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Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 7.3

7.3 The Hydrogen Electrode, standard electrode potential and standard conditions

What is, and how is, a standard hydrogen electrode is constructed? The standard electrode potential of a half–cell is based on an arbitrary zero for the hydrogen ion/hydrogen gas half–cell. What is a standard electrode potential? What do we mean by standard conditions?

(c) doc b GCSE/IGCSE reversible reactions & chemical equilibrium notes

and (c) doc b GCSE/IGCSE Notes on Electrochemistry

Part 7 sub–index: 7.1 Half cell equilibria, electrode potential * 7.2 Simple cells notation and construction * 7.3 The hydrogen electrode and standard conditions * 7.4 Half–cell potentials, Electrochemical Series and using Eθcell for reaction feasibility * 7.5 Electrochemical cells ('batteries') and fuel cell systems * 7.6 Electrolysis and the electrochemical series * 7.7 Exemplar Questions, Appendix 1. The Nernst Equation, Appendix 2 Free Energy, Cell Emf and K

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4. Partition, solubility product and ion–exchange * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, Acid–base titrations–indicators, pH curves and buffers * Part 8. Phase equilibria–vapour pressure, boiling points and intermolecular forces


7.3 The Hydrogen Electrode, standard electrode potential and standard conditions

  • You cannot measure or calculate absolute half–cell potentials, so, as in the case of ΔH values, where the elements in their normal stable states at 298K are arbitrarily given the enthalpy value of zero, in electrode potentials, the aqueous hydrogen ion/hydrogen gas half–cell is given the arbitrary standard value of EθH2(g)/H+(aq) = 0.00V under standard conditions.

    • The standard hydrogen half–cell is illustrated as the left–hand side of Fig.4 and Fig.5.

  • The half–cell equation is: 2H+(aq) + 2e (c) doc b H2(g)  for the standard hydrogen electrode.

    • The hydrogen electrode is quite difficult to set up accurately with a specially prepared inert platinum electrode, and in practice, other accurate reference electrodes are used e.g. the calomel electrode.

    • You should also remember that half–cell potentials are dependent on the concentration of any ion involved in any of the electrode reactions, whether a product or reactant.

  • θ Means standard conditions, here, for any half–cell electrode system

    • 298 K for temperature (25oC), for any standard electrode

    • 1 atmosphere pressure (101kPa) of hydrogen gas H2(g) for hydrogen electrode,

    • 1 mol dm–3 concentration of H+(aq) (hydrogen ions for hydrogen electrode),

      • e.g. 1 mol dm–3 HCl(aq) maybe quoted with caution, BUT they represent an unfortunate over–simplification, read on!

      • ... meaning the choice of acid requires some comment.

      • Hydrochloric acid is the most widely quoted (exam board mark schemes, textbooks, internet technical references etc), and at school/college pre–university level, it seems perfectly acceptable in answers to quote the electrolyte for the hydrogen electrode as 1.00 mol dm–3 HCl(aq) i.e. equal to 1.00 mol dm–3 H+(aq)

        • Hydrochloric acid is an uncomplicated monobasic/monoprotic acid, non–oxidising and is fully ionised in water.

        • With a pKa = –7, Ka = 1.0 x 107 mol dm–3, no problem on ionisation here! and a 1 molar solution theoretically gives a 1 molar solution of hydrogen ions.

        • However if you research this aspect of the hydrogen electrode on the internet, you find that a 1.8 molar solution of hydrochloric acid may be used to give an activity' of [H+(aq)] of 1.00 mol dm–3, but please do not worry about this, this is undergraduate stuff!!

          • I'm not going to the details of why, but unfortunately, in any equilibrium situation, solutions do NOT behave ideally and I'm afraid 1 molar HCl ain't 1 molar H+(aq).

        • However, in pH calculations at pre–university level, its acceptable to assume this, so why not in an answer to a hydrogen electrode question?

          • There is an electrode potential limit to the use of hydrochloric acid.

          • The half–cell potential for Cl2/Cl is +1.36V, therefore if the other cell has an electrode potential of > +1.36V, you may get oxidation of the chloride ion at the hydrogen electrode.

          • This is NOT the standard reaction required, i.e. you will not measure the correct cell voltage from which to calculate the standard electrode potential of the half–cell under investigation!

      • Sulfuric acid is sometimes quoted i.e. 0.50 mol dm–3 H2SO4(aq), should on complete ionisation give a 1 molar solution of hydrogen ions.

        • However, it is a dibasic/diprotic acid, the first ionisation is ~100% complete (pKa1 very negative, Ka extremely high), therefore ~100% ionisation.

        • However, the 2nd ionisation/dissociation is far from complete,

          • pKa2 = 1.92, Ka = 1.20 x 10–2 mol dm–3, so sulfuric acid is not fully ionised and a 0.50 molar solution of sulphuric acid will not give a 1.00 molar solution of hydrogen ions.

        • However, in pH calculations at pre–university level, it seems acceptable to assume this, so why not in an answer to a hydrogen electrode question?

          • BUT I've also done a calculation to show that a 0.500 molar solution of sulfuric acid only contains 0.511 mol dm–3 of hydrogen ions, which means technically, 0.50 molar sulphuric acid is not a good answer!

          • See bottom of equilibrium page 5b on strong acid calculations for more details on this.

      • Nitric acid may seem a good monobasic/monoprotic alternative to hydrochloric acid.

        • It's pKa = –1.4, Ka = 40 mol dm–3, so is pretty well fully ionised (~97.5%), but it is a strong oxidising agent and may theoretically oxidise the hydrogen gas and there are other possible complications too.

      • Phosphoric acid would be useless, its a tribasic/triprotic acid, too complicated, and the first ionisation is far from complete (see below), quoting 0.333 molar H3PO4 is not a good idea!

        • H3PO4(aq) (c) doc b H+(aq) + H2PO4(aq)

        • for H3PO4(aq) pK1 = 2.15, Ka1 = 7.08 x 10–3 mol dm–3, and the other two Ka values are much smaller!

      • Final comment and conclusion ...

        • As far as I can tell, 1.00 mol dm–3 HCl(aq) will be the more accurate to use than 0.5 mol dm–3 H2SO4(aq), but don't quote anything else, and there is an alternative and absolutely correct and universal answer!

        • The fact of the matter is, at this pre–university academic level, a simplified answer may suffice, as to which acid, and its concentration, to use in a hydrogen electrode, but my advice is, make sure quote in any answer ...

        • .. an acid solution equal to a 1 molar solution of hydrogen ions

          • i.e. [H+(aq)] = 1.00 mol dm–3, and that aught to satisfy any answer scheme, and if it doesn't, then chief examiners need to do some research for themselves!

    • 1 mol dm–3 concentration of metal ion in the other half–cells mentioned in these pages.

      • e.g. 1.0 mol dm–3 concentration of Zn2+(aq) or Cu2+(aq) ions.

      • Bearing in mind the above discussion on the hydrogen ion concentration in standard cells, it should be absolutely correct in an exam to quote molarities of all ions as 1 molar for any standard cell at 298K, 1 atm.

  • The standard electrode potential of system is defined as the e.m.f. of an electrochemical cell in which at 298K, the hydrogen gas (1 atm)/aqueous hydrogen ion (1M [H+(aq)] half–cell electrode is coupled with the other half–cell electrode system i.e. the half–cell potential of a system is its e.m.f. compared to the standard hydrogen electrode

  • In principle, any accurately known half–cell potential can be used in a cell system to obtain an unknown half–cell potential by measuring the voltage of the complete cell using a high resistance voltmeter at virtually zero current flow.

    • Why use a high resistance voltmeter?

    • By minimising the current flow to ~zero, inaccurate readings due to polarisation are avoided.

    • Polarisation is caused by the build up of the cell reaction products or the diminution of reactants.

    • Any change in concentrations will affect the half–cell potential, giving non–standard readings.

  • The diagram Fig.4 below shows the measurement of the standard copper atom/copper(II) ion half–cell potential using a standard hydrogen electrode.

  • The hydrogen electrode consists of an 'inlet' system for hydrogen gas to come into contact with hydrogen ions into which is dipped a platinum electrode.

    • The platinum electrode must specially prepared and beyond the scope of these notes, but more importantly, the Pt electrode allows electron transfer between the aqueous hydrogen ions and hydrogen gas, ideally creating true equilibrium conditions.

    • When connected to another half–cell and high resistance voltmeter a salt bridge (electrolyte solution/gel) is needed to complete the electrical circuit and allows the slow drift of ions in both directions (NOT electrons). 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.

cell2 Fig.4

  • The copper/aqueous copper(II) ion half–cell registers +0.34V with respect to the hydrogen electrode. The electron flow is from the most negative/least positive potential half–cell (–ve pole) to the most positive/least negative potential half–cell (–ve pole), i.e. H2/H+ to Cu/Cu2+.

    • TOP and LINKSCell notation (matching diagram and IUPAC convention)

    • Pt|H2(g)|H+(aq)||Cu2+(aq)|Cu(s)

      • 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

    • On setting up the cell it is found that the copper strip is the positive electrode i.e. where the reduction occurs.

    • Eθcell = +0.34V = Eθ(+ve/red) – Eθ(–ve/ox) = Eθ(Cu2+/Cu) – Eθ(H+/H2)

    • Eθ(Cu2+/Cu) = Eθcell + Eθ(H+/H2) = +0.34 + 0.00 = +0.34V

    • for the feasible cell reaction: Cu2+(aq) + H2(g) ==> Cu(s) + 2H+(aq)

    • Here, polarisation is caused by the build up of the cell reaction products e.g. H+ ions from the oxidation of the hydrogen gas and the diminution of the copper ions Cu2+ on their reduction to copper.

    • This is reduced by using a very high resistance voltmeter.

  • The diagram Fig.5 below shows the measurement of the standard zinc atom/zinc ion half–cell potential using a standard hydrogen electrode. The hydrogen electrode is kept on the left to compare the zinc/zinc ion with the copper/copper(II) ion half–cell above.

cell3 Fig.5

cell4 Fig.1

  • The zinc/aqueous zinc ion half–cell registers –0.76V with respect to the hydrogen electrode. In this case the electron flow is the opposite of the copper–hydrogen cell in Fig.4. the electron flow is always from the most negative/least positive potential half–cell (–ve pole) to the most positive/least negative potential (+ve pole), i.e. here it is Zn/Zn2+ to H2/H+. This is the complete opposite of the effect of the Cu/Cu2+ half–cell (see Fig.1 above).

    • TOP and LINKSCell notation (to match the diagram)

    • Pt|H2(g)|H+(aq)||Zn2+(aq)|Zn(s)

      • 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

    • On setting up the cell it is found that the zinc strip is negative electrode i.e. where the oxidation occurs.

    • Eθcell = –0.76V, so Eθ(Zn2+/Zn) = –0.76V

    • for the cell reaction: Zn2+(aq) + H2(g) ==> Zn(s) + 2H+(aq)

    • but since the cell voltage is negative, this is NOT the feasible, so

    • Eθcell = –0.76V +0.76V for the feasible cell reaction

    • Zn(s) + 2H+(aq) ==> Zn2+(aq) + H2(g)

    • IUPAC cell notation Zn(s)|Zn2+(aq)¦¦H+(aq)|H2(g)|Pt

    • Here, polarisation is caused by the build up of the cell reaction products e.g. Zn2+ ions from the oxidation of the zinc electrode and the diminution of hydrogen ions H+ on their reduction to hydrogen gas.

    • This is reduced by using a very high resistance voltmeter.

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

  • Fig.6 shows how to measure the standard potential of a half–cell consisting of an element in two different oxidation states in aqueous solution.

    cell5 Fig. 6

  • The diagram above illustrates in principle how the half–cell potential for the inter–conversion of iron(II)/iron(III) ions can be measured. The value obtained is +0.77 so the electrons move from negative pole (Pt) of the H2/H+ half–cell to the positive pole (Pt) of the Fe2+/Fe3+ half–cell.

    • Fe3+ (aq) + e (c) doc b Fe2+ (aq)  [Fe(III) => Fe(II)]

    • Cell notation (to match the diagram and IUPAC convention)

    • Pt|H2(g)|H+(aq)||Fe3+(aq),Fe2+(aq)|Pt

      • 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

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

  • You can do the same for many other systems mixing solutions of the two species in the 'half–cell beaker e.g.

    • manganese(II) and manganate(VII) ions in acidified solution

      • MnO4(aq) + 8H+(aq) + 5e (c) doc b Mn2+(aq) + 4H2O(l)   [Mn(VII) => Mn(II)]

    • aqueous chlorine molecule and the chloride ion

      • Cl2(aq) + 2e (c) doc b 2Cl(aq)  [Cl(0) => Cl(–1)]

    • oxo–vanadium(V) cation and the oxo–vanadium(IV) cation

      • VO2+(aq) + 2H+(aq) + 2e (c) doc b VO2+(aq) + H2O(l)  [V(V) => V(IV)]

    • aqueous iodine molecule and the iodide ion

      • I2(aq) + 2e (c) doc b 2I(aq)  [I(0) => I(–1)]

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

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Part 7 sub–index: 7.1 Half cell equilibria, electrode potential * 7.2 Simple cells notation and construction * 7.3 The hydrogen electrode and standard conditions * 7.4 Half–cell potentials, Electrochemical Series and using Eθcell for reaction feasibility * 7.5 Electrochemical cells ('batteries') and fuel cell systems * 7.6 Electrolysis and the electrochemical series * 7.7 Exemplar Questions, Appendix 1. The Nernst Equation, Appendix 2 Free Energy, Cell Emf and K

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibria and industrial processes * Part 4. Partition, solubility product and ion–exchange * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, Acid–base titrations–indicators, pH curves and buffers * Part 8. Phase equilibria–vapour pressure, boiling points and intermolecular forces

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