# Half-Reactions of Hydroxide [closed]

The following is the equation I have:

$$\ce{Ca(s) + 2H2O(l) -> Ca(OH)2 v + H2 ^}$$

I am tasked to write the half-reaction equations for both the oxidation and/or reduction process. So, currently I have: $$\ce{ Ca(s) -> Ca^{2+}(aq) + 2e-}\quad\text{(oxidation)}$$

but I'm not sure how do you write the reduction equation for the water and hydroxide.

Any help is appreciated :)

P.S.: I have no idea how to superscript or subscript in the forum.

## closed as off-topic by MaxW, Todd Minehardt, andselisk, Mithoron, airhuffDec 26 '18 at 4:38

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• Balance this for the reduction step: xH2O ---> H2 + y OH- – user55119 Dec 25 '18 at 0:14
• that would be: 2H2O --> H2+ 2OH- Is that correct? – Grimlock Dec 25 '18 at 1:05
• Right! since the oxidation is -2 electrons (Ca to Ca++) and the reduction (0 to -2) +2 electrons, just add the two reactions and simplify if needed. – user55119 Dec 25 '18 at 3:22
• Just so I'm getting this right: My Reduction equation reaction would be 2H2O --> H2 + 2OH-. Is there no e- required in this case? – Grimlock Dec 25 '18 at 5:39
• Of course e- is required, otherwise there would be no reduction. Besides, your equation must be balanced in charge as strict as it must be balanced in all elements. – Ivan Neretin Dec 25 '18 at 8:27

$$\ce{ 2H2O_{(l)} + 2e^- -> 2OH^-_{(aq)} + H_2_{(g)}}\quad\text{(Reduction)}$$