Which element is oxidized and which is reduced in the following reactions? (Part A) N2 (g) + 3H2 (g) →2NH3 (g) Express your answers as chemical symbols separated by a comma. Enter the oxidized element first. (Part B) 3Fe (NO3) 2 (aq) + 2Al (s) →3Fe (s) + 2Al (NO3) 3 (aq) Express your answers as chemical symbols separated by a comma. Enter the oxidized element first. (Part C) Cl2 (aq) + 2NaI (aq) → I2 (aq) + 2NaCl (aq) Express your answers as chemical symbols separated by a comma. Enter the oxidized element first. (Part D) PbS (s) + 4H2O2 (aq) →PbSO4 (s) + 4H2O (l) Express your answers as chemical symbols separated by a comma. Enter the oxidized element first.

N2 (g) + 3H2 (g) →2NH3 (g)

H, N

3Fe (NO3) 2 (aq) + 2Al (s) →3Fe (s) + 2Al (NO3) 3 (aq)

Al, Fe

Cl2 (aq) + 2NaI (aq) → I2 (aq) + 2NaCl (aq)

I, Cl

PbS (s) + 4H2O2 (aq) →PbSO4 (s) + 4H2O (l)

S, O

Explanation:

- Determinate the oxidation number in all compounds; the ox. number that increases means that the element oxidizes, if the number decreases, it is reduced.

N2 (g) + 3H2 (g) →2NH3 (g)

Both elements on the reactives has 0 as ox. number (0 as ground state)

In the ammonia, H acts with + 1 and N, with - 3

3Fe (NO3) 2 (aq) + 2Al (s) →3Fe (s) + 2Al (NO3) 3 (aq)

Fe acts with + 2 in reactives, it acts with 0 in products

Al acts with 0 in reactives, it acts with + 3 in products

Cl2 (aq) + 2NaI (aq) → I2 (aq) + 2NaCl (aq)

Cl acts with 0 in reactives, - 1 in products

I acts with - 1 in reactives, 0 in products

PbS (s) + 4H2O2 (aq) →PbSO4 (s) + 4H2O (l)

O acts with - 1 in reactives, - 2 in products

S acts with - 2 in reactives, + 6 in products

A. H, N

B. Al, Fe

C. I, Cl

D. S, O

Explanation:

To determine if an element is oxidized or reduced we have to consider the change in the oxidation number (ON).

If the ON increases, the element is oxidized. If the ON decreases, the element is reduced.

(Part A)

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

H is oxidized because its ON increases from 0 to + 1.

N is reduced because its ON decreases from 0 to - 3.

(Part B)

3 Fe (NO₃) ₂ (aq) + 2 Al (s) → 3 Fe (s) + 2 Al (NO₃) ₃ (aq)

Al is oxidized because its ON increases from 0 to + 3.

Fe is reduced because its ON decreases from + 2 to - 0.

(Part C)

Cl₂ (aq) + 2 NaI (aq) → I₂ (aq) + 2 NaCl (aq)

I is oxidized because its ON increases from - 1 to 0.

Cl is reduced because its ON decreases from 0 to - 1.

(Part D)

PbS (s) + 4 H₂O₂ (aq) → PbSO₄ (s) + 4 H₂O (l)

S is oxidized because its ON increases from - 2 to + 6.

O is reduced because its ON decreases from - 1 to - 2.