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 find the number of electrons needed for reduction, we analyze nitrogen's oxidation state change. (A) N in N₂ oxidation state: Elemental nitrogen (N₂) has an oxidation state of 0. (B) N in NH₃ oxidation state: Hydrogen usually has a +1 oxidation state when bound to non-metals. Let nitrogen's oxidation state be *x*. In a neutral molecule, the sum of oxidation states is 0.
*x* + 3(+1) = 0
*x* + 3 = 0
*x* = -3. Nitrogen in NH₃ has an oxidation state of -3. (C) Oxidation state change per N atom: The oxidation state changes from 0 (in N₂) to -3 (in NH₃).
Change = Final State - Initial State = -3 - 0 = -3. Each nitrogen atom gains 3 electrons during reduction. (D) Total electrons for N₂ molecule: N₂ has two nitrogen atoms. Reducing one N₂ molecule to ammonia (N₂ → 2NH₃) requires reducing both nitrogen atoms.
Total electrons needed = (Number of N atoms) × (Electrons gained per N atom)
Total electrons needed = 2 atoms × 3 electrons/atom = 6 electrons. Alternatively, consider 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^-):
N₂ + 6H⁺ + 6e^- → 2NH₃
The balanced half-reaction shows 6 electrons are needed to reduce one N₂ molecule.