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5.1Unit 5. Transition Metals and Organic Nitrogen Chemistry

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

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5.1 Redox Equilibria — electrode potentials, cells, feasibility, fuel cells and redox titrations — IAL EDEXCEL LINEAR (CHEMISTRY) notes: Define and measure standard electrode potentials, write and use cell diagrams, predict feasibility and its limits, describe fuel cells and carr…

  • Standard electrode potential, E⦵
  • What the symbols mean
  • Zn / Cu cell
  • Standard hydrogen electrode
  • Salt bridge
  • Standard cell potential

Part 1 of 10Oxidation and reduction in the blocks

OIL RIG
  • OIL RIG oxidation is loss of electrons and reduction is gain of electrons.
Oxidation is also an increase in oxidation number; reduction is a decrease.
s-block metals are oxidised to 1+ and 2+ ions: Na → Na+ + e−, Ca → Ca2+ + 2e−.
p-block metals are oxidised to positive ions.
  • Al → Al3+ + 3e− matches the group.
  • Sn → Sn2+ + 2e− does not match the group.
p-block non-metals are reduced to negative ions with charge = group number − 8.
  • F + e− → F− and O + 2e− → O2−.
d-block elements have variable oxidation states and are usually oxidised to positive ions, e.g. Cu2+, Cr3+, V5+.
In acidic conditions balance O with H2O and H with H+; combine half-equations so electrons cancel.

Part 2 of 10Standard electrode potential

Standard electrode potential
The position of equilibrium, and so the electrode potential, depends on temperature, gas pressure and concentration.
Standard conditions
  • Ion concentration 1.00 mol dm−3.
  • Temperature 298 K.
  • Pressure 100 kPa.
Standard electrode potential, E⦵
  • Standard electrode potential, E⦵ the potential difference when a standard half-cell is connected to a standard hydrogen electrode under standard conditions.
A high-resistance voltmeter is used so that no current flows and the maximum potential difference is obtained.
Half-equations are written as reductions; a more positive E⦵ means the species on the left is more easily reduced (a stronger oxidising agent).
Half-equationE⦵ / V
Na+ + e- ⇌ Na-2.71
Zn2+ + 2e- ⇌ Zn-0.76
2H+ + 2e- ⇌ H20.00
Cu2+ + 2e- ⇌ Cu+0.34
Fe3+ + e- ⇌ Fe2++0.77
Ag+ + e- ⇌ Ag+0.80
Br2 + 2e- ⇌ 2Br-+1.09
MnO4− + 8H+ + 5e- ⇌ Mn2+ + 4H2O+1.52

Part 3 of 10The standard hydrogen electrode

The standard hydrogen electrode
It is the reference half-cell, given E⦵ = 0.00 V.
Components
  • H2 gas at 100 kPa in equilibrium with H+ at 1.00 mol dm−3: 2H+ + 2e− ⇌ H2.
  • An inert platinum electrode in contact with both.
  • Platinum black (finely divided platinum) increases the surface area and the rate of attaining equilibrium.
Secondary standards such as a calomel electrode are easier to use than the SHE.

Part 4 of 10Measuring electrode potentials

1
Metal / metal ion half-cell (e.g. Ag+/Ag)
  • Connected to the SHE; E⦵cell = (+0.80) − (0.00) = +0.80 V.
  • The more positive half-cell is the positive pole (reduction); the SHE is the negative pole (oxidation).
2
Non-metal / non-metal ion half-cell (e.g. Br2/Br−)
  • A platinum electrode makes contact; platinum is inert and the equilibrium is set up on its surface.
3
Ion / ion half-cell (e.g. MnO4−/Mn2+)
  • A platinum electrode is used; H+ is also present because it is needed to convert MnO4− to Mn2+.
4
Keep the temperature at 298 K and gas pressure at 100 kPa, and measure with a high-resistance voltmeter.

Part 5 of 10The salt bridge

The salt bridge
A salt bridge completes the circuit by letting ions move between the half-cells, balancing 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 very soluble, unreactive with the half-cell ions and unlikely to form precipitates.

Part 6 of 10Cell potential and conventional cell diagrams

What the symbols mean:
E⦵cell = E⦵right − E⦵left = E⦵reduction − E⦵oxidation, using the more positive E⦵ for the reduction.
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)
Zn / Cu cell
  • (+0.34) − (−0.76) = +1.10 V; the Cu2+/Cu half-cell is the positive pole.
Conventional representation
  • Reduced form | oxidised form || oxidised form | reduced form.
  • The more negative half-cell (oxidation) is written on the left: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s).
  • | is a phase boundary, || the salt bridge, and commas separate species in the same phase.
  • Use a platinum electrode when both species are aqueous or one is a gas: Pt | Fe2+(aq), Fe3+(aq).
  • Cl2(g), 2Cl−(aq) | Pt.
  • Cell diagrams are not quantitative.
Left Is Oxidation; the order of species is Reduced/Oxidised || Oxidised/Reduced.

Part 7 of 10Feasibility, thermodynamics and limitations

1
A reaction is feasible when E⦵cell is positive; the reverse reaction has the opposite sign.
2
Cl2/Cl− (+1.36 V) and Cu2+/Cu (+0.34 V) give Cu + Cl2 → Cu2+ + 2Cl− with E⦵cell = +1.02 V.
3
E⦵cell is proportional to ΔStotal and to ln K.
  • ΔG = −nFE⦵cell and ΔG = −RT ln K.
  • A value above about +0.3 V often means the reaction goes almost to completion.
4
Limitations of predictions
  • E⦵ gives no information about rate;
    • a high activation energy can make a feasible reaction kinetically stable, e.g. Cu2+ + H2 → Cu + 2H+ (+0.34 V) does not occur.
  • Non-standard concentrations shift the equilibrium and change the electrode potential (Le Chatelier).
  • Many redox reactions are not aqueous.
5
For V3+ + e− ⇌ V2+
  • More V3+ shifts the equilibrium right, removing electrons, so the potential becomes less negative.
  • More V2+ shifts it left, so the potential becomes more negative.

Part 8 of 10Fuel cells

Fuel cell
  • Fuel cell a cell in which a fuel donates electrons at one electrode and oxygen gains electrons at the other;
    • fuel and oxidant are supplied continuously and the energy is not stored in the cell.
The anode is the negative electrode and the cathode is the positive electrode.
Hydrogen–oxygen cell
  • Porous platinum-coated electrodes separated by an electrolyte or membrane, with separate inlets for H2 and O2 and an outlet for water.
  • Alkaline
    • anode H2 + 2OH− → 2H2O + 2e− (−0.83 V);
      • cathode O2 + 2H2O + 4e− → 4OH− (+0.40 V).
  • Acidic
    • anode H2 → 2H+ + 2e− (0.00 V);
      • cathode O2 + 4H+ + 4e− → 2H2O (+1.23 V).
  • Overall 2H2 + O2 → 2H2O, E⦵cell = +1.23 V.
Advantages
  • Water is the only product.
  • More efficient than combustion because energy goes to electricity rather than heat and light, and no NOx forms.
  • Used on spacecraft, where the water can be drunk.
Disadvantages
  • Hydrogen is highly flammable, hard to store and needs thick-walled cylinders.
  • Production often depends on fossil fuels, so widespread use is limited until it is cheap.
  • Hydrogen has high energy per gram but low energy per unit volume, so larger containers are needed.
  • Platinum catalysts are expensive.
Ethanol and methanol fuel cells also exist.

Part 9 of 10Redox titrations

Redox titrations
An oxidising agent is titrated against a reducing agent, with electrons transferred; transition metal ions often change colour so no indicator is needed.
Potassium manganate(VII)
  • MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O, mole ratio 1 : 5.
  • Acidify with dilute sulfuric acid, not HCl, which would be oxidised to Cl2.
  • Self-indicatingpurple becomes colourless until the first permanent pale pink.
  • Used for the percentage purity of iron tablets and the formula of hydrated ethanedioic acid;
    • MnO4− : C2O42− is 2 : 5 (warm to about 60 °C).
Iodine–thiosulfate: 2S2O32− + I2 → 2I− + S4O62−
  • An oxidant (e.g. ClO−) liberates I2 from excess KI, and the I2 is titrated with thiosulfate.
  • Add starch when the solution is straw coloured; blue-black to colourless is the end point.
TitrantAnalyteMole ratioEnd point
KMnO4 (acidified)Fe2+1 : 5First permanent pale pink
KMnO4 (acidified)C2O42-2 : 5Pale pink
Na2S2O3I22 : 1Blue-black to colourless (starch)
K2Cr2O7Fe2+1 : 6Redox indicator colour change

Part 10 of 10Redox titration calculations and uncertainty

1
Write the half-equations and the overall equation.
2
Calculate the moles of titrant (concentration × volume in dm3).
3
Use the mole ratio to find the moles of the reductant or oxidant in the aliquot.
4
Scale up to the whole solution, e.g. × 10 if 25.0 cm3 of 250 cm3 was used.
5
Find the concentration, mass or percentage purity.
Redox titration calculations and uncertainty
Percentage purity = mass of pure substance ÷ mass of impure sample × 100.
Percentage uncertainty = uncertainty ÷ measured value × 100; readings taken twice double the absolute uncertainty.
Reducing uncertainty in a titration: make the burette readings larger by diluting the solution in the burette.

Key terms

TermMeaning
Standard electrode potentialThe emf of a half-cell measured against the standard hydrogen electrode under standard conditions.
Standard hydrogen electrodeH2 at 100 kPa over platinum in 1.00 mol dm−3 H+ at 298 K, with E⦵ = 0.00 V.
Salt bridgeA connection that allows ions to move between half-cells and completes the circuit.
Standard cell potentialE⦵cell = E⦵right − E⦵left.
Fuel cellA cell that converts the chemical energy of a continuously supplied fuel directly into electrical energy.
Redox titrationA titration in which the reaction involves electron transfer.
Oxidising agentA species that accepts electrons and is itself reduced.

Avoid these mistakes

1
Writing half-cells with the oxidation on the right — oxidation is on the left of the cell diagram.
2
Using HCl to acidify a manganate(VII) titration — it is oxidised to chlorine, so use dilute sulfuric acid.
3
Multiplying E⦵ by the number of electrons when balancing equations — E⦵ does not change with stoichiometry.
4
Saying a positive E⦵cell means the reaction will be fast — it only shows feasibility.
5
Forgetting platinum electrodes for ion/ion and gas half-cells.

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