Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

Chemistry Tip: Pay close attention to the wording! The total electron transfer for the balanced equation is 2, but the question asks "per mole of oxidizing agent" ($HCl$), which cuts the ratio in half.

Answer 1

The Correct Option is D

Concept: Chemistry (Redox Reactions) - Oxidizing Agents and Electron Transfer.

Step 1: Determine the oxidation states of the elements. In the reactants: Solid zinc ($Zn$) is in its elemental state, so its oxidation number is 0. In $HCl$, hydrogen is $+1$ and chlorine is $-1$. In the products: In zinc chloride ($ZnCl_2$), zinc is $+2$ and chlorine remains $-1$. Hydrogen gas ($H_2$) is elemental, so its oxidation number is 0.

Step 2: Identify the oxidation and reduction half-reactions. Oxidation (loss of electrons): Zinc's oxidation state increases from 0 to $+2$. The half-reaction is $Zn \rightarrow Zn^{2+} + 2e^-$. Reduction (gain of electrons): Hydrogen's oxidation state decreases from $+1$ to 0. The half-reaction is $2H^+ + 2e^- \rightarrow H_2$.

Step 3: Identify the oxidizing agent. An oxidizing agent is the substance that gets reduced (gains electrons) and causes the oxidation of another substance. Since the $H^+$ ions in $HCl$ are gaining electrons and being reduced to $H_2$, $HCl$ is the oxidizing agent.

Step 4: Analyze the electron transfer per mole of the reaction. From the balanced overall equation $Zn + 2HCl \rightarrow ZnCl_2 + H_2$ and the reduction half-reaction $2H^+ + 2e^- \rightarrow H_2$, we can see that it requires exactly 2 moles of electrons to reduce 2 moles of $HCl$.

Step 5: Calculate the electron gain for ONE mole of the oxidizing agent. The question specifically asks for the number of moles of electrons gained by exactly ONE mole of the oxidizing agent. Since 2 moles of $HCl$ gain 2 moles of electrons, we divide by 2. Therefore, 1 mole of $HCl$ gains $\frac{2}{2} = 1$ mole of electrons. $$ \therefore \text{One mole of the oxidizing agent gains } 1 \text{ mole of electrons.} $$