# Balancing OH- reduction to H2 [closed]

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?

• If you write NaOH as OH- why not towrite Na2O as O^2- ? Commented Jan 7, 2022 at 11:19
• Does this answer your question? Balancing reaction between copper and nitric acid given the ratio of formed nitrogen oxides Commented Jan 7, 2022 at 12:06
• I think it could be: 2$\ce{B}$ + 6$\ce{NaOH}$ $\rightarrow$ 2$\ce{NaBO2}$ + 3$\ce{H2}$ + 2 $\ce{Na2O}$ Commented Jan 7, 2022 at 14:53
• Per this source: "Boron reacts with fused sodium hydroxide to form sodium borate and hydrogen." chemistry.stackexchange.com/questions/67434/… Commented Jan 7, 2022 at 16:45
• It would be rather Na3B3O6, as metaborate is a cyclical trimer with alternating O and BO- Commented Jan 7, 2022 at 18:03

As suggested by Poutnik in the comments I considered $$\ce{O2-}$$ in the reduction half reaction which then became:
$$\ce{2OH- + 2e- -> H2 + 2O^2-}$$
$$\ce{14NaOH + 2B -> 2NaBO2 + 3H2 + 6Na2O + 4H2O}$$
• Subtract from it reaction $\ce{8 NaOH -> 4 Na2O + 4 H2O}$ that does not happen in a condensed phase and you will get $\ce{6 NaOH + 2 B -> 2 NaBO2 + 3 H2 + 2 Na2O}$ Commented Jan 7, 2022 at 16:30