Calculate oxidation number:
To calculate the oxidation number of an element in a compound or ion, you need to assign electrons to each atom according to certain rules. The most common rules are:
The oxidation number of an atom in an element is always 0.
The oxidation number of an ion is equal to its charge.
The sum of the oxidation numbers of all atoms in a neutral molecule is 0.
Example:
Calculate the oxidation number of sulfur in H₂SO₄.
Solution:
The oxidation number of hydrogen is +1. Since there are two hydrogen atoms, their total contribution is +2.
The oxidation number of oxygen is -2. Since there are four oxygen atoms, their total contribution is -8.
The sum of the oxidation numbers must be equal to the charge of the molecule, which is 0.
Therefore, the oxidation number of sulfur can be calculated as follows:
+2 - 8 + x = 0
where x is the oxidation number of sulfur.
Solving for x gives x = +6.
Thus, the oxidation number of sulfur in H₂SO₄ is +6.
Balance chemical equations:
Changes in oxidation numbers can be used to help balance chemical equations, particularly those involving redox reactions.
In a redox reaction, there is a transfer of electrons from one reactant to another.
The reactant that loses electrons is said to be oxidized, while the reactant that gains electrons is said to be reduced.
The overall charge must be conserved in a chemical reaction, so the oxidation numbers of the atoms must balance on both sides of the equation.
Example:
MnO₄⁻ + H₂C₂O₄ → Mn²⁺ + CO₂
1. Assign oxidation numbers to each element in the equation:
Mn: +7 in MnO₄⁻ and +2 in Mn²⁺
C: +3 in H₂C₂O₄ and +4 in CO₂
O: -2 in MnO₄⁻, -2 in H₂C₂O₄, and -2 in CO₂
H: +1 in H₂C₂O₄
2. Identify the elements undergoing oxidation and reduction:
Mn is being reduced (its oxidation number decreases from +7 to +2)
C is being oxidized (its oxidation number increases from +3 to +4)
3. Write half-reactions for oxidation and reduction:
Separate the oxidation and reduction reactions into two half-reactions.
In the oxidation half-reaction, the element that is oxidized loses electrons, and in the reduction half-reaction, the element that is reduced gains electrons.
4. Balance the atoms and charges in each half-reaction so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
For example, in the equation
MnO₄⁻ + H₂C₂O₄ → Mn²⁺ + CO₂ + H₂O,
The oxidation half-reaction is:
MnO₄⁻ → Mn²⁺
The oxidation state of Mn changes from +7 to +2, so it loses 5 electrons.
The balanced half-reaction is:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
To balance the number of oxygen, H₂O is added.
To balance the number of hydrogen H⁺ is added.
The reduction half-reaction is:
H₂C₂O₄ → CO₂
The oxidation state of C changes from +3 to +4, so it loses 1 electron. The balanced half-reaction is:
H₂C₂O₄ + 2H₂O + 2e⁻ → 2CO₂ + 6H⁺
To balance the number of oxygens, water(H₂O) is added.
To balance the number of hydrogens, H⁺ is added.
Balance the overall equation:
To balance the overall equation, multiply each half-reaction by an integer so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
5. Then add the two half-reactions together, cancelling out any spectator ions that appear on both sides of the equation.
After adding both of them the balanced equation is:
MnO₄⁻ + 8H⁺ + 5H₂C₂O₄ → Mn²⁺ + 5CO₂ + 13H₂O
Redox agents
Oxidizing agents
An oxidizing agent is a substance that causes another substance to lose electrons, and in the process, the oxidizing agent itself is reduced.
This means that the oxidizing agent gains electrons, as it accepts them from the substance it is oxidizing.
For example, when chlorine gas (Cl₂) reacts with sodium metal (Na), the chlorine is the oxidizing agent as it accepts an electron from sodium, forming sodium chloride (NaCl).
The oxidation state of chlorine decreases from 0 to -1, indicating that it has been reduced.
Reducing Agents
A reducing agent is a substance that causes another substance to gain electrons, and in the process, the reducing agent itself is oxidized.
This means that the reducing agent loses electrons, as it donates them to the substance it is reducing.
For example, in the reaction of zinc metal (Zn) with hydrochloric acid (HCl) to form hydrogen gas (H₂) and zinc chloride (ZnCl₂), the zinc metal is the reducing agent as it donates electrons to hydrogen ions from the acid.
The oxidation state of zinc increases from 0 to +2, indicating that it has been oxidized.
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