My book - NCERT Chemistry for Class XII -Page 218 Contains the following paragraph:

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Initially they mention that the Low value for Standard Electrode Potential of $\ce{Sc^3+/Sc^2+}$ is due to the higher stability of $\ce{Sc^3+}$ which makes sense as it would have a lesser tendency to move to the $+2$ oxidation state.

But later on, they mention that the Lower value of SEP of $\ce{V^3+/V^2+}$ is due to the stability of $\ce{V^2+}$.

I get that the half filled $\ce{t_{2g}}$ configuration would make $\ce{V^2+}$ more stable but then, shouldn't the SEP value become higher rather than being low? As $\ce{V^3+}$ would have more tendency to move to $\ce{V^2+}$

  • $\begingroup$ Do you see the fallacy of the word stability, so think a little deeper. When you say V2+ is more stable, but what is your reference point? Say, there is a lake X, which is deep but lake Y is deeper and I don't mention my reference starting point for measuring the depth. This textbook is exactly doing the same: repeating the word stability again and again but without telling you their reference point. $\endgroup$
    – AChem
    May 2, 2022 at 4:49
  • $\begingroup$ There are several problems in the textbook. It talks about the Sc half-cell for M3+/M2+ but it is not shown for Sc! "The low value for Sc reflects" ... there is no value there in Table 8.2. Did you see this value? The second biggest problem is the word stability. Stability with respect to what? Stability in chemistry can have different context and it can have so many meanings. $\endgroup$
    – AChem
    May 2, 2022 at 4:53
  • $\begingroup$ Yes I agree and did refer to the table for the values. Although there was no value given for Sc, if we still look at what they're trying to prove, the reason that $\ce{Sc^3+}$ is more stable justifies the statement made. But on the other hand, they've still given the value for V right? And the value given is negative... meaning $\ce{V^3+}$ has no tendency to reduce to $\ce{V^2+}$ but their relative stabilities(That $\ce{V^2+}$ is more stable than $\ce{V^3+}$) indicate the opposite tendency. That's where I'm stuck at! $\endgroup$ May 2, 2022 at 5:26
  • $\begingroup$ Again, you will have to think about the meaning of stability for Sc. Without any electrode potential value, they should not be talking about it. Perhaps this is causing confusion. We do not know the value of Sc half cells as per your Table. In electrochemical language, something can be stable with respect to oxidation, reduction or disproportionation. Just saying V2+ is more stable V3+ does not mean anything in an absolute sense. $\endgroup$
    – AChem
    May 2, 2022 at 5:48
  • 1
    $\begingroup$ I remember as a young-chemist to perform experiment "color chameleon" with NH4VO3 in diluted acid, reduced by zinc, causing colour transition for each step of $\ce{VO2^+ -> VO^2+ -> V^3+ -> V^2+}$. $\endgroup$
    – Poutnik
    May 3, 2022 at 7:34

2 Answers 2


The $\ce{E^°(M^3+/M^2+)}$ standard electrode potential for $\ce{V}$ has a low negative (that is, a relatively more positive) value. This is attributed to the greater stability of $\ce{V^2+}$ as given in the text.

This can be seen in the $\ce{E^°(M^3+/M^2+)}$ values for $\ce{Ti,V,Cr}$ being $\pu{-0.37, -0.26, -0.41}$ respectively.

The book is not very clear in its meaning of low in this case. Looking at the table of values and comparing the $\ce{E^°(M^3+/M^2+)}$ of nearby elements can help you understand the text better.


An independent source gives $-0.37$ for titanium, $-0.26$ for vanadium and $-0.42$ for chromium, showing that vanadium is relatively high compared with its immediate neighbors. The book should have said "less negative" rather than "comparatively low".

In any event the effect of half-filling the $t_{2g}$ orbitals is weak compared with half-filling all of the $3d$ subshell. The above source gives $+1.5$ for manganese versus the aforementioned $-0.42$ for chromium and $+0.77$ for iron. Half-filling the $t_{2g}$ orbitals alone would have more impact with ligands that have a stronger splitting than water.


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