# Balancing Oxidation-Reduction equations

Anode: $\ce{2 H2O_{(l)} -> O2 + 4 H+_{(aq)} + 4e^-}$

Cathode: $\ce{ 2 H2O_{(l)} + 2 e- -> H2_{(g)} + 2 OH-_{(aq)}}$

Can someone show me the steps for adding the two. For the overall reaction, it is $\ce{2 H2O_{(l)} -> 2 H2_{(g)} + O2_{(g)}}$

It is not entirely inconceivable that a Galvanic cell could be created with an acidic anode and basic cathode. Having both half-reactions occurring in the same beaker, however, is unlikely. That has implications when determining the thermodynamics (cell potential) of the reaction but not in balancing the half reactions. Given: \begin{align} \ce{2 H2O_{(l)} & -> O2 + 4 H+_{(aq)} + 4e^-} \tag{1}\\ \ce{ 2 H2O_{(l)} + 2 e- & -> H2_{(g)} + 2 OH-_{(aq)}} \tag{2} \end{align}

Multiply equation 2 by 2 and add:

\begin{align} \ce{2 H2O_{(l)} & -> O2 + 4 H+_{(aq)} + 4e^-} \tag{3}\\ \ce{4 H2O_{(l)} + 4 e- & -> 2 H2_{(g)} + 4 OH-_{(aq)}} \tag{4} \\ \ce{6 H2O_{(l)} + 4 e- & -> O2 + 2 H2 + 4 H+ + 4 OH- + 4 e^-} \tag{5} \end{align}

The key here is that the $\ce{4H+ + 4OH-}$ showing up on the same side equation 5 will yield $\ce{4H2O}$. Combining acid and base (to make $\ce{4 H2O}$ on the product side) and canceling the $\ce{4e-}$ and $\ce{4 H2O}$ that occur on both sides of the chemical equation gives the desired equation.

You cant really combine those because for the top one you're making an acid, and for the bottom one you're making a base. You've either got to make an acid on one side and consume it on the other, or consume a base on one side and make it on the other. Here are the two ways:

Anode: $\ce{2H2O(l) ⟶ O2(g) + 4H+(aq) + 4e-}$

Cathode: $\ce{2H+(aq) + 2e- ⟶ H2(g)}$

Multiply bottom by two and the $\ce{H+}$ and $\ce{e-}$ cancel out with the top leaving the overall reaction.

Or:

Anode: $\ce{4OH- (aq) ⟶ 2H2O(l) + O2(g) + 4e-}$

Cathode: $\ce{2H2O(l) + 2e- ⟶ H2(g) + 2OH- (aq)}$

Multiply bottom by two and again get the overall reaction.