Question

Observe the following reactions: \[ \text{I. } H_2O(l) + 2Na(s) \rightarrow 2NaOH(aq) + H_2(g) \] \[ \text{II. } 2H_2O(l) + 2F_2(g) \rightarrow 4H^+(aq) + 4F^-(aq) + O_2(g) \] Identify the correct statement:

• A. In both reaction I and reaction II, water is oxidized.
• B. In both reaction I and reaction II, water is reduced.
• C. In reaction I water is reduced and in reaction II water is oxidized.
• D. In reaction I water is oxidized and in reaction II water is reduced.

Hint:

Oxidation is the loss of electrons, while reduction is the gain of electrons.

Answer 1

The Correct Option is C

To identify the correct statement about the reactions, we need to analyze the oxidation and reduction processes in each reaction. Reaction I: \[ H_2O(l) + 2Na(s) \rightarrow 2NaOH(aq) + H_2(g) \] Oxidation: Sodium (\(Na\)) is oxidized from \(0\) to \(+1\). Reduction: Water (\(H_2O\)) is reduced. The hydrogen in \(H_2O\) goes from \(+1\) to \(0\) in \(H_2\). Thus, in Reaction I, water is reduced. Reaction II: \[ 2H_2O(l) + 2F_2(g) \rightarrow 4H^+(aq) + 4F^-(aq) + O_2(g) \] Oxidation: Water (\(H_2O\)) is oxidized. The oxygen in \(H_2O\) goes from \(-2\) to \(0\) in \(O_2\). Reduction: Fluorine (\(F_2\)) is reduced from \(0\) to \(-1\). Thus, in Reaction II, water is oxidized. Correct Statement:
In Reaction I, water is reduced.
In Reaction II, water is oxidized. The correct option is: \[ \boxed{\text{(3) In reaction I water is reduced and in reaction II water is oxidized.}} \]