d- and f-Block Elements · Lesson 4 of 12
Electrode Potentials and Oxidation-State Stability
“Combine ionisation, atomisation, hydration and reduction potentials to predict chemical change.”
• Interpret the sign and direction of standard reduction potentials. • Explain irregular M²⁺/M and M³⁺/M²⁺ trends through complete energy balances. • Predict reducing strength and distinguish thermodynamic tendency from rate. • Explain aqueous Cu⁺ disproportionation and ligand effects on cobalt.
Why a stable configuration is not a complete explanation
Cu⁺ has d¹⁰, yet uncomplexed copper(I) is unstable toward disproportionation in water. Cu²⁺ has d⁹, yet hydrated copper(II) is widespread. There is no contradiction: stability compares entire systems, including solvent and possible products. Removing an electron can be costly in a gas, while the resulting higher-charge ion can be strongly stabilised by hydration in water. The final balance decides the direction of a reaction.
A useful energy cycle for M(s)→M²⁺(aq)+2e⁻ separates atomisation M(s)→M(g), first and second ionisation of the gas atom, and hydration M²⁺(g)→M²⁺(aq). Atomisation and ionisation cost enthalpy; hydration usually releases enthalpy. Entropy and the electron-reference convention must also be included to obtain an electrode potential. Therefore an enthalpy sum explains contributions but is not by itself an exact calculation of E°.
| Metal | Atomisation | I₁+I₂ | Hydration of M²⁺ | E°(M²⁺/M) / V |
|---|---|---|---|---|
| Ti | 469 | 1965 | -1866 | −1.63 |
| V | 515 | 2064 | -1895 | −1.18 |
| Cr | 398 | 2245 | -1925 | −0.90 |
| Mn | 279 | 2226 | -1862 | −1.18 |
| Fe | 418 | 2323 | -1998 | −0.44 |
| Co | 427 | 2402 | -2079 | −0.28 |
| Ni | 431 | 2488 | -2121 | −0.25 |
| Cu | 339 | 2703 | -2121 | +0.34 |
| Zn | 130 | 2640 | -2059 | −0.76 |
The table uses a consistent thermochemical dataset; small differences from other tables reflect data selections and rounding. Copper requires a large first-plus-second ionisation input that hydration does not fully offset in the overall electrode comparison. Its positive reduction potential makes reduction of Cu²⁺ favourable relative to the standard hydrogen electrode. For Mn and Zn, favourable d⁵ Mn²⁺ and d¹⁰ Zn²⁺ help explain more negative M²⁺/M values than a smooth trend would suggest. Strongly exothermic hydration contributes importantly for Ni. None of these factors should be treated in isolation.
Reduction potentials always refer to a written direction
Problem
Use E°(Cu²⁺/Cu)=+0.34 V, E°(Fe²⁺/Fe)=−0.44 V and E°(H⁺/H₂)=0 V.
- 1.For M+2H⁺→M²⁺+H₂, hydrogen ions are reduced and the metal is oxidised. E°cell=0−E°(M²⁺/M).
- 2.For Cu, E°cell=−0.34 V: H⁺ alone does not favour ordinary copper dissolution to Cu²⁺ under these standard conditions. Nitric acid can dissolve Cu because nitrate is an additional oxidant.
- 3.For Fe, E°cell=+0.44 V, so dissolution is thermodynamically favoured. Surface oxides, concentration and temperature can still control the observed rate.
- 4.Titanium and vanadium can show passivity in dilute non-oxidising acids despite favourable potentials: a protective layer impedes the kinetic pathway.
Comparing M³⁺/M²⁺ couples
| Reduction couple | E° / V | Interpretation in water |
|---|---|---|
| Ti³⁺/Ti²⁺ | −0.37 | Ti²⁺ is a strong reductant |
| V³⁺/V²⁺ | −0.26 | V²⁺ is reducing |
| Cr³⁺/Cr²⁺ | −0.41 | Cr²⁺ readily oxidises to Cr³⁺ |
| Mn³⁺/Mn²⁺ | +1.57 | Mn³⁺ strongly tends to accept an electron |
| Fe³⁺/Fe²⁺ | +0.77 | Fe³⁺ oxidises suitable reductants |
| Co³⁺/Co²⁺ | +1.97 | Uncomplexed aqueous Co³⁺ is strongly oxidising |
Both Cr²⁺ and Mn³⁺ are formally d⁴, but they react in opposite directions. Cr²⁺ loses an electron to form d³ Cr³⁺, stabilised in octahedral hydration with three electrons in lower-energy t₂g orbitals. Mn³⁺ gains an electron to form high-spin d⁵ Mn²⁺. Their initial equality of d count does not make their energies, charges or reaction partners equal. Likewise Mn²⁺ resists oxidation to +3 more than Fe²⁺: Mn²⁺ starts d⁵, whereas Fe²⁺ can reach d⁵ Fe³⁺.
Problem
Can Fe³⁺ oxidise Cr²⁺? Compare Cr²⁺ and Fe²⁺ as reductants.
- 1.Write Fe³⁺+e⁻→Fe²⁺, E°=+0.77 V. Reverse Cr³⁺+e⁻→Cr²⁺, whose reduction value is −0.41 V.
- 2.The balanced reaction is Cr²⁺+Fe³⁺→Cr³⁺+Fe²⁺. E°cell=0.77−(−0.41)=+1.18 V.
- 3.Positive E°cell supports the forward standard-state reaction. Cr²⁺ is the stronger reductant: it is much more readily oxidised to +3 than Fe²⁺ is under comparable aqueous conditions.
The broad “+2 becomes more stable” statement for the first half of 3d refers to its increasing importance as removing additional d electrons becomes more difficult, culminating in d⁵ Mn²⁺. It is not a claim that every reduction potential changes monotonically. Sc³⁺ and Ti(IV) remain strongly favoured in much of their chemistry. A d¹ ion such as Ti³⁺ often tends toward d⁰ Ti(IV) with suitable oxidants, but d¹ is not universally unstable: V(IV) in vanadyl compounds is a clear persistent example.
Disproportionation and the role of ligands
A reaction in which one initial oxidation state is both oxidised and reduced, producing a higher and a lower oxidation state of the same element.
In water, 2Cu⁺→Cu²⁺+Cu(s). One Cu(I) is oxidised to Cu(II), the other reduced to metal. The strong hydration of Cu²⁺ helps outweigh the cost of producing a d⁹ ion from d¹⁰ Cu⁺. Ligands or precipitation can stabilise Cu(I), so the aqueous uncomplexed conclusion is not a ban on all Cu(I) compounds. Green manganate(VI) similarly disproportionates on acidification to Mn(VII) permanganate and Mn(IV) dioxide; its balanced equation is developed with permanganate preparation.
Cobalt provides another environment dependence. Water alone favours Co²⁺ relative to highly oxidising Co³⁺. Ligands that strongly stabilise Co(III) can shift the Co(III)/Co(II) reduction potential downward, making oxidation of Co(II) easier with a suitable oxidant. The ligand is part of the energy balance; this is why a statement about hydrated cobalt cannot be transferred unchanged to every cobalt complex.
In the specified acidic comparison, oxidising strength rises broadly from dioxovanadium(V), VO₂⁺, to dichromate and then permanganate reduced to Mn²⁺. The associated lower products and complete standard reduction potentials support this ranking; it should not be asserted for arbitrary pH, concentrations or alternative products. Greater formal oxidation number alone is not a universal measure of oxidising power.
Quick Quiz
Quiz
For Fe+2H⁺→Fe²⁺+H₂, what is E°cell if E°(Fe²⁺/Fe)=−0.44 V?
Explanation: Subtract the metal reduction potential from the hydrogen reduction potential, 0−(−0.44).
Why can d¹⁰ Cu⁺ disproportionate in water?
Explanation: The gas-phase configuration is only one contribution to aqueous reaction free energy.
Which is the stronger aqueous reductant?
Explanation: The Cr³⁺/Cr²⁺ reduction potential is much more negative, favouring Cr²⁺ oxidation.
A positive E°cell establishes what?
Explanation: It concerns thermodynamics under standard-state assumptions, not kinetics or all compositions.
Why can complexing agents facilitate oxidation of Co²⁺?
Explanation: Changing the relative energies of complexed Co(III) and Co(II) changes their redox balance.
Practice Problems
- With E°(Mn³⁺/Mn²⁺)=1.57 V and E°(Fe³⁺/Fe²⁺)=0.77 V, predict Mn³⁺+Fe²⁺.
- Compare ease of oxidation of Cr, Mn and Fe using −0.9, −1.2 and −0.4 V for M²⁺/M.
- Why is Mn³⁺ more strongly oxidising than Fe³⁺ or Cr³⁺ in the given aqueous comparison?
- Explain why thermochemical enthalpy sums cannot be used as exact electrode potentials.
- Give two disproportionations and identify all metal states.
- Is Fe³⁺ “more stable” than both Cr³⁺ and Mn³⁺?
Mn³⁺ accepts an electron and Fe²⁺ supplies it. Mn³⁺+Fe²⁺→Mn²⁺+Fe³⁺ has E°cell=1.57−0.77=+0.80 V. One electron is transferred, so ΔG°≈−96485×0.80=−77.2 kJ mol⁻¹ for the written reaction.
Key Takeaways
• E° values refer to reductions as written. • Cell potential is cathode reduction potential minus anode reduction potential. • Atomisation, ionisation and hydration jointly explain irregular metal trends. • Positive E° predicts a standard-state tendency, not a fast reaction. • Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising despite both being d⁴. • Aqueous Cu⁺ disproportionation shows why filled-shell arguments are insufficient alone. • Ligands and pH can change oxidation-state stability.