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BALNCING REDOX REACTIONS

在处理化学计量学问题时，通常必须编写平衡的化学方程式。在编写反应物和产品后，可以通过命中和试验方法进行平衡。但是，可以使用系统的方法来平衡氧化还原反应。在描述这些方法之前，以下介绍了元素氧化状态计算的规则。

Rules to Compute Oxidation Number

The oxidation number of an element is the number assigned to it by following the arbitrary rules given below:

(i) A free element (regardless of whether it exists in monatomic or polyatomic form, e.g. Hg, H₂, P₄and S₈) is assigned an oxidation number of zero.

(ii) A free monatomic ion is assigned an oxidation number equal to the charge it carries. For example, the oxidation number of Al³⁺+ S^2-and Cl^-are +3, -2 and -1 respectively.

(iii) In their compounds, the alkali and alkaline earth metals are assigned oxidation numbers of +1 and +2 respectively.

(iv) The oxidation number of hydrogen in its compounds is generally +1 except the ionic hydrides such as LiH. LiAIH₄where its 'oxidation number is -1.

(v) The oxidation number of fluorine in all its compounds is -1. The oxidation number of all other halogens is -1 in all compound except those with oxygen (e.g. CIO₄^-)和卤素原子序数较低(如。ICI₃^-). The oxidation number of the latter is determined via oxygen and halogen of lower atomic number.

(vi) The oxidation numbers of both oxygen and sulphur in their normal oxides (e.g. Na₂O) and sulphides (e.g. CS₂) is -2. The exception are the peroxides (e.g. H₂O₂and Na₂O₂), superoxides (e.g. KO2) and the compound OF₂。它们的氧化数由规则3、4和5确定。

(vii) The algebraic sum of oxidation numbers of atoms in a chemical species (compound or ion) is equal to the net charge on the species.

A few examples of the computation of oxidation number of atoms N in various compounds are given as follows.

If x is the oxidation number of N, we 'have

NH₃x+3(+1) = 0 whichgiv.esx = -3
HN₃+1 + 3（x）= 0，给出x = -1/3
N₂H₄2x + 4(+1) = 0 which gives x =-2
NO₂x + 2(-2) = 0 which gives x = 4
N₂O₄2x + 4（-2）= 0给出x = 4
NO₂2_-x + 2（-2）= -1，给出x = 3
NH₂OH x + 2(+1) + 1(-2) + 1 = 0 which gives x =-1
NO x + 1(-2) = 0 which gives x = 2
HNO₃+1 + x + 3（-2）= 0，给出x = 5
H₂O 2x + 1(-2) = 0 which gives x = 1
HCN + 1 + 4 + x = 0，给出x = -5


Balancing Redox Reactions Via Oxidation Numbers

The steps involved in this method are as follows:

1. Identify the elements in the unbalanced equation whose oxidation number are changed.
2. Balance the number of atoms of each element whose oxidation number is changed.
3. Find out the total change in oxidation number for each of oxidant and reductant and make them equal by multiplying by small coefficients.
4. Balance the remainder of atoms by inspection and add, if necessary H⁺(acidic medium) or OH^-(alkaline medium) or H2O (to balance oxygen) as the case may be.

Balancing Redox Reactions Via Oxidation Numbers

The steps involved in this method are as follows:

1. Identify the elements in the unbalanced equation whose oxidation number are changed.

2. Balance the number of atoms of each element whose oxidation number is changed.

3. Find out the total change in oxidation number for each of oxidant and reductant and make them equal by multiplying by small coefficients.

4.通过检查并在必要时添加H+（酸性介质）或OH来平衡其余的原子^-(alkaline medium) or H₂o（平衡氧）可能是。

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