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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.

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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.

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

enter image description here

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.

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    $\begingroup$ Complexes of Ni3+ and Cu3+ are known. $\endgroup$ – Mithoron Oct 15 '17 at 15:46
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    $\begingroup$ I’m sure they are. But that’s not relevant to this question, because the reduction potentials are those of the aquo ions. $\endgroup$ – orthocresol Oct 15 '17 at 15:58
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    $\begingroup$ Well, even +4 are known, but yeah, not aquocomplexes :D $\endgroup$ – Mithoron Oct 15 '17 at 16:02
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    $\begingroup$ @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. $\endgroup$ – orthocresol Apr 29 '18 at 9:54
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    $\begingroup$ @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. $\endgroup$ – orthocresol Mar 16 at 1:41

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