Question

Explain why the following reaction occurs:
Zn + 2HCl → ZnCl2 + H,

Answer 1

Answer:

The reaction $\rm Zn\, (s) + 2\, HCl \, (aq) \to ZnCl_2\, (aq) + H_2\, (g)$ is spontaneous (under standard conditions) because the cell potential is positive.

Explanation:

In this reaction:

• Metallic zinc $\rm Zn\, (s)$ is oxidized to produce zinc ions $\rm Zn^{2+}$.
• Hydrogen ions $\rm H^{+}$ is reduced to produce hydrogen gas $\rm H_2\, (g)$.

The anode is where the oxidation half-reaction takes place. In this case, the anode half-reaction is $\rm Zn\, (s) \to Zn^{2+} \, (aq) + 2\, e^{-}$.

The cathode is where the reduction half-reaction takes place. In this case, the cathode half-reaction is $\rm 2\, H^{+}\, (aq) + 2\, e^{-} \to H_2\, (g)$.

Look up the standard reduction potential for these two half-reactions:

• $E^{\circ}_{\text{reduction, anode}} = \rm -0.7618\; V$ for the reaction $\rm Zn^{2+} \, (aq) + 2\, e^{-} \rightleftharpoons Zn\, (s)$.
• $E^{\circ}_{\text{reduction, cathode}} = \rm 0\; V$ for the reaction $\rm H^{+} \, (aq) + e^{-} \rightleftharpoons \dfrac{1}{2}H_2\, (g)$.

The standard cell potential is equal to the standard reduction potential at the cathode minus that at the anode:

$E^{\circ}_\text{cell} = E^{\circ}_{\text{reduction, cathode}} - E^{\circ}_{\text{reduction, anode}} = \rm 0.7618\; V$.

The electrochemical reaction in the cell is spontaneous if and only if the cell potential is positive. Therefore, the reaction  $\rm Zn\, (s) + 2\, HCl \, (aq) \to ZnCl_2\, (aq) + H_2\, (g)$ is spontaneous under standard conditions.