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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 $\ce{2 H2O_{(l)} -> 2 H2_{(g)} + O2_{(g)}}$

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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.

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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.

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