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.

The oxidizing agent experiences a decrease in oxidation number whole the reducing agent experiences an increase in oxidation number.

Oxidation refers to loss of electrons and the process leads to an increase in oxidation number while reduction has to do with gain of electrons. The reducing agent in a reaction is oxidized while the oxidizing agent is reduced.

For the reaction;

N2(g) + 3H2(g) →2NH3(g)

The element that is oxidized is hydrogen while the element that is reduced is nitrogen.

For the reaction;

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

The element that is oxidized is aluminium while the element that is reduced is iron

For the reaction;

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

The element that is oxidized is iodine while the element that is reduced is chlorine.

For the reaction;

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

The element that is oxidized is hydrogen while the element that is reduced is oxygen

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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

c) should be the answer.