Question

Используя метод электронного баланса, расставьте коэффициенты и укажите окислитель и восстановитель в реакции KI + H2SO4 (конц.) ---> I2 +H2S + K2SO4 + H2O I - это йод, если что

Answer 1

+  -      + +6 -2         0     + -2      + +6-2      + -2
8KI + 5H2SO4(k) =4 I2 + H2S +4 K2SO4 + 4H2O
 -             0
2I -2*1e→I2      2          4  процесс окисления, восстановитель
 +6         -2              8
S + 8e→S        8           1 процесс восстановления, окислитель

Answer 2

2J - - 2e= J2 0          |4|

S +6 +8 e = S 2-       |1|

8KJ + 5H2SO4 = 4J2 + H2S + 4 K2SO4 + 4 H2O

Answer 3

Balancing the Equation using the Electron Balance Method

To balance the equation: KI + H2SO4 (conc.) -> I2 + H2S + K2SO4 + H2O, we can use the electron balance method. This method involves balancing the number of electrons gained and lost during the reaction.

Let's break down the reaction into half-reactions:

Oxidation Half-Reaction: KI -> I2

Reduction Half-Reaction: H2SO4 -> H2S + K2SO4 + H2O

Now, let's balance each half-reaction individually.

Balancing the Oxidation Half-Reaction

In the oxidation half-reaction, iodide ions (I-) are oxidized to form iodine (I2). To balance the number of iodine atoms, we need to add a coefficient of 2 in front of KI:

2KI -> I2

Balancing the Reduction Half-Reaction

In the reduction half-reaction, sulfuric acid (H2SO4) is reduced to form hydrogen sulfide (H2S), potassium sulfate (K2SO4), and water (H2O). To balance the number of sulfur atoms, we need to add a coefficient of 1 in front of H2S:

H2SO4 -> H2S + K2SO4 + H2O

Now, let's balance the hydrogen and oxygen atoms. We have 4 hydrogen atoms on the left side and 6 hydrogen atoms on the right side. To balance the hydrogen atoms, we need to add a coefficient of 4 in front of H2O:

H2SO4 -> H2S + K2SO4 + 4H2O

Next, let's balance the oxygen atoms. We have 4 oxygen atoms on the left side and 10 oxygen atoms on the right side. To balance the oxygen atoms, we need to add a coefficient of 6 in front of H2SO4:

6H2SO4 -> H2S + K2SO4 + 4H2O

Final Balanced Equation

Now that we have balanced both half-reactions, we can combine them to form the balanced equation:

2KI + 6H2SO4 -> I2 + H2S + K2SO4 + 4H2O

In this reaction, iodine (I2) is the oxidizing agent because it causes the oxidation of iodide ions (I-) to form iodine (I2). Conversely, sulfuric acid (H2SO4) is the reducing agent because it undergoes reduction to form hydrogen sulfide (H2S), potassium sulfate (K2SO4), and water (H2O).

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