# Why does the Co³⁺/Co²⁺ couple have such a high reduction potential?

Cobalt is changing from $$\mathrm{(3d)^6}$$ to $$\mathrm{(3d)^7}$$ electronic configuration. What's so stabilizing about that? Just by reduction potentials this couple is more oxidising than hydrogen peroxide and that just sounds amazing and crazy at the same time.

The electronic configuration has nothing to do with it. The reduction potentials of $$\ce{Ni^3+}/\ce{Ni^2+}$$, $$\ce{Cu^3+}/\ce{Cu^2+}$$ and $$\ce{Zn^3+}/\ce{Zn^2+}$$, if they have been/could be measured, would be even greater.

The reduction potential for $$\ce{M^3+}/\ce{M^2+}$$ is most dependent upon the third ionisation energy. If $$I_3$$ is large then it will be difficult to oxidise $$\ce{M^2+}$$ to $$\ce{M^3+}$$ (hence a high reduction potential). Likewise if $$I_3$$ is small then the reduction potential will be smaller.

$$\begin{array}{c|cc} \hline \text{Metal} & I_3\text{ / eV} & E^\circ(\ce{M^3+}/\ce{M^2+})\text{ / V} \\ \hline \ce{Sc} & 24.76 & (-2.6) \\ \ce{Ti} & 27.48 & (-1.1) \\ \ce{V} & 29.31 & -0.26 \\ \ce{Cr} & 30.96 & -0.41 \\ \ce{Mn} & 33.67 & +1.60 \\ \ce{Fe} & 30.65 & +0.77 \\ \ce{Co} & 33.50 & +1.93 \\ \ce{Ni} & 35.16 & (+4.2) \\ \ce{Cu} & 36.94 & (+4.6) \\ \ce{Zn} & 39.72 & (+7.0) \\ \hline \end{array}$$

Data from Weller et al. Inorganic Chemistry 6ed; Johnson Some Thermodynamic Aspects of Inorganic Chemistry. Numbers in parentheses indicate predicted values. Graph by me in Microsoft Office 365.

So your question is essentially: why is $$I_3$$ of $$\ce{Co}$$ so "large"?

Going across the 3d block, effective nuclear charge increases which leads to the general trend of IE3 and $$E^\circ(\ce{M^3+}/\ce{M^2+})$$ increasing.

Any stabilisation derived from "special" electronic configurations e.g. the half-filled subshell in $$\ce{Fe^3+}$$, or the ligand-field stabilisation energy of $$\ce{Cr^2+}$$, merely lead to relatively small deviations from the general trend.

Moral of the story is: look at the big picture first. Those $$\mathrm{d^5}, \mathrm{d^{10}}, \cdots$$ configurations can be relevant, but they aren’t responsible for the overall trend, and certainly shouldn’t be the first thing you consider when trying to rationalise this set of data.

• Complexes of Ni3+ and Cu3+ are known. – Mithoron Oct 15 '17 at 15:46
• I’m sure they are. But that’s not relevant to this question, because the reduction potentials are those of the aquo ions. – orthocresol Oct 15 '17 at 15:58
• Well, even +4 are known, but yeah, not aquocomplexes :D – Mithoron Oct 15 '17 at 16:02
• @AbhigyanC no, that cannot explain anything. If anything, the large CFSE of Co(3+) should lower the reduction potential (as it stabilises the oxidised form). Anyway, as described in the main text, these details are minor. – orthocresol Apr 29 '18 at 9:54
• @YUSUFHASAN, if you’re talking about Cr(II) being a good reducing agent, then fair enough; this is partly explained by CFSE/LFSE, and that is why in the graphs above there is a dip in E° although there is no dip in IE3. But this dip is relatively small, and as already mentioned, the big picture is that the trend in E° is determined primarily by the trend in IE3. If CFSE was really such a major factor, then why isn’t Co(II) as good a reducing agent as Cr(II), since Co(III) is typically low-spin (t2g)6? That’s precisely what this Q+A is about, so please re-read it carefully. – orthocresol Mar 16 '19 at 1:41