Balancing OH- reduction to H2 [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed last year.








[Improve this question](/posts/162020/edit) 



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?




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:


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


 0