Predicting feasibility and limitationsEdexcel A-Level Chemistry: Revision notes

To predict whether a redox reaction is thermodynamically feasible, write the two half-equations, take the more positive half-cell as the one reduced, and calculate $E^\ominus_{cell} = E^\ominus_{reduced} - E^\ominus_{oxidised}$.

• If $E^\ominus_{cell}$ is positive, the reaction is feasible under standard conditions.
• If it is negative, the reaction is not feasible (the reverse reaction is).

Worked example. Will acidified MnO₄⁻ oxidise Fe²⁺? $E^\ominus$(MnO₄⁻/Mn²⁺) = +1.51 V and $E^\ominus$(Fe³⁺/Fe²⁺) = +0.77 V. $E^\ominus_{cell} = +1.51 - 0.77 = +0.74$ V, so yes: $\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}$. The $E^\ominus$ values are not multiplied by the 5 electrons.

Listing half-equations in order of $E^\ominus$ gives the electrochemical series. Reading down from most negative to most positive: Li⁺/Li −3.04 V; Mg²⁺/Mg −2.37 V; Zn²⁺/Zn −0.76 V; Fe²⁺/Fe −0.44 V; H⁺/H₂ 0.00 V; Cu²⁺/Cu +0.34 V; I₂/I⁻ +0.54 V; Fe³⁺/Fe²⁺ +0.77 V; Ag⁺/Ag +0.80 V; Br₂/Br⁻ +1.07 V; Cl₂/Cl⁻ +1.36 V; MnO₄⁻/Mn²⁺ +1.51 V.

• The more positive the $E^\ominus$, the stronger the oxidising agent (the species on the left of the half-equation) and the more stable the reduced form.
• The more negative the $E^\ominus$, the stronger the reducing agent (the species on the right of the half-equation).

A species will oxidise any reduced form with a less positive $E^\ominus$ in the series.

The standard cell potential is directly proportional to the total entropy change and to $\ln K$:

$\Delta S_{total} = \frac{nFE^\ominus_{cell}}{T} \qquad \ln K = \frac{nFE^\ominus_{cell}}{RT}$

where $n$ is the number of electrons transferred, $F = 96\,500$ C mol⁻¹, $R = 8.31$ J K⁻¹ mol⁻¹ and $T$ is in kelvin. A positive $E^\ominus_{cell}$ means $\Delta S_{total} > 0$ and $K > 1$.

Worked example. Zn + Cu²⁺ → Zn²⁺ + Cu, $E^\ominus_{cell}$ = +1.10 V, n = 2, T = 298 K:

$\Delta S_{total} = (2 \times 96\,500 \times 1.10)/298 = +712$ J K⁻¹ mol⁻¹

$\ln K = (2 \times 96\,500 \times 1.10)/(8.31 \times 298) = 85.7$, so $K \approx 1.7 \times 10^{37}$: the reaction goes essentially to completion.

A positive $E^\ominus_{cell}$ does not guarantee that a reaction will be seen, for two reasons:

1. Kinetic inhibition. $E^\ominus$ gives only thermodynamic feasibility. If the reaction has a very high activation energy, the rate is negligible and no change is observed. For example, MnO₄⁻ in acid is feasibly able to oxidise water ($E^\ominus_{cell}$ = +0.28 V), but solutions stay purple for weeks.
2. Non-standard conditions. $E^\ominus$ values apply only at 1.00 mol dm⁻³, 298 K and 100 kPa. Changing concentration, temperature or pressure shifts the half-cell equilibria and alters the electrode potentials. If $E^\ominus_{cell}$ is only slightly positive or negative, a change in conditions can reverse the prediction.

Disproportionation is the simultaneous oxidation and reduction of the same species. It is feasible when the species is more easily reduced than it is oxidised, i.e. $E^\ominus$(reduction couple) is more positive than $E^\ominus$(oxidation couple).

Copper(I): Cu⁺ + e⁻ ⇌ Cu, +0.52 V; Cu²⁺ + e⁻ ⇌ Cu⁺, +0.15 V.

$\mathrm{2Cu^+ \rightarrow Cu^{2+} + Cu}$, $E^\ominus_{cell} = +0.52 - 0.15 = +0.37$ V: feasible, so Cu⁺ is unstable in water.

Iron(II): Fe²⁺/Fe is −0.44 V and Fe³⁺/Fe²⁺ is +0.77 V, so $E^\ominus_{cell} = -0.44 - 0.77 = -1.21$ V: not feasible, so Fe²⁺ is stable.