Oxidation-reduction reactions overall equation

The balanced overall equation for the reaction now can be written as shown in Figure 16.3. This equation is the same as the equation for the formation of zinc oxide that you read at the beginning of the discussion on redox reactions. Now you know that it represents the net oxidation-reduction reaction and is the sum of an oxidation half-reaction and a reduction half-reaction. 

The reduction of nitrobenzene to aniline is a typical oxidation-reduction reaction in which tin metal, Sn°, is oxidized to stannic ion, Sn, in the form of stannic chloride, SnCl4 hydrochloric acid serves as the source of protons. A plausible mechanism of this reaction is outlined in Scheme 21.3. Generally, the reduction occurs by a sequence of steps in which an electron is first transferred from a tin atom to the organic substrate to give an intermediate radical ion that is then protonated. The oxygen atoms on the nitro group are eventually removed as water molecules. It is left as an exercise to write a balanced equation for the overall reaction and to provide a mechanism for the reduction of N-phenylhydroxylamine (11) into aniline (see Exercises 13 and 14 at the end of this section). 

A transfer of electrons causes changes in the oxidation states of one or more elements. Any chemical process in which elements undergo changes in oxidation number is an oxidation-reduction reaction. This name is often shortened to redox reaction. An example of a redox reaction can be seen in Figure 1.4, in which copper is being oxidized and NO from nitric acid is being reduced. The part of the reaction involving oxidation or reduction alone can be written as a half-reaction. The overall equation for a redox reaction is the sum of two half-reactions. Because the number of electrons involved is the same for oxidation and reduction, they cancel each other out and do not appear in the overall chemical equation. 

Equations for oxidation-reduction (redox) reactions are often difiBcult to balance. The method to be described in this section is especially appropriate for reactions involving ions in solution. The method resolves the reaction into two parts, or partial reactions a loss of electrons (oxidation) and a gain of electrons (REDUCTION). In reality, most oxidation-reduction reactions probably do not proceed in this way. The method is essentially a bookkeeping device enabling us to deal with the oxidation and with ihe reduction separately, and finally to combine the results into a balanced equation. This artificial recipe, while giving the correct answer for the overall reaction, should not be interpreted as an explanation of the reaction. 

The reactions in Examples 4 and 5 are not oxidation-reduction reactions. Nevertheless, they can be resolved into partial equations that add up to the overall reaction, and the electrode potentials can then be used to calculate AG , and K. 

In the half-reaction method for balancing equations, an oxidation-reduction reaction is written as two half-reactions. As each half-reaction is balanced for atoms and charge, it becomes apparent which one is oxidation and which one is reduction. Once the loss and gain of electrons are equalized for the half-reactions, they are combined to obtain the overall balanced equation. The half-reaction method is typically used to balance equations that are written as ionic equations. Let us consider the reaction between aluminum metal and a solution of Cu as shown in Sample Problem 15.3. 

We can show that the reaction is an oxidation-reduction reaction by evaluating changes in oxidation state, but there is another especially useful way to establish this. Think of the reaction as involving two half-reactions occurring at the same time—an oxidation and a reduction. The overall reaction is the sum of the two half-reactions. We can represent the half-reactions by halfequations and the overall reaction by an overall equation. 

Representing an Oxidation-Reduction Reaction Through Half-Equations and an Overall Equation... 

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. 

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. 

The basic idea of this method is to split a complicated equation into two parts called half-reactions. These simpler parts are then balanced separately, and recombined to produce a balanced overall equation. The splitting is done so that one of the half-reactions deals only with the oxidation portion of the redox process, whereas the other deals only with the reduction portion. What ties the two halves together is the fact that the total electrons lost by the oxidation process MUST equal the total gained by the reduction process (step 6). 

The overall standard potential for the equation is the difference between the potential for the reduction half-reaction (+1.45 V) and that for the oxidation half-reaction ( + 0.94 V) = 1.45 — 0.94 = +0.51 V, which, being positive, means that the reaction is feasible. 

A second process that has been invoked is heterolytic fission of H2.49 Examples are shown in equations (15) and (16). Although the overall process certainly involves heterolysis, some or all of these reactions may well go via initial oxidative addition followed by fast deprotonation or reductive elimination (equation 17). A third variant closely related to heterolytic fission is the addition of hydrogen across a metal-ligand or metal-metal bond. Examples are shown in equations (18) and (19). Here too, homolytic addition via equation (1) followed by fast reductive elimination may be taking place in some cases. In d° systems, the homolytic fission reaction of equation (1) is unlikely as the metal is already at its maximum valency, so examples of heterolytic fission and addition to... 

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

All methods of balancing oxidation-reduction equations are based on the overall gain of oxidation numbers in a reaction being the same as the overall loss of oxidation numbers in the reaction (because the same number of electrons must be gained as lost). 

The general procedure is to balance the equations for the half-reactions separately and then to add them to obtain the overall balanced equation. The half-reaction method for balancing oxidation-reduction equations differs slightly depending on whether the reaction takes place in acidic or basic solution. 

For each of the following unbalanced equations, (i) write the half-reactions for oxidation and for reduction, and (ii) balance the overall equation in acidic solution using the half-reaction method. 

The individual steps of this overall reduction reaction produce HCIO2, HOCl, and CI2, which all behave as oxidizing agents. An acidic medium is required, as CIO2 disproportionates in alkaline solution, as shown in equation 6. 

Which of these two is a half reaction of oxidation Which one is a half reaction of reduction Write the equation for the overall reaction. Which reagent is the oxidizing agent (electron acceptor) Which reagent is the reducing agent (electron donor) ... 

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