Question

How many electrons are needed to reduce N$_2$ to NH$_3$?

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

Using a simple frame or just bolding for the box Key Points: Reduction involves a decrease in oxidation state (gain of electrons). Determine the oxidation state of the element in the reactant and product. Calculate the change in oxidation state per atom. Multiply the change per atom by the number of atoms of that element in the reactant molecule (N$_2$ has 2 N atoms) to find the total electrons transferred for the molecule. Oxidation State: N in N$_2$ = 0; N in NH$_3$ = -3.

Answer 1

The Correct Option is D

To determine the number of electrons needed for the reduction, we can examine the change in the oxidation state of nitrogen. (A) Oxidation state of N in N₂: In its elemental form (N₂), the oxidation state of nitrogen is 0. (B) Oxidation state of N in NH₃: Hydrogen usually has an oxidation state of +1 when bonded to non-metals. Let the oxidation state of nitrogen be x. The sum of oxidation states in a neutral molecule is 0.

x + 3(+1) = 0
x + 3 = 0
x = -3. The oxidation state of nitrogen in NH₃ is -3. (C) Change in oxidation state per N atom: The oxidation state changes from 0 (in N₂) to -3 (in NH₃).

Change = Final State - Initial State = -3 - 0 = -3. This change of -3 indicates that each nitrogen atom gains 3 electrons during the reduction. (D) Total electrons for N₂ molecule: The reactant is N₂, which contains two nitrogen atoms. To reduce one molecule of N₂ to form ammonia (which requires forming two NH₃ molecules: N₂ → 2NH₃), both nitrogen atoms must be reduced.

Total electrons needed = (Number of N atoms) × (Electrons gained per N atom)
Total electrons needed = 2 atoms × 3 electrons/atom = 6 electrons. Alternatively, we can write the balanced half-reaction (e.g., in acidic solution):
N₂ → 2NH₃
Balance H atoms by adding H⁺:
N₂ + 6H⁺ → 2NH₃
Balance charge by adding electrons (e^-) to the side with the higher positive charge:
N₂ + 6H⁺ + 6e^- → 2NH₃
The balanced half-reaction shows that 6 electrons are required to reduce one molecule of N₂.