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I have to balance the following redox equation:

$\ce{B + NaOH -> NaBO2 + H2 + Na2O}$

Here I identified B as the species which undrgoes oxidation and wrote the oxidation half-reaction:

$\ce{4OH- + B -> BO2- + 2H2O + 3e-}$

The only element whose oxidation number is reduced is H from +1 in $\ce{OH-}$ to 0 in $\ce{H2}$. So the (unbalanced) reduction half-reaction should be something like:

$\ce{2e- + 2OH- -> H2}$

How does one balance that?

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As suggested by Poutnik in the comments I considered $\ce{O2-}$ in the reduction half reaction which then became:

$\ce{2OH- + 2e- -> H2 + 2O^2-}$

After multiplying each half-reaction by the number of electrons exchanged by the other one and adding back the spectator ions the balanced equation becomes:

$\ce{14NaOH + 2B -> 2NaBO2 + 3H2 + 6Na2O + 4H2O}$

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    $\begingroup$ Subtract from it reaction $\ce{8 NaOH -> 4 Na2O + 4 H2O}$ that does not happen in a condensed phase and you will get $\ce{6 NaOH + 2 B -> 2 NaBO2 + 3 H2 + 2 Na2O}$ $\endgroup$
    – Poutnik
    Commented Jan 7, 2022 at 16:30

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