# Which is the Electrolyse Reaction?

$$\ce{NaOH +H2O->H2 +O2 +NaOH}$$ We know that $\ce{H+}$ and $\ce{OH-}$ go to the cathode and anode, but how can I write the total reaction? Is it: $$\ce{2NaOH +2H2O->2H2 +O2 +2NaOH}$$ I've put NaOH both sides because if the mass of NaOH is given, how can I find the mass of H2 for example?

• Why would you need to write $\ce{NaOH}$ on both sides? Commented Nov 1, 2015 at 9:04
• Because if I'm given the mass of NaOH, how should I find the mass of other substances? Commented Nov 1, 2015 at 9:09

Electrolyse of water is a redox reaction, meaning that there is a transfer of electrons between atoms. Now, every redox reaction can be written as a combination of 2 half-equations. One half-equation shows one species getting oxidised (losing electrons) and the other half-equation shows the other species getting reduced (gaining electrons). When you combine these two half-equations you get the overall reaction, which is what you are asking for. Here is guide that should help you find the overall equation of any redox equation, not just this one example.

Step 1: Determine what species gets oxidised and species gets reduced

Now, what I think the reaction you are trying to describe is the electrolyse of water. Correct me if I am wrong. If that is the case, there is no need for $\ce{NaOH}$ in the equation as it appears on both side, meaning that they cancel each other out (also $\ce{NaOH}$ has nothing to do with electrolyse of water). Secondly, it is $\ce{H2O}$ that gets oxidised and reduced, not $\ce{H+}$ and $\ce{OH-}$.

Step 2: Determine half-equations

These can be easily found by using your table of standard reduction potentials. Oxidation half-equation: $$\ce{2H2O -> O2 + 4H+ + 4e-}$$ Reduction half-equation: $$\ce{2H2O + 2e- -> H2 +2OH-}$$

Step 3: Balance each equation so that there is the same number of electrons

We can't immediately add the half-equations as the electrons won't cancel each other out. This means that we need to multiply the reduction half-equation by 2:$$\ce{4H2O + 4e- -> 2H2 +4OH-}$$

When you add each half-equation, you get: $$\ce{6H2O -> 2H2 + O2 + 4H+ + 4OH-}$$ The $\ce{H+}$ and $\ce{OH-}$ combine to form water which then cancel out with the water the other side. Therefore the overall reaction is: $$\ce{2H2O -> 2H2 + O2}$$