Question

what are the half reactions for the single displacement reaction of: mg(s) + hcl(aq) ==> mgcl2(aq) + h2(g) 2e⁻ + mg(s) ==> mg²⁺(aq) mg(s) ==> mg²⁺(aq) + 2e⁻ 2e⁻ + 2h⁺(aq) ==> h2(g) 2cl⁻(aq) ==> cl2(aq) + e⁻

Explanation:

Step1: Identify Oxidation Half - Reaction

In the reaction \( \text{Mg}_{(s)}+\text{HCl}_{(aq)}
ightarrow\text{MgCl}_2_{(aq)}+\text{H}_2_{(g)} \), Magnesium (Mg) goes from an oxidation state of 0 (in solid Mg) to + 2 (in \( \text{Mg}^{2 + } \)). Oxidation is the loss of electrons. So the oxidation half - reaction is \( \text{Mg}_{(s)}
ightarrow\text{Mg}^{2+}_{(aq)} + 2e^- \) because Mg loses 2 electrons to form \( \text{Mg}^{2+} \).

Step2: Identify Reduction Half - Reaction

Hydrogen in HCl has an oxidation state of + 1 (in \( \text{H}^+ \)). In \( \text{H}_2 \), the oxidation state of H is 0. Reduction is the gain of electrons. Each \( \text{H}^+ \) gains 1 electron, and since 2 \( \text{H}^+ \) form \( \text{H}_2 \), the reduction half - reaction is \( 2e^-+2\text{H}^+_{(aq)}
ightarrow\text{H}_2_{(g)} \). The half - reaction \( 2e^-+\text{Mg}_{(s)}
ightarrow\text{Mg}^{2+}_{(aq)} \) is incorrect because Mg should lose electrons (oxidation), not gain. The half - reaction \( 2\text{Cl}^-_{(aq)}
ightarrow\text{Cl}_2_{(aq)}+e^- \) is incorrect because Cl does not change its oxidation state in this reaction (it remains - 1 in HCl and in \( \text{MgCl}_2 \)).

Answer:

The correct half - reactions are:

• \( \text{Mg}_{(s)}\ ==>\ \text{Mg}^{2+}_{(aq)}\ +\ 2e^- \)
• \( 2e^-\ +\ 2\text{H}^+_{(aq)}\ ==>\ \text{H}_2_{(g)} \)