1
$\begingroup$

This question already has an answer here:

$\ce{Co^3+(aq) + e- → Co^2+(aq)}$ +1.81 V

$\ce{Mn^3+(aq) + e- → Mn^2+(aq)}$ +1.51 V

This is the data of Standard reducing potentials I have found on internet. Why is reducing potential of $\ce{Mn^3+}$ is less than that of $\ce{Co^3+}$. $\ce{Mn^3+}$ converts to $\ce{Mn^2+}$ which is half filled configuration and hence more stable. Am I missing something?

$\endgroup$

marked as duplicate by Mithoron, Jan, Jon Custer, pentavalentcarbon, andselisk Oct 15 '17 at 17:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ The answer is simple, you are overlooking the most important trend in the 3d block. Everybody likes to think about the fancy stuff like exchange energy, Jahn-Teller effects, blah blah blah but they forget that effective nuclear charge increases going across the 3d block. $\endgroup$ – orthocresol Mar 22 '16 at 23:30