Video: Novi's final assignment (Equalization of Redox Reaction Equations)

Equalization of Redox Reaction Equations

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https://www.youtube.com/watch?v=aryN7NGmdz8

  

How to equalize the redox equation can be done in two ways, ie
• half way reaction, and
• How to change the oxidation number.
Way of half reaction
How to equalize the equation of a redox reaction by means of a half-reaction, ie by looking at the electrons received or released. Equivalence is done by equalizing the number of electrons. This preferred way for reaction with reaction atmosphere is known. Equalization mode:

Stage 1       : Write down half the reaction for the two substances to be reacted.
Stage 2       : Fix the elements that change the oxidation state.
Stage 3       : Add one H2O molecule to:
                     - Acid atmosphere: on the deficiency of the O atom.
                     - The atmosphere of base: on the excess of the O atom.
Stage 4       : Resolve the hydrogen atom by:
                     - Acid atmosphere: by adding H + ions.
                     - Atmosphere of base: by adding OH- ions.
Stage 5       : Raise the charge by adding electrons.
Stage 6       : Simulate the number of electrons received with the released, then sum it up.

Here is a brief explanation of the half-reaction method: unequal redox equations are converted into ionic equations and then split into two half-reactions, ie oxidation and reduction reactions; Each of these half-reactions is synchronized separately and then combined to produce equalized ions; Finally, the observing ions are reintroduced into equated ionic equations, changing the reaction to its molecular form.
For example, the steps to equalize the following redox equations:

          Fe²⁺_(aq) + Cr₂O₇^2-_(aq) ——> Fe³⁺_(aq) + Cr³⁺_(aq)

1.   Write down the overall reaction equation

Fe²⁺+ Cr₂O₇^2- ——> Fe³⁺+ Cr³⁺

2.   Divide the reaction into two half-reactions

Fe²⁺ ——> Fe³⁺

Cr₂O₇^2- ——> Cr³⁺

3.   Equalize the type of atom and the number of atoms and charge in each half-reaction; In an acidic atmosphere, add H2O to equal the atoms O and H + to equalize the H atom

Fe²⁺ ——> Fe³⁺+ e^–

6 e^– + 14 H⁺ + Cr₂O₇^2- ——> 2 Cr³⁺ + 7 H₂O

4.   Summing up the two half-reactions; Electrons on both sides must eliminate each other; If oxidation and reduction have different number of electrons, it must be equated first

6 Fe²⁺ ——> 6 Fe³⁺ + 6 e^– ……………… (1)

6 e^– + 14 H⁺ + Cr₂O₇^2- ——> 2 Cr³⁺ + 7 H₂O ……………… (2)

6 Fe²⁺ + 14 H⁺ + Cr₂O₇^2- ——> 6 Fe³⁺ + 2 Cr³⁺ + 7 H₂O ………………… [(1) + (2)]

5.   Check again and believe that both sides have the same type of atom and number of atoms, and have the same charge on both sides of the reaction equation. For reactions that take place in an alkaline atmosphere, add the OH ions - in an amount equal to the H + ions in each segment to remove the H + ions. The equation of the reaction turns into the following:

6 Fe²⁺ + 14 H⁺ + 14 OH^– + Cr₂O₇^2- ——> 6 Fe³⁺ + 2 Cr³⁺ + 7 H₂O + 14 OH^–

6 Fe²⁺ + 14 H₂O + Cr₂O₇^2- ——> 6 Fe³⁺ + 2 Cr³⁺ + 7 H₂O + 14 OH^–

6 Fe²⁺ + 7 H₂O + Cr₂O₇^2- ——> 6 Fe³⁺ + 2 Cr³⁺ + 14 OH^–

Reference :
https://renideswantikimia.wordpress.com/kimia-kelas-xii-3/semester-i/2-reaksi-redoks-dan-elektrokimia/1-persamaan-reaksi-redoks/
https://www.academia.edu/5965686/REAKSI_REDUKSI-OKSIDASI_REAKSI_REDOKS_