Question

Расставьте коэффициенты в окислительно-восстановительных реакци-ях методом электронного баланса. Укажите процесс окисления и восстановления,окислитель и восстановитель и определите их эквиваленты: K2Cr2O7 + HCl → CrCl3 + KCl + Cl2 + H2O;

Answer 1

 2K2Cr2O7 + 16HCl →4CrCl3 +4KCl + Cl2 + 8H2O советую зайти на химик ру там есть онлайн прога которая уравнивает

Answer 2

 2K2Cr2O7 + 16HCl →4CrCl3 +4KCl + Cl2 + 8H2O;

Answer 3

K2Cr2O7 + 14HCl → 2CrCl3 + 2KCl + 7Cl2 + 7H2O

Процесс окисления: Cr6+ → Cr3+ Процесс восстановления: Cl- → Cl2

Окислитель: K2Cr2O7 Восстановитель: HCl

Эквивалент окислителя (K2Cr2O7) = 1 Эквивалент восстановителя (HCl) = 14

Answer 4

Balancing the Redox Reaction: K2Cr2O7 + HCl → CrCl3 + KCl + Cl2 + H2O

To balance the given redox reaction using the method of electron balance, we need to assign oxidation numbers to each element in the reaction and then balance the number of electrons transferred.

The reaction is: K2Cr2O7 + HCl → CrCl3 + KCl + Cl2 + H2O

Let's assign oxidation numbers to each element in the reaction:

- The oxidation number of K is +1. - The oxidation number of Cr is +6. - The oxidation number of O is -2. - The oxidation number of H is +1. - The oxidation number of Cl is -1.

Now, let's identify the elements that are being oxidized and reduced:

- The oxidation state of Cr changes from +6 to +3, so it is being reduced. - The oxidation state of Cl changes from -1 to 0, so it is being oxidized.

To balance the reaction, we need to make sure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.

Balancing the Reduction Half-Reaction:

The reduction half-reaction involves the reduction of Cr from +6 to +3. To balance the reduction half-reaction, we need to add 3 electrons to the left side:

Cr2O7^2- + 14H^+ + 6e^- → 2Cr^3+ + 7H2O [[1]]

Balancing the Oxidation Half-Reaction:

The oxidation half-reaction involves the oxidation of Cl from -1 to 0. To balance the oxidation half-reaction, we need to add 6 electrons to the right side:

2Cl^- → Cl2 + 2e^- [[2]]

Combining the Half-Reactions:

To combine the half-reactions, we need to multiply each half-reaction by a suitable factor so that the number of electrons transferred is the same in both reactions. In this case, we need to multiply the reduction half-reaction by 6 and the oxidation half-reaction by 3:

6Cr2O7^2- + 42H^+ + 18e^- → 12Cr^3+ + 21H2O (reduction half-reaction) 6Cl^- → 3Cl2 + 6e^- (oxidation half-reaction)

Now, we can add the two half-reactions together:

6Cr2O7^2- + 42H^+ + 18e^- + 6Cl^- → 12Cr^3+ + 21H2O + 3Cl2

Final Balanced Equation:

Finally, we can simplify the equation by canceling out the common species on both sides:

K2Cr2O7 + 14HCl → 2CrCl3 + 2KCl + 3Cl2 + 7H2O

In this balanced equation, the process of oxidation occurs to chlorine (Cl), which changes its oxidation state from -1 to 0, forming chlorine gas (Cl2). The process of reduction occurs to chromium (Cr), which changes its oxidation state from +6 to +3, forming chromium(III) chloride (CrCl3).

The equivalents of the oxidizing agent (K2Cr2O7) and the reducing agent (HCl) can be determined by comparing the coefficients in the balanced equation. In this case, the equivalents of K2Cr2O7 and HCl are both equal to their respective coefficients, which are 1.

Please note that the above explanation is based on the information available at the time of the response.