$2 ,matop {Mg}limits_{}^{0} ,,+,, mathop {O_2}limits_{}^{0} ,, longrightarrow ,, 2 mathop {Mg}limits_{}^{+2} mathop {O }limits_{}^{-2} ,,,,,,,, (1)$

The oxidation number of $Mg$ increases from $0$ to $+2$ $,Rightarrow ,$ $Mg$ gives up electrons:

$mathop {Mg}limits_{}^{0} ,, longrightarrow ,, mathop {Mg}limits_{}^{+2} ,,+,, 2e$

Oxygen accepts electrons:

$mathop {O}limits_{}^{0} ,,+,, 2e ,, longrightarrow ,, mathop {O}limits_{}^{-2}$

$Longrightarrow ,$ The process of $Mg$ giving up electrons is the oxidation of $Mg$.

$Longrightarrow ,$ In the reaction $(1)$, the oxidizing agent is $Oxi$, the reducing agent is $Mg$.

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$ullet$ Example 2:

$mathop {Cu}limits_{}^{+2} mathop {O}limits_{}^{-2} ,,+,, mathop {H_2}limits_{}^{0} ,, longrightarrow ,, mathop {Cu} limits_{}^{0} ,,+,, mathop {H_2}limits_{}^{+1} mathop {O}limits_{}^{-2} ,,,,,,, (2)$

Oxidation number of $Cu$ decreased from $+2$ to $0$ $,Rightarrow ,$ $Cu$ in $CuO$ get an extra $2e$:

$mathop {Cu}limits_{}^{+2} ,,+,, 2e ,, longrightarrow ,, mathop {Cu}limits_{}^{0}$

The oxidation number of $H$ increases from $0$ to $+1$ $,Rightarrow ,$ $H$ gives way to $1e$:

$mathop {H}limits_{}^{0} ,, longrightarrow ,, mathop {H}limits_{}^{+1} ,,+,, 1e$

$Longrightarrow ,$ The process of $mathop {Cu}limits_{}^{+2}$ getting an extra $2e$ is called the reduction of $mathop {Cu}limits_{}^{+2}$ (the reduction of $mathop { Cu}limits_{}^{+2}$).

$Longrightarrow ,$ In the reaction $(2)$, the oxidizing agent is $CuO$, the reducing agent is $Hydro$.

$ullet ,$ In short:

A reducing agent (a substance that is oxidized) is a substance that donates electrons.

An oxidizing agent (a substance that is being reduced) is an electron acceptor.

Oxidation (oxidation) is the process of losing electrons.

Reduction (reduction) is the process of gaining electrons.

2. Consider the reaction without oxygen participating

$ullet ,$ Example 3:

The reaction $(3)$ has an oxidation number change, electron donating – accepting:

$mathop {Na}limits_{}^{0} ,, longrightarrow ,, mathop {Na}limits_{}^{+1} ,,+,, 1e$

$mathop {Cl}limits_{}^{0} ,,+,, 1e ,, longrightarrow ,, mathop {Cl}limits_{}^{-1}$

$ullet ,$ Example 4:

$mathop {H_2}limits_{}^{0} ,,+,, mathop {Cl_2}limits_{}^{0} ,, longrightarrow ,, 2mathop {H}limits_{}^{+1} mathop {Cl}limits_ {}^{-1} ,,,,,,,, (4)$

The reaction $(4)$ has a change in the oxidation numbers of the substances, because the shared electron pair shifts towards $Cl$.

$ullet ,$ Example 5:

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$mathop {N}limits_{}^{-3} {H_4} mathop {N}limits_{}^{+5} {O_3} ,, {overset{t^0}{longrightarrow}} ,, mathop {N_2} limits_{}^{+1} O ,,+,, 2, {H_2}O ,,,,,,,, (5)$

Reaction $(5)$ atom $mathop {N}limits_{}^{-3}$ gives $e$, $mathop {N}limits_{}^{+5}$ gets $e$

$longrightarrow ,$ has the oxidation number change of an element.

3. Oxidation – reduction reaction

– Oxidation – reduction reaction is a chemical reaction in which there is the transfer of electrons between reactants, or oxidation – reduction is a chemical reaction in which there is a change in the oxidation number of an integer. element.

II. CHEMICAL ENGINEERING OF OXIDATION – REACTION

$ullet ,$ The electron balance method is based on the principle that the total number of electrons lost by the reducing agent is equal to the total number of electrons gained by the oxidizing agent.

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– Step 1: Determine the oxidation numbers of the elements to find the oxidizing agent and the reducing agent.

– Step 2: Write the oxidation and reduction process, balance each process.

– Step 3: Find the appropriate coefficients for the oxidizing agent and the reducing agent so that the total number of electrons given is equal to the total number of electrons accepted.

– Step 4: Put the coefficients of the oxidizing and reducing agents in the reaction diagram, then calculate the coefficients of other substances in the equation. Check the atomic number balance of the elements and balance the charge on both sides to complete the chemical equation.

$ullet ,$ Example 1: Write a chemical equation for the following redox reaction:

$N{H_3} ,,+,, {Cl_2} ,,longrightarrow ,, {N_2} ,,+,, HCl$

– Step 1:

$mathop {N}limits_{}^{-3} mathop {H_3}limits_{}^{+1} ,,+,, mathop {Cl_2}limits_{}^{0} ,, longrightarrow ,, mathop {N_2} limits_{}^{0} ,,+,, mathop {H}limits_{}^{+1} mathop {Cl}limits_{}^{-1}$

+ Oxidation number of $N$ increased from $-3$ to $0,$: Reducing agent

+ Oxidation number of $Cl$ decreases from $0 to $-1,$: Oxidizing agent

– Step 2:

+ Oxidation: $,,2, mathop {N}limits_{}^{-3} ,, longrightarrow ,, mathop {N_2}limits_{}^{0} ,,+,, 6e$

+ Reduction process: $,,mathop {Cl_2}limits_{}^{0} ,,+,, 2e ,, longrightarrow ,, 2, mathop {Cl}limits_{}^{-1}$

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– Step 3:

+ Oxidation: $,,(2, mathop {N}limits_{}^{-3} ,, longrightarrow ,, mathop {N_2}limits_{}^{0} ,,+,, 6e) ,, imes 1$

+ Reduction process: $,,(mathop {Cl_2}limits_{}^{0} ,,+,, 2e ,, longrightarrow ,, 2, mathop {Cl}limits_{}^{-1}) ,, imes 3 $

$Longrightarrow , 2, mathop {N}limits_{}^{-3} ,,+,, 3, mathop {Cl_2}limits_{}^{0} ,, longrightarrow ,, mathop {N_2}limits_{}^{0 } ,,+,, 6, mathop {Cl}limits_{}^{-1}$

– Step 4:

$2,N{H_3} ,,+,, 3, {Cl_2} ,,longrightarrow ,, {N_2} ,,+,, 6, HCl$

$ullet ,$ Example 2: Write a chemical equation for the following redox reaction:

$mathop {Mg}limits_{}^{0} ,,+,, mathop {Al}limits_{}^{+3} {Cl_3} ,, longrightarrow ,, mathop {Mg}limits_{}^{+2} { Cl_2} ,,+,, mathop {Al}limits_{}^{0}$

$Mg$ is reducing agent; $mathop {Al}limits_{}^{+3}$ (in $AlCl_3$) is the oxidizing agent.

$(mathop {Mg}limits_{}^{0} ,, longrightarrow ,, mathop {Mg}limits_{}^{+2} ,,+,, 2e) ,, imes 3$

$(mathop {Al}limits_{}^{+3} ,,+,, 3e ,, longrightarrow ,, mathop {Al}limits_{}^{0}) ,, imes 2$

$Longrightarrow , 3, mathop {Mg}limits_{}^{0} ,,+,, 2, mathop {Al}limits_{}^{+3} ,, longrightarrow ,, 3, mathop {Mg}limits_{}^ {+2} ,,+,, 2, mathop {Al}limits_{}^{0}$

The equation would be:

$3,Mg ,,+,, 2,Al{Cl_3} ,,longrightarrow ,, 3, Mg{Cl_2} ,,+,, 2,Al$

$ullet ,$ Example 3: Write a chemical equation for the following redox reaction:

$K mathop {Cl}limits_{}^{+5} {O_3} ,, longrightarrow ,, K mathop {Cl}limits_{}^{-1} ,,+,, K mathop {Cl}limits_{}^{ +7} {O_4}$

$mathop {Cl}limits_{}^{+5}$ (in $KCl{O_3}$) is both a reducing agent and an oxidizing agent.

$(mathop {Cl}limits_{}^{+5} ,,+,, 6e ,, longrightarrow ,, mathop {Cl}limits_{}^{-1}) ,, imes 1$

$(mathop {Cl}limits_{}^{+5} ,, longrightarrow ,, mathop {Cl}limits_{}^{+7} ,,+,, 2e) ,, imes 3$

$Longrightarrow , 4, mathop {Cl}limits_{}^{+5} ,, longrightarrow ,, 1, mathop {Cl}limits_{}^{-1} ,,+,, 3, mathop {Cl}limits_{} ^{+7}$

The equation would be:

$4,KCl{O_3} ,,longrightarrow ,, KCl ,,+,, 3, KCl{O_4}$

$ullet ,$ Example 4: Write a chemical equation for the following redox reaction:

$K mathop {Cl}limits_{}^{+5} {O_3} ,, longrightarrow ,, K mathop {Cl}limits_{}^{-1} ,,+,, mathop {O_2}limits_{}^{0 }$

$mathop {Cl}limits_{}^{+5}$ (in $KCl{O_3}$) is the oxidizing agent; $mathop {O}limits_{}^{-2}$ (in $KCl{O_3}$) is the reducing agent.

$(mathop {Cl}limits_{}^{+5} ,,+,, 6e ,, longrightarrow ,, mathop {Cl}limits_{}^{-1}) ,, imes 2$

$(2,mathop {O}limits_{}^{-2} ,, longrightarrow ,, mathop {O_2}limits_{}^{0} ,,+,, 4e) ,, imes 3$

$Longrightarrow , 2, mathop {Cl}limits_{}^{+5} ,,+,, 6, mathop {O}limits_{}^{-2} ,, longrightarrow ,, 2, mathop {Cl}limits_{} ^{-1} ,,+,, 3, mathop {O_2}limits_{}^{0}$

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The equation would be:

$2,KCl{O_3} ,,longrightarrow ,, 2,KCl ,,+,, 3, {O_2}$

$ullet ,$ Example 5: Write a chemical equation for the following redox reaction:

$mathop {Fe}limits_{}^{+2} mathop {S_2}limits_{}^{-1} ,,+,, mathop {O_2}limits_{}^{0} ,, longrightarrow ,, mathop {Fe_2} limits_{}^{+3} mathop {O_3}limits_{}^{-2} ,,+,, mathop {S}limits_{}^{+4} mathop {O_2}limits_{}^{-2}$

$mathop {Fe}limits_{}^{+2},$, $mathop {S}limits_{}^{-1}$ (in $Fe{S_2}$) is the reducing agent, $mathop {O_2}limits_{ }^{0}$ is the oxidizing agent.

$mathop {Fe}limits_{}^{+2} ,, longrightarrow ,, mathop {Fe}limits_{}^{+3} ,,+,, 1e$

$2,mathop {S}limits_{}^{-1} ,, longrightarrow ,, 2,mathop {S}limits_{}^{+4} ,,+,, 10e$

$(mathop {Fe}limits_{}^{+2} mathop {S_2}limits_{}^{-1} ,, longrightarrow ,, mathop {Fe}limits_{}^{+3} ,,+,, 2, mathop {S}limits_{}^{+4} ,,+,, 11e) ,, imes 4$

$(mathop {O_2}limits_{}^{0} ,,+,, 4e ,, longrightarrow ,, 2,mathop {O}limits_{}^{-2}) ,, imes 11$

$Longrightarrow , 4, mathop {Fe}limits_{}^{+2} mathop {S_2}limits_{}^{-1} ,,+,, 11,mathop {O_2}limits_{}^{0} ,, longrightarrow ,, 4, mathop {Fe}limits_{}^{+3} ,,+,, 8, mathop {S}limits_{}^{+4} ,,+,, 22,mathop {O}limits_{}^ {-2}$

The equation would be:

$4,Fe{S_2} ,,+,, 11,{O_2} ,,longrightarrow ,, 2,{Fe_2}{O_3} ,,+,, 8, S{O_2}$

$ullet ,$ Example 6: Write a chemical equation for the following redox reaction:

$mathop {Mn}limits_{}^{+4} {O_2} ,,+,,H mathop {Cl}limits_{}^{-1} ,, longrightarrow ,, mathop {Mn}limits_{}^{+2 } {Cl_2} ,,+,, mathop {Cl_2}limits_{}^{0} ,,+,, {H_2}O$

$mathop {Mn}limits_{}^{+4}$ (in $Mn{O_2}$) is the oxidizing agent, $mathop {Cl}limits_{}^{-1}$ (in $HCl$) is the reduce.

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$(mathop {Mn}limits_{}^{+4} ,,+,, 2e ,, longrightarrow ,, mathop {Mn}limits_{}^{+2}) ,, imes 1$

$(2,mathop {Cl}limits_{}^{-1} ,, longrightarrow ,, mathop {Cl_2}limits_{}^{0} ,,+,, 2e) ,, imes 1$

$Longrightarrow , mathop {Mn}limits_{}^{+4} ,,+,, 2,mathop {Cl}limits_{}^{-1} ,, longrightarrow ,, mathop {Mn}limits_{}^{+2 } ,,+,, mathop {Cl_2}limits_{}^{0}$

The equation would be:

$Mn{O_2} ,,+,, 4,HCl ,,longrightarrow ,, Mn{Cl_2} ,,+,, {Cl_2} ,,+,, 2,{H_2}O$

III. MEANING OF OXIDATION – REMOVATION IN PRACTICE

Oxidation – reduction reaction is a type of chemical reaction that is quite common in nature and is of great importance in production and life.

1. In life

– Oxidation – reduction reactions create energy such as: combustion of gasoline in internal combustion engines, combustion of charcoal, electrolysis processes …

2. In production

Many oxidation – reduction reactions are the basis of chemical production processes such as iron, steel, aluminum smelting…