Question

Observe the following unbalanced equations:
(I) $H_2O(l) + Na(s) \rightarrow$
(II) $H_2O(l) + F_2(g) \rightarrow$
Identify the correct statement.

• A. In (I), water is oxidized to $H_2$ and in (II) water is reduced to $O_2$.
• B. In both (I) and (II), water is oxidized to $O_2$.
• C. In both (I) and (II), water is reduced to $H_2$.
• D. In (I), water is reduced to $H_2$ and in (II) water is oxidized to $O_2$.

Hint:

Water is amphoteric in redox reactions. With active metals (Na, K), it acts as an {oxidizing agent} ($H^+ \to H_2$). With strong non-metals (F), it acts as a {reducing agent} ($O^{2-} \to O_2$).

Answer 1

The Correct Option is D

Step 1: Analyze Reaction (I) - Sodium with Water:
Balanced Equation: $2Na + 2H_2O \rightarrow 2NaOH + H_2$.

• Oxidation State of H in $H_2O$: $+1$.
• Oxidation State of H in $H_2$: $0$.
• Change: $+1 \rightarrow 0$ (Decrease in oxidation number = Gain of electrons).
• Conclusion: Water acts as an oxidizing agent and gets reduced to Hydrogen gas ($H_2$). Sodium is a strong reducing agent.

Step 2: Analyze Reaction (II) - Fluorine with Water:
Balanced Equation: $2F_2 + 2H_2O \rightarrow 4HF + O_2$.

• Oxidation State of O in $H_2O$: $-2$.
• Oxidation State of O in $O_2$: $0$.
• Change: $-2 \rightarrow 0$ (Increase in oxidation number = Loss of electrons).
• Conclusion: Water acts as a reducing agent and gets oxidized to Oxygen gas ($O_2$). Fluorine is a very strong oxidizing agent.

Step 3: Final Answer:
In (I) water is reduced to $H_2$, and in (II) water is oxidized to $O_2$. This matches Option (D).