Question

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.

Answer 1

Answer:

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.

Answer 2

Answer:

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