Appendix 5. E^ø Half–cell potentials/reactions, full redox equations and calculating feasibility via E^ø_reaction

Database of Standard Electrode Potentials, E^ø values

See also Appendix 11 for more on electrode potential charts

• For those mentioned on this web page for aqueous systems under standard conditions,

• i.e. at 298K, 1 mol dm^–3 concentration _(aq), 1 atm. reactant gas pressure (if appropriate),

• and compared with the half–cell potential for the standard hydrogen gas–hydrogen ion electrode (via Pt electrode interface),

  • which is assigned the arbitrary convention value of E^ø_H+(aq)/H2(g) = 0.00 V

    • 2H⁺ _(aq) + 2e^–  H_{2 (g)} 

  • Details on Part 7. Equilibria – Redox systems (opens in new window)

  • Half–cell electrode potential equations are usually quoted as a reduction (as above and list below)

• The half–cell potentials are listed downwards from the strongest reducing agent system (most negative E^ø/V) to the strongest oxidising agent system (the most positive E^ø/V):

  • The (oxidation state changes) are also shown for each half–cell redox potential.

  • –0.76 for Zn²⁺_(aq) + 2e^– Zn_(s)  [Zn(II) ==> Zn(0)]

  • –0.56 for Fe(OH)_3(s) + e^– Fe(OH)_2(s) + OH^–_(aq)  [Fe(III) ==> Fe(II), in alkali]

  • –0.44 for Fe²⁺_(aq) + 2e^– Fe_(s)  [Fe(II) ==> Fe(0)]

  • –0.41 for Cr³⁺_(aq) + e^– Cr²⁺_(aq)  [Cr(III) ==> Cr(II), in acid]

  • –0.26 for V³⁺_(aq) + e^– V²⁺_(aq)  [V(III) ==> V(II), in acid]

  • –0.10 for [Co(NH₃)₆]³⁺_(aq) + e^– [Co(NH₃)₆]²⁺_(aq)   [Co(III) ==> Co(II) for NH₃ ligand]

  • 0.00 for 2H⁺_(aq) + 2e^–  H_2(g)  [the arbitrary assumed standard value, H(+1) ==> H(0)]

  • +0.34 for VO²⁺_(aq) + 2H⁺_(aq) + 2e^– V³⁺_(aq) + H₂O_(l)  [V(IV) ==> V(III)]

  • +0.40 for ¹/₂O_2(g) + H₂O_(l) + 2e^–  2OH^–_(aq)  [O(0) ==> O(–2), in alkali]

  • +0.54 for I_2(aq) + 2e^– 2I^–_(aq)  [I(0) ==> I(–1)]

  • +0.68 for O_2(g) + 2H⁺_(aq) + 2e^– ==> H₂O_2(aq)  [O(0) ==> O(–1)

  • +0.77 for Fe³⁺_(aq) + e^– Fe²⁺_(aq)  [Fe(III) ==> Fe(II), in acid]

  • +0.80 for Ag⁺_(aq) + e^– ==> Ag_(s) (Ag(1) ==> Ag(0)]

  • +1.00 for VO₂⁺_(aq) + 2H⁺_(aq) + 2e^– VO²⁺_(aq) + H₂O_(l)  [V(V) ==> V(IV) in acid]

  • +1.23 for ¹/₂O_2(g) + 2H⁺_(aq) + 2e^–   H₂O_(l)  [O(0) ==> O(–2), in acid???]

  • +1.33 for Cr₂O₇^2–_(aq) + 14H⁺_(aq) + 6e^– 2Cr³⁺_(aq) + 7H₂O_(l)  [Cr(VI) ==> Cr(III)]

  • +1.36 for Cl_2(aq) + 2e^– 2Cl^–_(aq)  [Cl(0) ==> Cl(–1)]

  • +1.51 for MnO₄^–_(aq) + 8H⁺_(aq) + 5e^– Mn²⁺_(aq) + 4H₂O_(l)  [Mn(VII) ==> Mn(II)]

  • +1.52 for Mn³⁺_(aq) + e^– Mn²⁺_(aq) + H₂O_(l)  [Mn(III) ==> Mn(II)]

  • +1.77 for H₂O_2(aq) +  2H⁺_(aq) + 2e^– 2H₂O_(l)  [O(–1) ==> O(–2), in acid?]

  • +1.82 for Co³⁺_(aq) + e^– Co²⁺_(aq)  [Co(III) ==> Co(II) for H₂O ligand]

  • +2.01 for S₂O₈^2–_(aq) + 2e^– 2SO₄^2–_(aq)  [2O(–1) ==> 2O(–2)]

  • –

How standard electrode potentials are determined

Electrode Potential Chart for the 3d–block transition metals

Calculation of E^ø for a redox reaction

• In principle, any accurately known half–cell potential can be used in a cell system to obtain an unknown half–cell potential which can be used to theoretically predict the feasibility of a reaction.

• The electrochemical series and electrode potential charts, know how to construct, read and use them.

• Other half–cells, they don’t have to simple metal/ metal ions, all you need is two interchangeable oxidation states eg Cl_2(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.

• One way of working out E^ø values for a complete reaction:

  • E^ø_{cell (reaction)} =  E^ø_(red) – E^ø_(ox)   ... where ...

  • E^ø_(red) is the half–cell potential of the reduction 'half–reaction' of the oxidising agent.

  • E^ø_(ox) is the half–cell potential of the oxidation 'half–reaction' of the reducing agent.

  • which amounts to the difference between the half–cell potentials on an electrode potential chart.

  • If you consider the copper–zinc cell for the overall reaction

    • Cu²⁺_(aq) + Zn_(s) ==> Cu_(s) + Zn²⁺_(aq)

  • E^ø_(red) is the most positive or the least negative = the strongest oxidising agent or electron acceptor of the two half–cell systems.

    • It is the +ve battery pole, eg Cu/Cu²⁺ (+0.34V) compared to Zn/Zn²⁺ (–0.76V).

    • so the Cu²⁺_(aq) + 2e^– ==> Cu_(s) reduction occurs rather than reduction of Zn²⁺ to Zn.

  • E^ø_(ox) is the least positive or the most negative = the strongest reducing agent or electron donor of the two half–cell potentials.

    • It is the –ve battery pole eg Zn/Zn²⁺ compared to Cu/Cu²⁺,

    • so the Zn_(s) – 2e^– ==> Zn²⁺_(aq) oxidation happens rather than oxidation of Cu to Cu²⁺.

  • For overall cell redox reaction: Cu²⁺_(aq) + Zn_(s) ==> Cu_(s) +Zn²⁺_(aq)  

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

    • E^ø_(red) = E^ø_{Cu(s)/Cu2+(aq)} = +0.34V, E^ø_(ox) = E^ø_{Zn(s)/Zn2+(aq)} = –0.76V

    • E^ø_cell =  E^ø_(red) – E^ø_(ox)= +0.34V – (–0.76) = + 1.10 V (feasible!)

    • E^ø_{overall cell reaction} must be >0 for the reaction to be feasible

For more details and examples see

Equilibrium Part 7 Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series