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lucy107

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专业: 环境化工

[交流] Oxidation-Reduction Reactions已有1人参与

Oxidation-Reduction Reactions
Oxidation is the removal of electrons from, and reduction is the addition of electrons to an atom. Consider the galvanizing of iron, that is coating from with zinc to prevent rusting. The two competing reactions are:
Fe2++ 2e==Fe    E°=-0.44volts Zn2++ 2e=Zn    E°=-0.76volts Since the zinc reaction in the forward reaction produces a larger negative potential, that is liberates more energy, the spontaneous reaction is
Zn+ Fe2+==Fe+Zn2+
氧化反应是原子失去电子，还原反应是原子得到电子。就锌镀铁而言，就是一层锌的涂层防止铁生锈。两个竞争的反应是
Fe2++ 2e==Fe    E°=-0.44volts Zn2++ 2e=Zn    E°=-0.76volts 由于锌的正向反应产生较大的负电势，释放更多的能量，这个自发反应是
Zn+ Fe2+==Fe+Zn2+
The coating of zinc serves two purposes: firstly it covers the iron and prevents its oxidation  (rather like a coat of paint) and secondly it provides anodic protection. If the galvanized steel is scratched, allowing the air to oxidize some iron, the Fe2+ produced is immediately reduced to iron by the zinc, and rusting does not occur. Similar applications in which one metal is sacrificed to protect another are the attaching of sacrificial blocks of magnesium to underground steel pipelines and the hulls of ships to prevent the rusting of iron.
表面镀锌有两个目的：一是覆盖在铁表面可以保护铁不被氧气，二是其提供阳极保护。如果镀锌钢铁被划破，让空气氧化一些铁，Fe2+在锌下还原成铁，以而生锈不会发生。牺牲一种金属以保护另一种金属的类似应用是把一块作牺牲用的镁锭同下部的钢管和船体联接起来以防止铁的生锈。 Reduction Potential Diagrams
The reactions and stability of the various oxidation states of an element can be shown by the appropriate half reactions and reduction potentials. Consider iron:
Fe2++ 2e = Fe                   E°=-0.44volts
Fe3++ 3e=Fe                   E°=-0.04 Fe3++ e = Fe2+                   E°=+0.77
FeO2-4 +3e+8H+ = Fe3+ + 4H2O    E°=+2.20 This information is consolidted into a single diagram, in which the highest oxidation state is written at the left, and the lowest state at the right。
还原电势表
一种元素大量氧化态的反应性和稳定性可以从适当的一半反应和电势的还原反应表现出来，观察铁：
Fe2++ 2e = Fe                   E°=-0.44volts
Fe3++ 3e=Fe                   E°=-0.04 Fe3++ e = Fe2+                   E°=+0.77 FeO2-4 +3e+8H+ = Fe3+ + 4H2O    E°=+2.20
这些信息整理成一个单独图解，把最高氧化态写在左边，而把最低氧化态写在右边。
From the potentials it is apparent that reduction of FeO2-4 to Fe3+ releases a lot of energy, so FeO2-4 is a strong oxidizing agent. Similarly, Fe3+ is a weaker oxidizing agent going to Fe2+, but neither Fe3+nor Fe2+has any tendency to reduce to Fe.
Standard electrode potentials are measured on a scale with
H++e = H    E°=0.0volts
Since hydrogen is normally regarded as a reducing agent, reactions with negative values for E° are more strongly reducing than hydrogen, that is they are strongly reducing. Materials which  are generally accepted as oxidizing agents have E°values above +0.8 volts, those such as Fe3+→ Fe2+ of about 0.8 volts are stable (equally oxidizing and reducing), and those below +0.8 volts becoming increasingly reducing.
从电势可以明显看出FeO2-4 到Fe3+的反应释放出大量的能量，所以FeO2-4是强氧化剂。相似的，Fe3+与Fe2+相比是较弱的氧化剂，但是不管是Fe3+还是Fe2+都没有任何还原成Fe的趋势。标准电极电势是以（如下）标度（为准）来测量的。
H++e = H    E°=0.0volts
由于氢通常是被当作还原剂，具有负电势的反应，它比H还原性强，从而说明它们是强还原性。元素材料的电势超过0.8v通常被作为氧化剂所接受，那些像Fe3+    Fe2+有约0.8v的是稳定的，而那些低于0.8v的则有逐渐增强还原性。
At first sight the potential of -0.04 V for Fe3+-Fe seems wrong since the potentials for Fe3+-Fe2+, and Fe2+-Fe are 0.77 V and -0.44V respectively. Potentials are not thermodynamic functions, and may not be added, but the potential may be "calculated from the free energy G, using the equation △G= —nFE0 where n is the number of electrons involved and F the Faraday.
起先看到的 Fe3+-Fe 的电势是-0.04v时是错误的，因为Fe3+-Fe2+, and Fe2+-Fe分别是0.77 V and -0.44V时，电势不是热力学函数，所以不能相加，但是电势可能以自由能G计算出来，使用公式△G= —nFE0，这里n是相关的电子数，F是法垃第（常数）。
The reduction potential diagram for copper in acid solution is oxidation state. The potential, and hence the energy released when Cu2+is reduced to Cu+are both very small, hence Cu2+is stable. On moving from left to right the potentials Cu'+ - Cu+ - Cu become more positive. Whenever this is found, the species in the middle (Cu+ in this case) disproportionates, that is it behaves as both a self-oxidizing and a self-reducing agent because it is energetically favourable for the following two changes to occur together.Thus Cu+ disproportionates in solution, and is only found in the solid state. 酸溶液中铜的还原铜是氧化态。Cu2+被还原成Cu+时的电势及后来释放的能量都是很少的，因此Cu2+是稳定的。Cu'+ - Cu+ - Cu以左侧到右侧电势变得更正：当存在这种情况时，处于中间的物质便会发生歧化反应，即它起自身氧化剂和自身还原剂的作用。因为下列两种变化同时在能量上是有利的。因而Cu 能在溶液中歧化，(故)它只能在固态中找到。

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