Question

K2CrO4+Kl+H2O-Ce2O3+I2+KOH Расставьте коэффициенты методом электронного баланса. Укажите окислитель и восстановитель,процессы окисления и восстановления.

Answer 1

2k2cro4+6ki+5H2O=cr2o3+3i2+10koh
окислитель: cr(+6)+3e=cr(+3) процесс восстановления
восстановитель: 2I(-)-2e=i2(0) процесс окисления

Answer 2

Balancing the Chemical Equation

To balance the chemical equation: K2CrO4 + Kl + H2O → Ce2O3 + I2 + KOH, we can use the method of electron balance. This method involves balancing the number of electrons gained and lost in the redox reactions.

First, let's identify the oxidation states of the elements involved in the reaction:

- K2CrO4: The oxidation state of K is +1, the oxidation state of Cr is +6, and the oxidation state of O is -2. - Kl: The oxidation state of K is +1 and the oxidation state of I is -1. - H2O: The oxidation state of H is +1 and the oxidation state of O is -2. - Ce2O3: The oxidation state of Ce is +3 and the oxidation state of O is -2. - I2: The oxidation state of I is 0. - KOH: The oxidation state of K is +1, the oxidation state of O is -2, and the oxidation state of H is +1.

Now, let's balance the equation step by step:

1. Balance the atoms other than hydrogen and oxygen: - K2CrO4 + 2Kl + H2O → Ce2O3 + I2 + KOH

2. Balance the oxygen atoms by adding water molecules: - K2CrO4 + 2Kl + H2O → Ce2O3 + I2 + KOH + 4H2O

3. Balance the hydrogen atoms by adding hydrogen ions (H+): - K2CrO4 + 2Kl + 2H+ + H2O → Ce2O3 + I2 + KOH + 4H2O

4. Balance the charges by adding electrons: - K2CrO4 + 2Kl + 2H+ + H2O → Ce2O3 + I2 + KOH + 4H2O + 6e-

5. Balance the electrons by multiplying the half-reactions: - 6K2CrO4 + 12Kl + 12H+ + 6H2O → 6Ce2O3 + 6I2 + 6KOH + 24H2O + 36e-

6. Simplify the equation by canceling out common terms: - 6K2CrO4 + 12Kl + 12H+ + 6H2O → 6Ce2O3 + 6I2 + 6KOH + 24H2O + 36e- - 6K2CrO4 + 12Kl + 6H+ → 6Ce2O3 + 6I2 + 6KOH + 18H2O + 36e-

Oxidation and Reduction Processes

In this reaction, the oxidation state of chromium (Cr) in K2CrO4 changes from +6 to +3 in Ce2O3. This indicates that chromium is being reduced and is the reducing agent (or the reductant). The reduction process involves the gain of electrons.

On the other hand, the oxidation state of iodine (I) in Kl changes from -1 to 0 in I2. This indicates that iodine is being oxidized and is the oxidizing agent (or the oxidant). The oxidation process involves the loss of electrons.

To summarize: - The oxidizing agent is Kl. - The reducing agent is K2CrO4. - The oxidation process occurs at Kl. - The reduction process occurs at K2CrO4.

Please note that the balanced equation and the identification of the oxidizing and reducing agents are based on the information provided.