Topic 16: Redox Equilibria
- Syllabus
- 2017
- Topic
- —
- Level
- A2
| Process | Electron definition | Oxidation-number change |
|---|---|---|
| oxidation | loss of electrons | increases |
| reduction | gain of electrons | decreases |
Assign oxidation numbers before and after, name the element whose value changes, and state both the direction and values. In Hg + nitrate to Hg(II) + NO, Hg changes 0 to +2 and is oxidised, while N changes +5 to +2 and is reduced.
The same rules apply to s-, p- and d-block species. Oxidation and reduction occur together because electrons lost by one species are gained by another.
Name the element, not merely the whole compound, when explaining an oxidation-number change. An oxidising agent is itself reduced; a reducing agent is itself oxidised.
The standard electrode potential, E°, of a half-cell is the emf measured when that half-cell is connected to the standard hydrogen electrode under standard conditions, with no current flowing. It is quoted as a reduction potential in volts.
A single half-cell potential cannot be measured in isolation: a voltmeter measures a potential difference. Assigning the standard hydrogen electrode E° = 0.00 V supplies the common reference.
A more positive E° means the written reduction has a greater thermodynamic tendency relative to H+/H2; a more negative value means a weaker tendency to be reduced under standard conditions.
E° is not the voltage of an isolated electrode and does not by itself give a reaction rate.
| Quantity | Standard value |
|---|---|
| temperature | 298 K |
| gas pressure | 100 kPa |
| concentration of aqueous ions | 1.00 mol dm^-3 |
These fixed conditions make tabulated E° values comparable. Every ion participating in a half-equation must have the stated concentration; stoichiometry may require choosing solution concentrations carefully.
The superscript ° asserts standard conditions. A measured E without ° may differ when concentration, pressure or temperature differs.
| Feature | Requirement |
|---|---|
| electrode | platinised platinum, inert and catalytic |
| gas | H2 at 100 kPa |
| solution | H+(aq) at 1.00 mol dm^-3 |
| temperature | 298 K |
| half-equation | 2H+ + 2e- ⇌ H2 |
Connect it to the test half-cell through a salt bridge and a high-resistance voltmeter. The sign and magnitude of the measured emf relative to the 0.00 V reference define the other half-cell's E°.
Platinum is not consumed and is not the source of hydrogen ions; it provides an electrical contact and catalytic surface.
| Half-cell | Electrode arrangement | Example |
|---|---|---|
| metal/metal ion | the metal dips into its aqueous ions | Cu(s) in Cu2+(aq) |
| non-metal/ion | inert Pt conducts; relevant phases contact it | Pt with Cl2(g)/Cl-(aq) |
| two aqueous oxidation states | inert Pt contacts both ions | Pt with Fe3+(aq), Fe2+(aq) |
Use a conducting solid that participates when the redox couple contains a metal, but use inert platinum when every redox species is gaseous or aqueous. Include all species and standard conditions.
An iron electrode is unsuitable for Fe3+/Fe2+ because it introduces another redox species; platinum transfers electrons without changing the intended couple.
| Component | Function |
|---|---|
| two correct electrodes and solutions | form the half-cells |
| salt bridge soaked in an inert electrolyte | permits ion movement between half-cells |
| high-resistance voltmeter | measures emf while drawing negligible current |
| complete external circuit | permits electron transfer through the wire |
Clean electrodes, prepare stated concentrations, connect the salt bridge so it touches both solutions, attach the voltmeter, record polarity and a stable voltage, and repeat while controlling temperature. For a Zn/Cu cell, use Zn in Zn2+ and Cu in Cu2+.
A practical bridge may be filter paper soaked in saturated KNO3. Choose ions that do not react with either half-cell.
Electrons travel through the external wire; ions travel through the salt bridge. A power supply would drive the cell rather than measure its emf.
E^\circ_{cell}=E^\circ_{reduction}-E^\circ_{oxidation}=E^\circ_{right}-E^\circ_{left}
Keep both tabulated half-equations written as reductions. The more positive couple operates as reduction; reverse the other half-equation for oxidation, balance electrons, then subtract its tabulated reduction potential.
For Ag+/Ag, +0.80 V, paired with Ti3+/Ti2+, -0.37 V, E°cell = +0.80 - (-0.37) = +1.17 V.
Multiply half-equations to balance electrons, but never multiply E° values: potential is not an amount-dependent quantity.
| Symbol | Meaning |
|---|---|
| , | species in the same phase |
| Pt | inert conducting electrode when needed |
Write the oxidation half-cell on the left and reduction half-cell on the right, placing each electrode at an outer end. Include state symbols and every redox species.
\mathrm{Pt(s)|Ti^{2+}(aq),Ti^{3+}(aq)||Ag^+(aq)|Ag(s)}
Do not put electrons in a cell diagram. Use a single line only for a phase boundary and a double line only for the salt bridge.
E depends on temperature, gas pressure and the activities approximated by solution concentrations. Changing them shifts the half-cell equilibrium and changes the potential, so an E value cannot automatically be compared with E°.
For M^z+ + ze- ⇌ M, diluting M^z+ favours the left side and makes the reduction potential more negative. Apply the same equilibrium reasoning to every species in the actual half-equation.
Record actual conditions and allow the cell to equilibrate. A concentration change in either half-cell can alter Ecell and may even change a borderline prediction.
Do not apply a memorised 'dilution lowers E' rule to every equation; the direction depends on where each changed species appears.
Write the proposed oxidation and reduction half-reactions, use their tabulated reduction potentials, calculate E°cell, and accept the proposed direction as thermodynamically feasible under standard conditions when E°cell is positive.
E^\circ_{cell}>0:\ \text{feasible};\qquad E^\circ_{cell}<0:\ \text{reverse direction favoured}
Test sequential reductions separately. A reagent may give positive E°cell for conversion to one oxidation state but negative E°cell for further reduction, explaining selective products.
A positive E°cell predicts thermodynamic direction under standard conditions, not observable speed or complete conversion.
\Delta S^\circ_{total}=\frac{nFE^\circ_{cell}}{T}\qquad\ln K=\frac{nFE^\circ_{cell}}{RT}
n is electrons transferred in the balanced overall equation, F = 96500 C mol^-1, R = 8.31 J mol^-1 K^-1 and T is kelvin. A positive E°cell gives positive total entropy change and K greater than 1.
Balance the redox equation before choosing n, use volts as J C^-1, and exponentiate ln K only after evaluating the full expression.
n is not automatically 1 and is not the sum of electrons in both half-equations; it is the number cancelled in the balanced reaction.
| Limitation | Consequence |
|---|---|
| activation energy or slow mechanism | feasible reaction may be kinetically stable and appear not to occur |
| non-standard concentration/pressure/temperature | actual E values differ from E° |
| current flows and composition changes | cell emf falls as equilibrium is approached |
A small positive or negative E°cell is especially sensitive to concentration changes. Concentrated reactants can shift both relevant half-cell equilibria enough to reverse the standard prediction.
Thermodynamic feasibility is not a promise of rapid reaction. Conversely, a negative standard value does not settle behavior under strongly non-standard conditions.
Standard electrode potentials are standard reduction potentials because every listed half-equation is written in the reduction direction. Ordering them by E° produces the electrochemical series.
| Position/value | Meaning for written reduction |
|---|---|
| more positive E° | stronger tendency to gain electrons; oxidised form is a stronger oxidising agent |
| more negative E° | weaker tendency to gain electrons; reduced form is a stronger reducing agent |
The electrochemical series is not identical to a simple metal reactivity series: it includes non-metals, ions and molecular couples and assumes standard conditions.
In disproportionation, the same intermediate oxidation state is oxidised in one half-reaction and reduced in another, producing both a higher and a lower oxidation state.
Select the two couples that share the intermediate species, reverse the oxidation branch, balance and combine them, then calculate E°cell = E°(reduction branch) - E°(oxidation branch). A positive value predicts feasibility.
Verify that the shared starting species appears on the reactant side of both branches after directions are chosen; otherwise the calculation is not for disproportionation.
Do not merely subtract adjacent values without first fixing the proposed reaction directions and oxidation states.
| Step | Operation |
|---|---|
| 1 | write/balance the ionic redox equation |
| 2 | calculate titrant moles with n = cV, V in dm3 |
| 3 | apply the equation's mole ratio to analyte |
| 4 | scale aliquot to original flask if required |
| 5 | convert to concentration, mass or percentage and round suitably |
\mathrm{MnO_4^-+8H^++5Fe^{2+}\rightarrow Mn^{2+}+5Fe^{3+}+4H_2O}
\mathrm{I_2+2S_2O_3^{2-}\rightarrow 2I^-+S_4O_6^{2-}}
Do not use coefficients from an unbalanced equation or forget the aliquot-to-flask scale factor. Preserve unrounded values until the final answer.
%\ uncertainty=\frac{absolute\ uncertainty}{measured\ value}\times100
For a burette titre formed from two readings, include the uncertainty of both readings. Add percentage uncertainties for quantities combined by multiplication or division to estimate the total percentage uncertainty.
Compare the total percentage uncertainty with the reported precision and with differences between results. An answer quoted to digits far smaller than the experimental uncertainty implies unjustified precision.
Low random uncertainty supports precision but does not remove systematic error, reaction incompleteness or bias; those can make a precise result invalid.
| Titration | Indicator/endpoint |
|---|---|
| acidified MnO4- into Fe2+ | MnO4- is self-indicating; first permanent pale pink after swirling |
| S2O3^2- into iodine | add starch when iodine is pale straw-yellow; blue-black changes to colourless |
Rinse apparatus appropriately, pipette the analyte, acidify permanganate work with sulfuric acid, titrate while swirling, add reagent dropwise near the endpoint, and obtain concordant titres.
Adding starch only near the iodine endpoint avoids a strongly bound iodine-starch complex that can make the endpoint slow or unclear.
Do not acidify Fe2+/MnO4- with HCl or HNO3: chloride can be oxidised by manganate(VII), while nitrate can oxidise Fe2+.
A fuel cell separates oxidation of a continuously supplied fuel from reduction of oxygen. Electrons released at the negative electrode travel through the external circuit to the positive electrode, generating a voltage while ions cross the electrolyte.
Hydrogen, methanol and other hydrogen-rich fuels can be used. Hydrogen-oxygen cells form water at point of use; methanol is easier to store but its oxidation forms carbon dioxide.
| Advantage | Limitation |
|---|---|
| operates while reactants are supplied; no recharging pause | fuel production and storage infrastructure are required |
| hydrogen use produces water locally | overall environmental impact depends on how hydrogen is produced |
| fewer moving parts | catalysts can be costly and fuels may be flammable |
A fuel cell is not an energy source independent of fuel: its sustainability depends on production, storage and the complete lifecycle.
| Electrolyte | Negative electrode: oxidation | Positive electrode: reduction |
|---|---|---|
| acidic | H2 -> 2H+ + 2e- | O2 + 4H+ + 4e- -> 2H2O |
| alkaline | H2 + 2OH- -> 2H2O + 2e- | O2 + 2H2O + 4e- -> 4OH- |
\mathrm{2H_2+O_2\rightarrow2H_2O}
Hydrogen is oxidised at the negative electrode and oxygen is reduced at the positive electrode in both electrolytes. Multiply the hydrogen half-equation by two before adding; H+ or OH- then cancels.
Electrons move through the external circuit from negative to positive; they do not cross the electrolyte membrane. Acidic and alkaline half-equations differ even though the overall reaction is the same.