Question

Подберите коэффициенты в схемах следущих окислительно-восстановительных реакции, укажите окислитель и восстановитель: a)Cu + HNO3=> Cu(NO3)2 + NO2 + H2O б)KI + U2 => KSE+ I2 в)H2O2 + H2S =>H2SO4 + H2O г)ZnS + O2 => ZnO + SO2 д)HI + H2SO4 => I2 + S + H2O

Answer 1

А) Cu + 4HNO3 --> 1Cu(NO3)2 + 2NO2 + 2H2O
Cu(0) -2e-->Cu(2+) 2 1 во-ель
N(5+) +1e-->N(4+). 2 ок-ель

Б). я не знаю

В)4H2O2 + 1H2S --> H2SO4 + 4H2O
S(-2) -8e ---> S(+6) 8 1
O²(-) +2e ---> O(-2). 4
1 -- вос-ель
4 -- окислитель

Answer 2

Oxidation-Reduction Reactions and Balancing Equations

Oxidation-reduction reactions, also known as redox reactions, involve the transfer of electrons between species. In these reactions, one species loses electrons (oxidation) while another species gains electrons (reduction). To balance redox equations, it is important to identify the oxidizing agent (the species that gets reduced) and the reducing agent (the species that gets oxidized).

Let's analyze the given reactions and determine the coefficients and the oxidizing and reducing agents for each one.

a) Cu + HNO3 -> Cu(NO3)2 + NO2 + H2O

In this reaction, copper (Cu) is oxidized from its elemental state (Cu) to Cu(II) in Cu(NO3)2. Nitric acid (HNO3) is reduced from its +5 oxidation state to +4 in NO2. The balanced equation for this reaction is:

Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O

The oxidizing agent is HNO3, which gains electrons and is reduced, while Cu is the reducing agent, which loses electrons and is oxidized.

b) KI + U2+ -> KSE + I2

In this reaction, iodide (I-) is oxidized from -1 to 0 in I2, while uranium (U2+) is reduced from +2 to 0 in U. The balanced equation for this reaction is:

2KI + U2+ -> KSE + I2

The oxidizing agent is KI, which gains electrons and is reduced, while U2+ is the reducing agent, which loses electrons and is oxidized.

c) H2O2 + H2S -> H2SO4 + H2O

In this reaction, hydrogen peroxide (H2O2) is reduced from its -1 oxidation state to 0 in H2O, while hydrogen sulfide (H2S) is oxidized from 0 to +6 in H2SO4. The balanced equation for this reaction is:

H2O2 + H2S -> H2SO4 + H2O

The oxidizing agent is H2S, which loses electrons and is oxidized, while H2O2 is the reducing agent, which gains electrons and is reduced.

d) ZnS + O2 -> ZnO + SO2

In this reaction, zinc sulfide (ZnS) is oxidized from 0 to +2 in ZnO, while oxygen (O2) is reduced from 0 to -2 in SO2. The balanced equation for this reaction is:

2ZnS + 3O2 -> 2ZnO + 2SO2

The oxidizing agent is O2, which gains electrons and is reduced, while ZnS is the reducing agent, which loses electrons and is oxidized.

e) HI + H2SO4 -> I2 + S + H2O

In this reaction, hydrogen iodide (HI) is oxidized from -1 to 0 in I2, while sulfuric acid (H2SO4) is reduced from +6 to +4 in S. The balanced equation for this reaction is:

8HI + H2SO4 -> 4I2 + H2S + 4H2O

The oxidizing agent is H2SO4, which gains electrons and is reduced, while HI is the reducing agent, which loses electrons and is oxidized.

I hope this helps! Let me know if you have any further questions.