Observe the following unbalanced equations:
(I) $H_2O(l) + Na(s) \rightarrow$
(II) $H_2O(l) + F_2(g) \rightarrow$
Identify the correct statement.
Show 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} ($O^{2-} \to O_2$).
In (I), water is oxidized to $H_2$ and in (II) water is reduced to $O_2$.
In both (I) and (II), water is oxidized to $O_2$.
In both (I) and (II), water is reduced to $H_2$.
In (I), water is reduced to $H_2$ and in (II) water is oxidized to $O_2$.
Show Solution
The Correct Option isD
Solution and Explanation
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).