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The CRC Elecrochemical Series and the solution to our exercise suggest that the oxidation states of borohydride are -V for Boron and +I for Hydrogen.

Since H (2.2) is more electronegativ than B (2.0), I would have expected the oxidation states to be +III for B and -I for H.

What am I missing here?

The solution to our exercise lists this (WARNING: it turned out that the solution was wrong):

And the CRC Elecrochemical Series lists this (also suggesting -V for B):

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I would have expected the oxidation states to be +III for B and -I for H.

You're right, boron is in the (+3) oxidation state in these equations.

In $\ce{B(OH)_3}$ each $\ce{OH}$ is (-1) and the boron is (+3), overall the molecule is neutral

In $\ce{BH_{4}^{-}}$ each $\ce{H}$ is (-1) and the boron is (+3), overall the ion is (-1)

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  • $\begingroup$ Thanks for answering, I am still skeptical, however. Usually our assignments are prepared quite carefully, and how would you explain that redox potential? Boron(+III) taking up 8 electrons would also result in B(-V). $\endgroup$ – Reto Höhener Jul 1 '14 at 14:29
  • $\begingroup$ The boron does not take up any electrons. All of the electrons are used by the hydrogens. On the left side of the equation we have 10 $\ce{H^{+}}$, on the right side of the equation we have 4 $\ce{H^{-}}$ and 6 $\ce{H^{+}}$. The 8 electrons reduced 4 $\ce{H^{+}}$ to 4 $\ce{H^{-}}$. $\endgroup$ – ron Jul 1 '14 at 15:06
  • $\begingroup$ Sorry for having been skeptical ;-). It turns out that the solution was wrong after all (info from the professor): Hydrogen is being reduced, not boron. I'm glad... (voted and accepted). $\endgroup$ – Reto Höhener Jul 3 '14 at 13:16
  • $\begingroup$ @Zalumon Glad we got it figured out! $\endgroup$ – ron Jul 3 '14 at 19:52

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