Oxidation and Reduction?

Question

Identify which reagent gets oxidized in each of the following redox reactions

(HIO3 or H2SO3) -- HIO3 + 3H2SO3 --> HI + 3H2SO4

(HNO3 or H2S) -- 2HNO3 + 3H2S --> 3S + 2NO + 4H2O

(Si or Cr2O3) -- 3Si + 2Cr2O3 --> 4Cr + 3SiO2

(PbS or O2) -- 2PbS + 3O2 --> 2PbO + 2SO2


Identify which reagent gets reduced in each of the following redox reactions

(N2 or H2) -- N2 + 3H2 --> 2NH3

(H2SO3 or HNO3) -- 3H2SO3 + 2HNO3 --> 2NO + H2O + 3H2SO4

(Cr2O3 or Al) -- Cr2O3 + 2Al --> 2Cr + Al2O3

(CuCl2 or Zn) --CuCl2 + Zn --> ZnCl2 + Cu


Can you also explain to me how to figure this question out please.. It will benefit me in understanding others like it.

Answer 1

First: Which reagant oxidizes:

Oxidize means losing electrons getting a higher number of oxidation.

HIO3 or H2SO3) -- HIO3 + 3H2SO3 --> HI + 3H2SO4
H2SO3 oxidizes because S(+4) increases to S(+6) in H2SO4

(HNO3 or H2S) -- 2HNO3 + 3H2S --> 3S + 2NO + 4H2O
H2S oxidizes because S(-2) incrases to S(0) in S

(Si or Cr2O3) -- 3Si + 2Cr2O3 --> 4Cr + 3SiO2
Si oxidizes because Si(0) jumps to Si(+2) in SiO2

(PbS or O2) -- 2PbS + 3O2 --> 2PbO + 2SO2
PbS oxidizes just in the sense that gets oxygen but oxidation number remains th same.

Second, which reagant gets reduced?


Reduction is the gaining of electrons so, oxidation number gets lower.

(N2 or H2) -- N2 + 3H2 --> 2NH3
H2 gets reduced because H2(0) jumps down to H(-1) in ammonia

(H2SO3 or HNO3) -- 3H2SO3 + 2HNO3 --> 2NO + H2O + 3H2SO4
HNO3 gets reduced because N(5+) jumps down to N(+2) in NO

(Cr2O3 or Al) -- Cr2O3 + 2Al --> 2Cr + Al2O3
Cr2O3 gets reduced because Cr(3+) jumps down to Cr(0) in Cr

(CuCl2 or Zn) --CuCl2 + Zn --> ZnCl2 + Cu
CuCl2 gets reduced because Cu(2+) jumps down to Cu(0) in Cu

That´s it!

God luck!

Answer 2

Oxidation is the loss of e- from a substance. Reduction is the gaining of e- in a substance. The only way to be sure which is oxidized and which is reduced is to assign oxidation numbers to each element in the reactants and again in the products:

N2 + 3 H2 ---> 2 NH3

1) Oxidation number of any element in its free state is 0. Therefore the ox # of N2 = 0 and H2 = 0 The reactants all have ox #s of 0.
2) In a compound with a nonmetal H has an ox # of 1+ Since there are 3 H 1+ in NH3, the total is 3+ for the compound. That means that the N must have an ox # of 3- to counteract the 3+ total of the H.
3) Half reactions show which is oxidized and which is reduced.
N2 ---> N 3- this means it must gain 3 e- and reduction is gaining electrons.
H2 ---> H 1+ this means it must lose 1 e- and oxidation is the loss of e-.

Answer 3

Yes. Generally, a reagent that acquires oxygen is said to be oxidized. A reagent that loses oxygen is said to be reduced.

In the first of the 8 reactions, you can see that the H2SO3 picks up oxygen, and is oxidized.

In the second of the 8, the H2S picks up oxygen in a sense. It is just the H that picks up oxygen and becomes water, and the sulfur is freed.

In the third, Si picks up the oxygen lost by the Cr.

In the fourth, PbS is the only reagent that can be oxidized, because the other reagent is oxygen. Indeed, the Pb and the S each pick up oxygen.

In the sixth reaction, you can see that the HNO3 loses oxygen, and the H2SO3 picks it up.

That ammonia reaction is a mystery to me.