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Doc Brown's A Level Chemistry Revision Notes

Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 7.4

7.4 Half–cell potentials, Electrochemical Series and calculating & using Eθcell for reaction feasibility

Using Half–cell potentials , the electrochemical series and the relative oxidising power or reducing power of a half–reaction.  How to calculate and use Eθcell (Emf of cell) to determine the feasibility of a redox reaction.

Chemical Equilibrium Notes Part 7 Index


 

 

cell67.4 Half–cell potentials, Electrochemical Series and using Eθcell for reaction feasibility

  • From the experiments described in 7.2 and 7.3 you can measure a half–cell potential for many redox systems.

  • From half–cell potential data you can theoretically calculate the Eθfull cell reaction and hence the thermodynamic feasibility of the reaction.

  • The electrode potential is defined as the Emf of a cell in which the electrode on the left is a standard hydrogen electrode (0.00V under standard conditions, see 7.3 The hydrogen electrode and standard conditions), and that on the right is the electrode in question.

  • Therefore in IUPAC cell notation for the definition and measurement of a half–cell electrode potential

  • Zn2+(aq)|Zn(s)||Pt|H2(g)|H+(aq) gives an Eθcell of –0.76V for the zinc electrode, and for the copper electrode

  • Pt|H2(g)|H+(aq)||Cu2+(aq)|Cu(s) gives an Eθcell of +0.34V, so the arithmetical sign associated with the electrode potential of the electrode in question is the polarity of this electrode when constituting the right–hand electrode, these can be represented in the form of an electrode potential chart (above right).

    • Note the standard conventions in common use

      • the | notation indicates a phase boundary

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

          • the oxidation state increases 'towards' it

  • Any half–cell with a negative potential of less than 0.00V can theoretically reduce aqueous hydrogen ions.

  • Any half–cell with a positive potential of over 0.00V can theoretically oxidise hydrogen molecules.

  • HALF–CELL POTENTIALS are listed below in relative order and the half–cell reactions shown as reductions with oxidation state changes in (), and this constitutes what is called The Electrochemical Series i.e from the most negative half–cell electrode potentials to the most positive half–cell potentials.

    • Note that these electrode potentials help to explain the reactivity series of metals and non-metals e.g.

      • The more reactive a metal the more negative its half-reaction electrode potential, the greater its potential to lose electrons and form a positive ion, Na > Mg > Al > Fe > Cu > etc.

      • The more reactive the non-metal, exemplified by the halogens, the more positive its electrode potential, the greater its potential to accept electrons to form a negative ion, F > Cl > Br > I

  • –2.71 for Na+(aq) + e reversible Na(s)

    • [Na(+1) == Na(0)]

  • –2.37 for Mg2+(aq) + 2e reversible Mg(s)

    • [Mg(+2) ==> (Mg(0)]

  • –1.66 for Al3+(aq) + 3e reversible Al(s)

    • [Zn(II) ==> Zn(0)]

  • –0.76 for Zn2+(aq) + 2e reversible Zn(s)

    • [Zn(II) ==> Zn(0)]

  • –0.56 for Fe(OH)3(s) + e reversible Fe(OH)2(s) + OH(aq)

    • [Fe(III) ==> Fe(II), in alkali]

  • –0.44 for Fe2+(aq) + 2e reversible Fe(s)

    • [Fe(II) ==> Fe(0)]

  • –0.10 for [Co(NH3)6]3+(aq) + e reversible [Co(NH3)6]2+(aq)

    • [Co(III) ==> Co(II) for NH3 ligand]

  • 0.00 for 2H+(aq) + 2e reversible H2(g)  the arbitrary assumed standard

    • [H(+1) ==> (H(0)]

  • +0.34 for Cu2+(aq) + 2e (c) doc b Cu(s)

    • [Cu(+2) ==> Cu(0)]

  • +0.40 for 1/2O2(g) + H2O(l) + 2e reversible 2OH(aq)

    • [O(0) ==> O(–2)]

  • +0.54 for I2(aq) + 2e reversible 2I(aq)

    • [I(0) ==> I(–1)]

  • +0.77 for Fe3+(aq) + e reversible Fe2+(aq)

    • [Fe(III) ==> Fe(II), in neutral or acid solution]

  • +1.09 for Br2(aq) + 2e reversible 2Br(aq)

    • [Br(0) ==> Br(–1)]

  • +1.23 for 1/2O2(g) + 2H+(aq) + 2e reversible  H2O(l)

    • [O(0) ==> O(–2)]

  • +1.33 for Cr2O72–(aq) + 14H+(aq) + 6e reversible 2Cr3+(aq) + 7H2O(l)

    • [Cr(VI) ==> Cr(III)]

  • +1.36 for Cl2(aq) + 2e reversible 2Cl(aq)

    • [Cl(0) ==> Cl(–1)]

  • +1.51 for MnO4(aq) + 8H+(aq) + 5e reversible Mn2+(aq) + 4H2O(l)

    • [Mn(VII) ==> Mn(II)]

  • +1.77 for H2O2(aq) +  2H+(aq) + 2e reversible 2H2O(l)

    • [O(–1) ==> O(–2), in acid?]

  • +1.82 for Co3+(aq) + e reversible Co2+(aq)

    • {Co(III) ==> Co(II) for H2O ligand, [Co(H2O)6]3+ ==> 2+}

  • +2.01 for S2O82–(aq) + 2e reversible 2SO42–(aq)

    • [2O(–1) ==> 2O(–2)]

  • +2.87 for F2(aq) + 2e reversible 2F(aq)

    • [F(0) ==>F(–1)]

  • Relative oxidising and reducing power

    • Down the  list the more positive/less negative the electrode potential the stronger the oxidising power of the half–cell system.

    • Up the list the more negative/less positive the electrode potential the stronger the reducing power of the half–cell system.

    • So at the top of the list above you get the powerful reducing reactive metals like sodium, (–2.71V), and magnesium (–2.37V) with very negative half–cell potentials.

    • At the bottom of the list you get the powerful oxidising agents like potassium manganate(VII), (+1.51V), hydrogen peroxide, (+1.77), peroxodisulphate ion (+2.01V) and fluorine (+2.87).

  • The Electrochemical Series

    • The list of half–cell reactions and half–cell potentials involving the elements is often referred to as the 'Electrochemical Series', though in reality, it is the whole list of all of them whether an element in its elemental state is involved at all.

    • Has as been mentioned already, it gives an accurate prediction of (i) oxidising/reducing power, (ii) reactivity trends for metals or non–metals and (iii) which ions are likely to be preferentially discharged in electrolysis.

  • Electrode potential and patterns of 'reactivity' e.g.

    • The reactivity series of metals: 'Up' the metal reactivity series the half–cell potential voltages becomes more negative and the metal becomes 'more reactive' e.g. Na (–2.71V) > Mg (–2.37) > Zn (–0.76V) > Fe (–0.44) > Cu (+0.34) etc. Metallic elements react by electron loss (ox. state increase) to form a positive cation (e.g. magnesium ion Mg2+), so, as the electron loss potential increases, so the metallic element's reactivity increases. A metallic element more –ve/least +ve potential

    • The Group 7 Halogen reactivity series: Down group 7 the reactivity decreases as the oxidising power decreases. The half–cell potential decreases down the group e.g. F (+2.87) > Cl (+1.36) > Br (+1.09) > I (+0.54V). Halogen elements react by electron gain (ox. state decrease) to form a single covalent bond (e.g. HCl) or the negative anion (e.g. chloride ion Cl in NaCl), so, as the electron accepting capacity power decreases, so does the element's reactivity.

  • Other half–cells, they don’t have to simple metal/ metal ions, all you need is two interchangeable oxidation states

    • e.g. Cl2(aq)/Cl(aq) or Mn2+(aq)/MnO4(aq) etc. but both components of the half–cell must be in the same solution and in contact with a platinum electrode that connects to the rest of the circuit.

  • Relating Eθcell to the direction of overall chemical change and feasibility of reaction.

    • If you calculate a –ve cell voltage for a given reaction, that is not the way cell reaction goes! please reverse the equation and re–calculate!

    • Examples of predicting the feasibility of a reaction by doing a theoretical calculation using the electrode potential data from above. I've included the oxidation state changes too.

      • Remember though that if feasible, a reaction may not take place to any noticeable extent due to several reasons e.g. too low a temperature or a very high activation energy.

    • (i) You probably know that aqueous chlorine can readily oxidise iron(II) ions to iron(III) ions, BUT can aqueous bromine also oxidise iron(ii) ions to iron(III)?

      • First set up the equation is: 2Fe2+(aq) + Br2(aq) ==> 2Fe3+(aq) + 2Br-(aq)

      • Reduction: Br2 ==> Br- (0 to -1); Oxidation: Fe2+ ==> Fe3+ (+2 to +3)

      • Eθreaction =  Eθ(reduced) – Eθ(oxidation) = (+1.09V) - (+0.77V) = +0.32V, >0V so reaction feasible

        • I/I2 half-cell potential is only +0.54V and so iodine cannot oxidise Fe2+ to Fe3+,

        • The equation is: 2Fe2+(aq) + I2(aq) ==> 2Fe3+(aq) + 2I-(aq)

        • Eθreaction =  Eθ(red) – Eθ(ox) = (+0.54V) - (+0.77V) = -0.23V, so NOT feasible

        • Without doing a second calculation you can immediately deduce that iron(III) ions can oxidise iodide ions, the reverse reaction must be feasible!

    • (ii) Can cheap iron displace more expensive zinc from an aqueous solution of a zinc salt?

      • Reaction: Fe(s) + Zn2+(aq) ===> Zn(s) + Fe2+(aq)

      • Reduction: Zn2+ ==> Zn (+2 to 0); Oxidation: Fe ==> Fe2+ (0 to +2)

      • Eθreaction =  Eθ(red) – Eθ(ox) = (-0.76V) - (-0.44V) = -0.32V, so NOT feasible

      • However, cheap (but more 'reactive') iron CAN displace more valuable copper from an aqueous solution of copper(II) sulfate.

      • Reaction: Fe(s) + Cu2+(aq) ===> Cu(s) + Fe2+(aq)

      • Reduction: Cu2+ ==> Cu (+2 to 0); Oxidation: Fe ==> Fe2+ (0 to +2)

      • Eθreaction =  Eθ(red) – Eθ(ox) = (+0.34V) - (-0.44V) = -+0.78V, so feasible

    • (iii) It doesn't matter how complicated the balanced equation, the Eθreaction calculation is quite simple as long as you use the correct electrode potential data!

      • e.g. can potassium manganate(VII) oxidise hydrochloric acid?

      • Balanced equation: 2MnO4(aq) + 16H+(aq) + 10Cl(aq) ==> 2Mn2+(aq) + 8H2O(l) + 5Cl2(aq)

      • Reduction: MnO4 ==> Mn2+ (+7 to +2);  Oxidation: Cl ==> Cl2 (-1 to 0)

      • Eθreaction =  Eθ(red) – Eθ(ox) = (1.51V) - (+1.36V) = +0.15V, so feasible

    • (iv)

  • The Ebattery–cell must be >0.00V for the cell, and any other redox reaction, to be theoretically feasible.

    • The Gibbs Free Energy change ΔG is only on some A level syllabuses.

    • The free energy change must be negative <0 for the cell reaction to be feasible (matching the Ecell rule of >0V for feasibility).

    • ΔGθcell = –nEθF (J)

    • n = number of moles of electrons transferred in the reaction per mol of reactants involved,

    • Eθ is the Emf for the overall reaction in volts i.e. for the theoretical cell,

    • F = the Faraday constant (96500 C mol–1).

    • e.g. for the zinc–copper Daniel cell producing +1.10V,

    • 2 electrons transferred (M2+(aq) + 2e <=> M(s))

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

    • ΔGθcell = –2 x 1.10 x 96500 = –212300 Jmol–1 or –212.3 kJmol–1

    • See also Appendix 2 Free Energy

  • These theoretical calculations can be used for any redox reaction BUT there are limitations:

    • You can’t say the reaction will definitely spontaneously happen (go without help!) because there may be rate limits especially if the reaction has a high activation energy or a very low concentration of an essential reactant.

    • However you can employ a catalyst, raise reactant concentrations or raise the temperature to get the reaction going! There is usually a way of getting most, but not all, feasible reactions to actually occur.

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


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