Respuesta :
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
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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