Question

подберите коэффициенты к уравнениям окислительно-восстановительных реакций, используя метод электронно-ионного баланса, укажите окислитель и восстановитель, процессы окисления и восстановления.KMnO4 + C6H12O6 + H2SO4 ® MnSO4 + CO2 + K2SO4 + H2O

Answer 1

Mn⁺⁷ +5e⁻=Mn⁺²  5            24      -восстановление
                               120
C°₆ - 4*6e⁻=6C⁺⁴ 24           5      - окисление
24KMnO4+5C6H12O6+36H2SO4=24MnSO4+30CO2+66H2O+12K2SO4
KMnO4 окислитель
C6H12O6 восстановитель

Answer 2

Balancing the Redox Reaction: KMnO4 + C6H12O6 + H2SO4 -> MnSO4 + CO2 + K2SO4 + H2O

To balance the given redox reaction using the electron-ion method, we need to follow a step-by-step process. Let's break it down:

Step 1: Assign oxidation numbers to each element in the reaction. Step 2: Identify the elements that are being oxidized and reduced. Step 3: Balance the atoms and charges in the half-reactions. Step 4: Balance the number of electrons transferred. Step 5: Combine the half-reactions and simplify the equation.

Let's go through each step in detail:

Step 1: Assign oxidation numbers

To assign oxidation numbers, we need to know the rules: - The oxidation number of an element in its elemental form is always zero. - The sum of oxidation numbers in a neutral compound is zero. - The oxidation number of a monatomic ion is equal to its charge.

In KMnO4: - The oxidation number of K is +1. - The oxidation number of Mn is +7. - The oxidation number of O is -2.

In C6H12O6: - The oxidation number of C is -2. - The oxidation number of H is +1. - The oxidation number of O is -2.

In H2SO4: - The oxidation number of H is +1. - The oxidation number of S is +6. - The oxidation number of O is -2.

Step 2: Identify the elements being oxidized and reduced

In this reaction, Mn is being reduced from +7 to +2, while C is being oxidized from -2 to +4.

Step 3: Balance the atoms and charges in the half-reactions

We will split the reaction into two half-reactions: the reduction half-reaction and the oxidation half-reaction.

Reduction half-reaction: MnO4- -> Mn2+ To balance the atoms, we need to add 4 H+ ions to the right side: MnO4- + 4H+ -> Mn2+ + 2H2O

Oxidation half-reaction: C6H12O6 -> CO2 To balance the atoms, we need to add 6 H2O molecules to the left side: C6H12O6 + 6H2O -> 6CO2

Step 4: Balance the number of electrons transferred

To balance the number of electrons transferred, we need to multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2:

5(MnO4- + 4H+ -> Mn2+ + 2H2O) 2(C6H12O6 + 6H2O -> 6CO2)

Step 5: Combine the half-reactions and simplify the equation

Now, we can combine the two half-reactions and cancel out the common species:

5MnO4- + 20H+ + 2C6H12O6 -> 5Mn2+ + 10H2O + 12CO2

Finally, we can simplify the equation by dividing all coefficients by 2:

5KMnO4 + 8C6H12O6 + 6H2SO4 -> 5MnSO4 + 12CO2 + 3K2SO4 + 6H2O

In this balanced equation, KMnO4 is the oxidizing agent, and C6H12O6 is the reducing agent. The oxidation process occurs to C6H12O6, while the reduction process occurs to MnO4-.