Ethanoic acid can be converted into ethanol using reagent

$LiAlH_4$ in $H_2O$

The correct answer is Option (3) → $LiAlH_4$ in $H_2O$

Conversion of carboxylic acid $\rightarrow$ primary alcohol is a reduction reaction. A strong hydride donor is required to reduce the $-\text{COOH}$ group to $-\text{CH}_2\text{OH}$.

Option Analysis

Option 1: $\text{Sn} + \text{HCl}$: This system is mainly used to reduce nitro compounds ($-\text{NO}_2$) to amines. It does not reduce carboxylic acids to alcohols. Hence it is unsuitable.

Option 2: $\text{Na}$ and $\text{C}_2\text{H}_5\text{OH}$: Sodium reacts with ethanol to form sodium ethoxide and hydrogen gas. No reduction of carboxylic acid occurs here.

Option 3: $\text{LiAlH}_4$ in $\text{H}_2\text{O}$: Lithium aluminium hydride is a strong reducing agent. It reduces carboxylic acids to primary alcohols by converting $-\text{COOH}$ to $-\text{CH}_2\text{OH}$. This is the correct reagent.

Option 4: $\text{H}_2$ and $\text{Pt}$: Catalytic hydrogenation generally reduces double bonds and some carbonyls, but carboxylic acids require harsher conditions and are not efficiently converted to alcohols this way in standard reactions.