Balancing Redox Equations - Half Reactions How do I balance this using the half-reaction method: $$\ce{Zn(s) + HCl(aq) -&gt; Zn^{2+} (aq) + H_2 (g)}$$

I would first split it up in its ionic components:
$\ce{Zn(s) + H^+ + Cl^- -&gt; Zn^{2+} (aq) + H_2 (g)}$. 

Then we have first the oxidation-reaction: $\ce{Zn(s) -&gt; Zn^{2+} (aq)}$. Balancing this with electrons gives $\ce{Zn(s) -&gt; Zn^{2+} + 2e^-}$. 

The reduction-reaction is $\ce{H^+ + Cl^- -&gt; H_2 (g)}$. I need to balance this with $\ce{Cl}$? So that would get me $\ce{H^+ + Cl^- -&gt; H_2 (g) + Cl}$. Then adding hydrogen to the left side with $\ce{H^+}$ gives $\ce{2H^+ + Cl^- -&gt; H_2 (g) + Cl}$. 

However, I feel like I made a mistake somewhere. Any help please?
You are right for the half reaction for oxidation is:
$$\ce{Zn(s) \rightarrow Zn^{2+} + 2e^-}$$
Now, you have to write properly the  half reaction for reduction:
 $$\ce{2H^+(\mathrm{aq}) + 2e^- \rightarrow H_2(g)}$$

This means, that the electrons given by zinc are taken by ion hydrogen.
So, if we add the two half reactions:
$$\ce{Zn(s) + 2H^+(\mathrm{aq}) \rightarrow Zn^{2+}(\mathrm{aq}) +  H_2 (g)}$$
Notice, that ion chloride doesn't change its oxidation state. Its role is to keep the medium neutral electrically. So, we'll add two ions chloride to the two members of the equation:
$$\ce{Zn(s) + 2(H^+_{\mathrm{aq}} + Cl^{-}_{\mathrm{aq}}) \rightarrow (Zn^{2+}_{\mathrm{aq}} +2 Cl^{-}_{\mathrm{aq}}) + H_2(g)}$$
Remember, redox reaction are reactions of electrons exchange between an oxidant and an reducer.