How do I balance this using the half-reaction method: $$\ce{Zn(s) + HCl(aq) -> Zn^{2+} (aq) + H_2 (g)}$$

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

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

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

However, I feel like I made a mistake somewhere. Any help please?

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    $\begingroup$ Try to blanace electrons, like you did: Zn(s)→Zn2+ +2e−, where these two electrons are going to? $\endgroup$ – Jaroslav Kotowski Apr 10 '15 at 12:56
  • $\begingroup$ Keep the chlorine ionic and don't react it with anything. Cl- should stay Cl-. $\endgroup$ – Burak Ulgut Apr 10 '15 at 13:04
  • $\begingroup$ I don't know where the electrons are going. I'm confused because in the equation $H^+ + Cl^- \rightarrow H_2 (g)$ there is a $Cl^-$ ion the left, but then it dissapears? Where does it go then? $\endgroup$ – Kamil Apr 10 '15 at 13:45
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    $\begingroup$ The mistake is right at the beginning in the initial equation. The chlorine has disappeared. So you need to rethink this equation and the subsequent reduction equation. $\endgroup$ – bon Apr 10 '15 at 15:02
  • $\begingroup$ I recently answered a question like this. It's not the top rated answer, though I feel it may be helpful: math.stackexchange.com/questions/624/… $\endgroup$ – ringo Apr 11 '15 at 2:54

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.


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