Question

Путём электронного баланса составьте уравнения следующих окислительно востановительных реакций С+HNO3=CO2+NO+H2O, H2S+H2SO3=S+H2O, KI+KIO3+H2SO4= I2+K2SO4+H2OПомогите пожалуйста очень надо.... Пожалуйста помогите

Answer 1

1. 3C + 4 HNO3 = 3CO2 + 4 NO+ 2 H2O

C - 4 e = C

N +3e = N
2. 2H2S +H2SO3 = S + 3H2O

S(-2) - 2e = S(0)

S(+4) + 4e = S(0)
3. KIO3 + 3H2SO4 + 5KI = 3I2↓ + 3H2O + 3K2SO4

Восстановление:
2IO3− + 12H+ + 10e− = I2 + 6H2O

Окисление:
2I− − 2e− = I2
2IO3− + 12H+ + 10I− = 6I2 + 6H2O

Answer 2

Balancing Redox Reactions

Balancing redox reactions involves assigning oxidation numbers to the elements in the reactants and products and then balancing the number of electrons transferred. Here are the balanced equations for the given redox reactions:

1. C + HNO3 = CO2 + NO + H2O:

To balance this equation, we need to assign oxidation numbers to each element. The oxidation number of carbon (C) is 0 in CO2, and the oxidation number of nitrogen (N) is +5 in HNO3. The oxidation number of oxygen (O) is -2 in both CO2 and HNO3.

The balanced equation is:

C + 4HNO3 = CO2 + 2NO + 2H2O [[1]]

2. H2S + H2SO3 = S + H2O:

The oxidation number of sulfur (S) is -2 in H2S and +4 in H2SO3. The oxidation number of oxygen (O) is -2 in H2O and -2 in H2SO3.

The balanced equation is:

H2S + 2H2SO3 = S + 3H2O [[2]]

3. KI + KIO3 + H2SO4 = I2 + K2SO4 + H2O:

The oxidation number of iodine (I) is -1 in KI and +5 in KIO3. The oxidation number of sulfur (S) is +6 in H2SO4 and +6 in K2SO4. The oxidation number of oxygen (O) is -2 in H2O, -2 in KIO3, and -2 in H2SO4.

The balanced equation is:

2KI + KIO3 + 3H2SO4 = 3I2 + K2SO4 + 3H2O [[3]]

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