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5.1Unit 5. Transition Metals and Organic Nitrogen Chemistry

Redox Equilibria — IAL Edexcel Modular (Chemistry) Revision Notes & Topical Questions

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5.1 Redox Equilibria — IAL EDEXCEL MODULAR (CHEMISTRY) notes: Define and measure standard electrode potentials, use E° values to predict feasibility, and carry out redox titration calculations.Definitions — oxidation is loss of electrons (increase in oxidation number) and reducti…

  • Half-equations
  • s-block metals
  • p-block metals
  • p-block non-metals
  • d-block elements
  • Why a reference is needed

Section 1 of 12Oxidation and reduction

Definitions
  • Definitions oxidation is loss of electrons and reduction is gain of electrons (decrease in oxidation number), remembered as OIL RIG.
    • (increase in oxidation number)
Half-equations
  • Half-equations are combined so that the electrons cancel.
  • In acidic conditions balance O with H2O and then H with H+.
s-block metals
  • s-block metals are oxidised to 1+ and 2+ ions.
  • Na → Na+ + e- and Ca → Ca2+ + 2e-.
p-block metals
  • p-block metals are oxidised to positive ions, usually with a charge matching the group.
  • Al → Al3+ + 3e-.
  • Some form ions that do not match the group, such as Sn2+.
p-block non-metals
  • p-block non-metals are usually reduced to negative ions with charge = group number - 8.
  • F + e- → F- and O + 2e- → O2-.
d-block elements
  • d-block elements are oxidised to positive ions with variable oxidation states such as Cu2+, Cr3+ and V5+.

Section 2 of 12Standard electrode potential and the standard hydrogen electrode

Standard electrode potential and the standard hydrogen electrode
Why a reference is needed
  • Why a reference is needed the position of equilibrium depends on temperature, gas pressure and concentration,
    • so potentials are compared against a common standard.
Standard conditions
  • Standard conditions ion concentration 1.00 mol dm-3, temperature 298 K and gas pressure 100 kPa.
Standard electrode potential (E°)
  • Standard electrode potential (E°) the potential difference produced when a standard half-cell is connected to a standard hydrogen electrode under standard conditions.
  • A high-resistance voltmeter is used so no current flows and the maximum potential difference is obtained.
Standard hydrogen electrode (SHE)
  • H2 gas at 100 kPa in equilibrium with 1.00 mol dm-3 H+ on an inert platinum electrode, 2H+ + 2e- ⇌ H2, assigned E° = 0.00 V.
  • Platinum black (finely divided platinum) increases the surface area and so the rate at which equilibrium is reached.
Half-equations
  • Half-equations are written as reductions.
  • A more positive E° means the species on the left is more easily reduced and is a stronger oxidising agent.
  • A more negative E° means the species on the right is more easily oxidised and is a stronger reducing agent.
  • Br2 + 2e- ⇌ 2Br- has E° = +1.09 V and Na+ + e- ⇌ Na has E° = -2.71 V.

Section 3 of 12Measuring a standard electrode potential

Measuring a standard electrode potential
Three types of half-cell
  • Three types of half-cell each is connected to the SHE.
  • Metal / metal ion, such as Ag+/Ag (E° = +0.80 V).
  • Non-metal / non-metal ion, such as Br2/Br- (E° = +1.09 V), using a platinum electrode.
  • Ion / ion in different oxidation states, such as MnO4-/Mn2+ (E° = +1.52 V) or Fe3+/Fe2+, using a platinum electrode.
Platinum electrode
  • Platinum electrode is inert like graphite, conducts, and provides the surface where equilibrium is established.
Poles
  • Poles the half-cell with the more positive E° is the positive pole where reduction occurs;
    • the other is the negative pole where oxidation occurs.
1
Set up the SHE as the reference half-cell.
2
Set up the test half-cell with the metal in 1.00 mol dm-3 of its ions, or platinum in equal concentrations of both ions.
3
Keep the temperature at 298 K and gas pressures at 100 kPa.
4
Join the solutions with a salt bridge such as KNO3(aq).
5
Connect the electrodes through a high-resistance voltmeter.
6
Read the voltage, which is E° of the test half-cell.

Section 4 of 12The salt bridge and cell potential

Salt bridge
  • Salt bridge completes the circuit by letting ions move between the half-cells and balances charge as electrons flow through the wire.
  • A metal wire cannot be used because it carries electrons, not ions.
  • KNO3 or KCl is used because the ions are highly soluble, unreactive and do not form precipitates.
Standard cell potential
  • E°cell = E°right - E°left = E°reduction - E°oxidation.
    What the symbols mean: Ecell = standard cell potential (V) · Eright = electrode potential of the positive (reduction) half-cell · Eleft = potential of the negative half-cell (V)
  • The half-cell with the more positive E° is the positive pole and is written on the right.
  • The half-cell with the less positive E° is the negative pole and is written on the left.
  • Zn/Cu cell+0.34 - (-0.76) = +1.10 V.
Memory aid
  • Left Is Oxidation, so the order is reduced/oxidised | | oxidised/reduced.

Section 5 of 12Conventional cell representation

Cell diagram
  • Cell diagram describes a cell without drawing it, with the more negative (oxidation) half-cell on the left of the salt bridge.
  • Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s).
Symbols
  • | is a phase boundary, || is the salt bridge and a comma separates species in the same phase.
Order of species
  • Order of species the reduced form is written next to the electrode on the left and the oxidised form next to the salt bridge, reversed on the right.
Platinum electrode
  • Platinum electrode is written at the outside of the diagram when both species are aqueous or when a non-metal is used.
  • Pt | Fe2+(aq), Fe3+(aq) || Cl2(g), 2Cl-(aq) | Pt.
Not quantitative
  • Not quantitative diagrams show the materials and processes, so balancing coefficients are optional.

Section 6 of 12Feasibility, entropy and equilibrium constant

Feasibility, entropy and equilibrium constant
Feasible reaction
  • Feasible reaction occurs when E°cell is positive, with the more positive half-cell reduced and the other oxidised.
  • Cl2/Cl- (+1.36 V) with Cu2+/Cu (+0.34 V) gives Cu + Cl2 → Cu2+ + 2Cl- with E°cell = +1.02 V.
  • The reverse reaction has E°cell = -1.02 V and is not feasible.
Link to entropy
  • E°cell is proportional to ΔStotal.
Link to equilibrium constant
  • E°cell is proportional to ln K.
  • These follow from ΔG = -nFE°cell and ΔG = -RT ln K.
  • A value above about +0.3 V often means the reaction goes almost to completion.

Section 7 of 12Limitations of electrode potential predictions

No rate information
  • E° shows thermodynamic feasibility only, so a reaction may be too slow to see.
High activation energy
  • Cu2+ + H2 → Cu + 2H+ has E°cell = +0.34 V but does not occur because the reactants are kinetically stable.
Non-standard conditions
  • Non-standard conditions changing a concentration shifts the equilibrium (Le Chatelier) and so changes the electrode potential.
  • For V3+ + e- ⇌ V2+, extra V3+ shifts the equilibrium right, removing electrons and making E less negative.
  • Extra V2+ shifts it left, adding electrons and making E more negative.
Non-aqueous reactions
  • Non-aqueous reactions many redox reactions do not take place in aqueous solution, so E° values do not apply.

Section 8 of 12Fuel cells

Fuel cells
Fuel cell
  • Fuel cell an electrochemical cell in which a fuel donates electrons at one electrode and oxygen gains electrons at the other.
  • Fuel and oxygen are supplied continuously, so energy is not stored in the cell.
Structure
  • Structure separate inlets for H2 and O2, platinum-coated electrodes to catalyse the reactions, an electrolyte, a semi-permeable membrane and a water outlet.
Alkaline cell
  • Alkaline cell anode (negative): H2 + 2OH- → 2H2O + 2e-, E = -0.83 V;
    • cathode (positive): O2 + 2H2O + 4e- → 4OH-, E = +0.40 V.
Acidic cell
  • Acidic cell anode: H2 → 2H+ + 2e-, E = 0.00 V; cathode: O2 + 4H+ + 4e- → 2H2O, E = +1.23 V.
Overall
  • 2H2 + O2 → 2H2O with E°cell = +1.23 V in both cells.
Advantages
  • Advantages water is the only product, no nitrogen oxides form and energy converts to electricity more efficiently than in combustion.
  • Used on spacecraft where the water is drunk.
Disadvantages
  • Disadvantages hydrogen is flammable and costly to store in thick-walled cylinders.
  • Production often relies on crude oil, a finite resource.
  • Energy per unit volume is low, so large containers are needed.

Section 9 of 12Redox titrations

1Principle
  • Principle an oxidising agent is titrated against a reducing agent and electrons are transferred.
  • Most transition metal ions change colour when their oxidation state changes, so an indicator is often unnecessary.
2Manganate(VII) with iron(II)
  • MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O.
  • Acidify with dilute sulfuric acid because it does not react with MnO4-.
  • Hydrochloric acid is unsuitable because Cl- would be oxidised to Cl2.
  • KMnO4 is self-indicating: purple turns colourless, and the end point is the first permanent pale pink.
3Iodine with thiosulfate
  • 2S2O32- + I2 → S4O62- + 2I-.
  • Iodine is yellow-brown and fades to straw colour.
  • Add starch near the end point to give a blue-black colour that disappears at the end point.
4Finding an oxidising agent
  • Finding an oxidising agent the oxidant liberates I2 from excess KI, which is titrated with thiosulfate.
  • ClO- + 2I- + 2H+ → Cl- + I2 + H2O.

Section 10 of 12Redox titration calculations and uncertainty

TitrantAnalyteMole ratioEnd point
KMnO4 (acidified)Fe2+1 : 5first permanent pale pink
KMnO4 (acidified)C2O42-2 : 5pale pink (warm to about 60 °C)
Na2S2O3I22 : 1blue-black to colourless (starch)
K2Cr2O7Fe2+1 : 6redox indicator colour change
1
Write the half-equations for the oxidant and reductant and deduce the overall equation.
2
Calculate moles of titrant = concentration x volume (dm3).
3
Use the equation ratio to find moles of analyte in the sample solution.
4
Scale up to the whole solution, for example x10 if 25.0 cm3 was taken from 250 cm3.
5
Convert to concentration, mass or percentage purity = mass of pure substance / mass of sample x 100.
  • Percentage uncertainty
    • x 100.
      • (absolute uncertainty / measured value)
    • A reading taken twice (burette start and end) doubles the absolute uncertainty.
    • Using a more dilute solution in the burette gives a larger titre and a smaller percentage uncertainty.

Section 11 of 12Key terms

TermMeaning
Standard electrode potential (E°)The emf of a half-cell measured against the standard hydrogen electrode under standard conditions (298 K, 100 kPa, 1.00 mol dm⁻³).
Standard hydrogen electrodeH₂ gas at 100 kPa over platinum in 1.00 mol dm⁻³ H⁺ at 298 K, assigned E° = 0.00 V.
Salt bridgeA connection (for example filter paper soaked in KNO₃) that lets ions move between half-cells to complete the circuit.
E°cellThe standard emf of a cell: E°right - E°left.
Fuel cellA cell that converts the chemical energy of a continuously supplied fuel and oxidant directly into electrical energy.
Oxidising agentA species that accepts electrons and is itself reduced.
Redox titrationA titration in which titrant and analyte react by electron transfer, for example MnO4− with Fe2+.

Section 12 of 12Exam technique

1
Using HCl to acidify a manganate(VII) titration — Cl⁻ is oxidised to Cl₂, so dilute sulfuric acid must be used.
2
Predicting that a positive E°cell means a fast reaction — it shows feasibility only and a high activation energy can stop it happening.
3
Reversing the subtraction for E°cell — it is E°right - E°left, that is E°reduction - E°oxidation.
4
Using a metal wire as a salt bridge — the bridge must carry ions, not electrons.
5
Forgetting to scale up from the aliquot to the whole solution in a titration calculation.

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