Chemical equations oxidation-reduction

The last definition has widespread use in the volumetric analysis of solutions. If a fixed amount of reagent is present in a solution, it can be diluted to any desired normality by application of the general dilution formula V,N, = V N. Here, subscripts 1 and 2 refer to the initial solution and the final (diluted) solution, respectively V denotes the solution volume (in milliliters) and N the solution normality. The product VjN, expresses the amount of the reagent in gram-milliequivalents present in a volume V, ml of a solution of normality N,. Numerically, it represents the volume of a one normal (IN) solution chemically equivalent to the original solution of volume V, and of normality N,. The same equation V N, = V N is also applicable in a different context, in problems involving acid-base neutralization, oxidation-reduction, precipitation, or other types of titration reactions. The justification for this formula relies on the fact that substances always react in titrations, in chemically equivalent amounts. 

We consider oxidation first. To show the removal of electrons from a species that is being oxidized in a redox reaction, we write the chemical equation for an oxidation half-reaction. A half-reaction is the oxidation or reduction part of the reaction considered alone. For example, one battery that Volta built used silver and zinc plates to carry out the reaction... 

Balancing the chemical equation for a redox reaction by inspection can be a real challenge, especially for one taking place in aqueous solution, when water may participate and we must include HzO and either H+ or OH. In such cases, it is easier to simplify the equation by separating it into its reduction and oxidation half-reactions, balance the half-reactions separately, and then add them together to obtain the balanced equation for the overall reaction. When adding the equations for half-reactions, we match the number of electrons released by oxidation with the number used in reduction, because electrons are neither created nor destroyed in chemical reactions. The procedure is outlined in Toolbox 12.1 and illustrated in Examples 12.1 and 12.2. 

The chemical equation for a reduction half-reaction is added to the equation for an oxidation half-reaction to form the balanced chemical equation for the overall redox reaction. 

Redox reactions are more complicated than precipitation or proton transfer reactions because the electrons transferred in redox chemishy do not appear in the balanced chemical equation. Instead, they are hidden among the starting materials and products. However, we can keep track of electrons by writing two half-reactions that describe the oxidation and the reduction separately. A half-reaction is a balanced chemical equation that includes electrons and describes either the oxidation or reduction but not both. Thus, a half-reaction describes half of a redox reaction. Here are the half-reactions for the redox reaction of magnesium and hydronium ions ... 

After oxidation and reduction half-reactions are balanced, they can be combined to give the balanced chemical equation for the overall redox process. Although electrons are reactants in reduction half-reactions and products in oxidation half-reactions, they must cancel in the overall redox equation. To accomplish this, multiply each half-reaction by an appropriate integer that makes the number of electrons in the reduction half-reaction equal to the number of electrons in the oxidation half-reaction. The entire half-reaction must be multiplied by the integer to maintain charge balance. Example illustrates this procedure. 

If you know the reactants and products of a chemical reaction, you should be able to write an equation for the reaction and balance it. In writing the equation, first write down the correct formulas for all reactants and products. After they are written down, only then start to balance the equation. Do not balance the equation by changing the formulas of the substances involved. For simple equations, you should balance the equation by inspection. (Balancing oxidation-reduction equations will be presented in Chap. 13.) The following rules will help you to balance simple equations. 

The amide functionality plays an important role in the physical and chemical properties of proteins and peptides, especially in their ability to be involved in the photoinduced electron transfer process. Polyamides and proteins are known to take part in the biological electron transport mechanism for oxidation-reduction and photosynthesis processes. Therefore studies of the photochemistry of proteins or peptides are very important. Irradiation (at 254 nm) of the simplest dipeptide, glycylglycine, in aqueous solution affords carbon dioxide, ammonia and acetamide in relatively high yields and quantum yield (0.44)202 (equation 147). The reaction mechanism is thought to involve an electron transfer process. The isolation of intermediates such as IV-hydroxymethylacetamide and 7V-glycylglycyl-methyl acetamide confirmed the electron-transfer initiated free radical processes203 (equation 148). 

You already know that some metals are more reactive than others. You may also have carried out an investigation on the metal activity series in a previous course. In Investigation 10-A, located on page 470, you will discover how this series is related to oxidation and reduction. You will write chemical equations, ionic equations, and half-reactions for the single displacement reactions of several metals. 

The third chemical equation, involving nitric oxide, represents a termolecular reaction. Therefore, the overall order of the reaction is expected to exceed that of the second-order reaction generally assumed in the pre-mixed gas burning model. The high exothermicity accompanying the reduction of NO to N2 is responsible for the appearance of the luminous flame in the combustion of a double-base propellant, and hence the flame disappears when insufScient heat is produced in this way, i. e., during fizz burning. 

The reduction of tin(IV) oxide to (white) tin metal by graphite proceeds at moderately low temperatures. More commonly, tin(IV) oxide is reduced to the metal by an excess of carbon monoxide at temperatures above 980 K. (a) Write the two chemical equations for the reduction of SnOz. 

Arsenic(III) sulfide is oxidized by acidic hydrogen peroxide solution to the arsenate ion As043. Write the chemical equation and the reduction and oxidation half-reactions for the reaction. 

The element was obtained by the reduction of an acidic solution of the iodate ion with sodium hydrogen sulfite, (a) Write the chemical equation for the reaction, assuming the oxidized product to be HS04. (b) Calculate the mass of sodium hydrogen sulfite needed to produce 50.0 g of iodine. 

Every chemical reaction has a driving force—a reason why it proceeds as it does. We can say that the reason why oxidation-reduction reactions proceed is because one atom is giving up electrons and another is accepting them. We can also make the statement a bit more forcefully by saying that an element is grabbing electrons from another. There is some terminology, discussed below, that is based on these ideas. Consider this equation... 

Half-reactions Chemical equations that show oxidation and reduction separately and can be combined to give the overall equation for a redox reaction. 

Balancing Chemical Equations Involving Oxidation Reduction Before a chemical equation can represent correctly a chemical reaction, it must satisfy the following conditions ... 

These reactions are part of a larger category of reactions known as redox reactions (redox is short for oxidation-reduction). Sometimes these are called displacement reactions. These are chemical reactions in which atoms of one element replace the atoms of a second element in a compound. A general equation for a single-replacement reaction involving a metal (A), replacing a metallic cation in solution (B) is ... 

Enamines have been recognized in organic chemistry as useful synthetic reagents since the early reports from Stork s laboratory1. At almost the same time similar chemical moieties were being implicated in biochemical systems. Because of their intrinsic instability in water, the biochemical enamines exist primarily as intermediates, although, some well-known coenzymes that participate in oxidation-reduction reactions also incorporate enamine structures in one of their oxidation states. The electronic structure of enamines involves two extreme resonance contributions as shown in equation 1. 

Our first major task of this chapter is to learn to balance equations for chemical reactions. Balancing simple equations will be covered in this chapter equations for more complicated oxidation-reduction reactions will be considered in Chapter 16. 

Writing the chemical equation in the reverse direction requires changing the sign of the cell or half-cell potential. Note that all the equations in Table 17.2 refer to reduction half-reactions, but each complete cell requires one oxidation and one reduction. Thus one of the half-cell equations must be reversed (and the sign of its potential changed) to add to the other to make a complete cell equation. 

Indeed the combination of the reactive intermediates in equations (la) and (2) forms the chemical basis of electron-transfer oxidation (reduction) of organic compounds in both stoichiometric and catalytic processes. ... 

The most of chemical reactions accompanied by electron transfer from an atom of one reagent (reducer) to an atom of another reagent (oxidizer). Each element can have some oxidation states. The standard oxidation-reduction potential between two oxidation states of element is bonded with standard thermodynamic free energy of the transition from one state to another by the following equation ... 

The first step in writing the equation for an oxidation-reduction reaction is the same as for any other chemical reaction be sure that you know what the reactants are and what the products are. 

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