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(D) In the given reaction: $Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_2(aq) + H_2(g)$$1$. The oxidation state of $Zn$ changes from $0$ to $+2$ (loss

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(D) In the given reaction: $Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_2(aq) + H_2(g)$$1$. The oxidation state of $Zn$ changes from $0$ to $+2$ (loss of $2$ electrons,oxidation).$2$. The oxidation state of $H$ in $HCl$ changes from $+1$ to $0$ in $H_2$ (gain of electrons,reduction).$3$. Since $H^+$ is being reduced,$HCl$ acts as the oxidizing agent.$4$. The reduction half-reaction is: $2H^+ + 2e^- \longrightarrow H_2$.$5$. This shows that $2$ moles of $H^+$ (which corresponds to $2$ moles of $HCl$) gain $2$ moles of electrons.$6$. Therefore,$1$ mole of the oxidizing agent $(HCl)$ gains $1$ mole of electrons.

What is the number of moles of electrons gained by one mole of oxidizing agent in the following redox reaction?
$Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_2(aq) + H_2(g)$

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What is the number of moles of electrons gained by one mole of oxidizing agent in the following redox reaction?$Zn_{(s)} + 2HCl_{(aq)} \longrightarrow ZnCl_2(aq) + H_2(g)$