Question 2 of 31intermediate🔧 ApplyShort Answer2 marks

Justify that the reaction: 2Na(s) + H₂(g) → 2NaH(s) is a redox change.

Correct Answer

In the given reaction, the compound formed is an ionic compound, which can be represented as Na+H−\text{Na}^+\text{H}^- (s). This suggests that the reaction can be split into two half reactions.

The first half reaction is: 2Na(s)→2Na+(g)+2e−2\text{Na}(s) \rightarrow 2\text{Na}^+(\text{g}) + 2\text{e}^-. Here, sodium loses electrons, meaning it undergoes oxidation. The second half reaction is: H2(g)+2e−→2H−(g)\text{H}_2(\text{g}) + 2\text{e}^- \rightarrow 2\text{H}^-(\text{g}). Here, hydrogen gains electrons, meaning it undergoes reduction.

Since sodium is oxidised and hydrogen is reduced simultaneously in this process, the complete reaction is a redox change.

Exercise: Problem 7.2 | Q: Problem 7.2 | (Page Page 3)
For More Understanding

Explanation

The textbook solution splits the reaction into two half-reactions based on the ionic nature of NaH (Na+H-). Sodium loses electrons (oxidation) and hydrogen gains electrons (reduction), fulfilling the definition of a redox reaction.

Solution Steps

  1. Step 1: Represent the ionic compound formed as Na+H−\text{Na}^+\text{H}^- (s).

  2. Step 2: Write the oxidation half-reaction: 2Na(s)→2Na+(g)+2e−2\text{Na}(s) \rightarrow 2\text{Na}^+(\text{g}) + 2\text{e}^-.

  3. Step 3: Write the reduction half-reaction: H2(g)+2e−→2H−(g)\text{H}_2(\text{g}) + 2\text{e}^- \rightarrow 2\text{H}^-(\text{g}).

  4. Step 4: Conclude that since both oxidation and reduction occur, it is a redox change.

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