • e.g.  Fe(s)|Fe2+(aq), Fe3+(aq)    (two interchangeable oxidation states of iron).

  • Introduction to Ecell calculations. As explained in 7.1 Half cell equilibria, electrode potential the chemical potential of zinc to form a zinc(II) ion is much greater than the chemical potential of copper to form a copper(II) ion and the difference in 'tendency' or chemical potential can be exploited in the construction of a simple cell (battery) which demonstrates that half–cell redox equilibria exist and when connected to form a circuit the equilibria head in the direction dictated by the chemical potentials. In electrolysis these reversible equilibria are forced in a particular direction by an external applied potential difference (p.d., volts).

  • A simple cell or battery is made from combining two half–cells and the system can be used to determine electrode potentials and the data provided can be used to theoretically 'test' the feasibility of a redox reaction.

    • The Daniel Cell, shown above in Fig.2, is a simple 'battery' system for producing electrical energy from chemical potential energy.

  • The half–cell potential is a measure of the tendency or chemical potential (in a thermodynamic sense) of a species to lose/gain electrons in the context of a half–cell equation.

  • Half–cell potentials are measured in volts compared to the standard hydrogen electrode (EθH+(aq)/H2(g) = 0.00V, fully described and explained in 7.3 The hydrogen electrode and standard conditions.

  • The more positive or less negative the potential, the greater the tendency of the half–cell to act as reduction change. The less positive or more negative the electrode potential, the greater the tendency for the half–cell reaction to be one of oxidation.

  • Zn(s) ==> Zn2+(aq) + 2e  (EθZn2+(aq)/Zn(s)* = –0.76V (most –ve or least +ve, oxidation, electron loss)

  • and  Cu2+(aq) + 2e ==> Cu(s)  (EθCu2+(aq)/Cu(s)* = +0.34V (most +ve or least –ve, reduction, electron gain)

    • * the oxidised form is written 1st

  • To obtain standard measurement of Eθcell (cell Emf) the metal strips of the half–cells are wired in series with a high resistance voltmeter for accuracy.

    • This avoids significant current flow causing polarisation, that is the build up of products or the diminution of reactants, either of which will cause a change in the cell Emf.

  • To complete the circuit a 'salt bridge' connects the two solutions of the half–cells.

    • This consists of a conducting  electrolyte solution of 'inert' ions e.g. potassium chloride.

    • The solution is supported e.g. crudely with soaked filter paper or a gel held in a U tube.

    • The ions are free to move and can carry current in any direction, but must not chemically react with anything.

    • We now need to consider how to work out theoretically the Emf the cell in operation (in crude terms, the theoretical 'battery voltage', but best to talk about cell Emf or Eθcell in doing calculations from standard electrode potential date.

  • cell6One way of working out Eθcell values

    • Strictly speaking the IUPAC convention for cell notation would require the cell notation to be

    • Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)

    • with the positive pole on the right so that for a feasible reaction Eθright – Eθleft gives a value of >0V.

    • Eθcell =  Eθ(right) – Eθ(left)

    • Eθcell =  Eθ(+ve/red) – Eθ(–ve/ox)

      • Eθcell =  Eθ(reduction) – Eθ(oxidation)

        • Eθ(reduction) is the most positive or the least negative of the two electrode potentials.

        • Eθ(oxidation) is the least positive or the most negative of the two electrode potentials.

    • which amounts to the difference between the half–cell potentials on an electrode potential chart (see Fig 3. below).

      • Eθ(+ve/red)  is the half-reaction electrode potential for the half-cell reduction i.e. Cu2+ + 2e- ==> Cu

      • Eθ(–ve/ox)  is the half-reaction electrode potential for the half-cell oxidation i.e. Zn ==> Zn2+ + 2e-

    • Eθ(+ve/red) is the most positive or the least negative half–cell potential, the strongest oxidising agent or electron acceptor of the two half–cell systems, and the +ve pole of the cell, e.g. Cu2+/Cu compared to Zn2+/Zn,

    • so the reduction (red) Cu2+(aq) + 2e ==> Cu(s), rather than reduction of Zn2+ to Zn.

    • Eθ(–ve/ox) is the least positive or the most negative half–cell potential, the strongest reducing agent or electron donor of the two half–cell potentials, and the –ve battery pole e.g. Zn2+/Zn compared to Cu2+/Cu,

    • so the oxidation (ox) Zn(s) – 2e ==> Zn2+(aq) happens rather than oxidation of Cu to Cu2+,

    • overall cell redox reaction: Cu2+(aq) + Zn(s) ==> Cu(s) + Zn2+(aq)

    • Calculating the voltage–Emf for the copper–zinc cell: 

      • Eθ(+ve pole/red) = EθCu2+(aq)/Cu(s) = +0.34V

      • Eθ(–ve pole/ox) = EθZn2+(aq)/Zn(s) = –0.76V

      • Eθcell =  Eθ(+ve pole/red) – Eθ(–ve pole/ox) = (+0.34V) – (–0.76) = + 1.10 V

      • If the voltage cell Emf is positive the cell reaction is feasible! as calculated based on the cell reaction with the matching oxidation (–ve pole) and reduction (+ve pole) potentials.

      • If in the calculation a negative voltage results, the reaction is NOT feasible, but will in fact go in the reverse direction.

  • A note on cell electrode potential measurements (cell EMF) and polarisation.

    • Half–cell potentials are dependent on the concentration of any ion involved in any of the electrode reactions, whether a product or reactant.

    • Polarisation is caused by the build up of the cell reaction products e.g. Zn2+ ions from oxidation of zinc electrode, or, the diminution of reactants e.g. copper ions Cu2+ on their reduction to copper.

    • Any changes in concentration will affect the half–cell potential of any electrode i.e. creating different equilibrium non–standard conditions.

    • This is reduced by using a very high resistance voltmeter to give the most accurate readings because if very little, if any, current flows, this will minimise the extent of the cell reaction and hence minimise any concentration changes.

cell6Fig.3 An example of a simple electrode potential chart (right).

The basis of the calculation can be illustrated by way of a simple electrode potential chart like the one shown on the right. The Emf produced by the cell is the difference between the two half–cell potentials. Note the relative reducing/oxidising power potential trends.

See also the list of half–cell potentials in 7.4 which constitutes a 'chart' when set out in relative potential order.

The most positive (or least negative) half–cell potential has the stronger the oxidising power of the two half–cells in a complete cell/reaction, so it will constitute the reduction change in the overall cell reaction.

The more negative (or least positive) half–cell has the stronger reducing power of the two half–cells, so it will constitute the oxidation change in the full cell reaction. The full cell potential is expressed as the difference between the two half–cell potentials ...

Eθcell = Eθhalf–cell of positive pole = most +ve or least –ve Eθ for the reduction half–cell reaction

– Eθhalf–cell of negative pole = least +ve or most –ve Eθ for the oxidation half–cell reaction

which I've expressed simply in the Daniel Zn–Cu cell above, and subsequent examples below,

as Eθcell = Eθ(+ve/red) – Eθ(–ve/ox)

e.g. Eθcell = EθCu2+(aq)/Cu(s) – EθZn2+(aq)/Zn(s)

so  Eθcell = (+0.34V) – (–0.76V) = +1.10V

For the effect of changing concentration on the value of the electrode potential see Appendix 1. The Nernst Equation


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