Oxidation-reduction equations acidic solution

Perhaps even more important is the effect of hydrogen ion concentration on the emf of a half-reaction of a particular species. Consider the permanganate ion as an oxidizing agent in acid solution (as it often is). From the Latimer diagram above we can readily see that the reduction emf is 1.51 V when all species have unit activity. What is not shown is the complete equation ... 

Write balanced ionic half-reaction equations for the reduction of each of the following oxidizing agents in acid solution. 

The more diflicult oxidation-reduction equations cun often be written more easily by use of tile Stock system ol oxidation numbers, which are positive or negative valences or charges. Consider the reaction of potassium diehromale. K Cr 0-. with potassium sultile. KjSOi. in acid solution to lorni chromiuntlllll sulfate. Cr SOj i. and potassium sullale. K S04. The unbalanced expression for the ionic reaction ts... 

Balance the oxidation-reduction equation for the oxidation of H2S(aq) by HN03(aq) to produce NO(g) and S(s) in aqueous acidic solution (thus H+ and H20 may be involved). 

Balancing oxidation-reduction equations for reactions occurring in aqueous acidic solutions is usually fairly straightforward since we can use H20 to balance O, and then H+ to balance H. In basic solution,... 

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. 

Balancing Oxidation-Reduction Equations Occurring in Acidic Solution by the Half-Reaction Method... 

When this oxidation-reduction equation is balanced in acidic solution, using only whole number coefiTicients, what is the coefficient for S(s l... 

Complete and balance the following oxidation-reduction equations using the method of half-reactions. Both reactions occur in acidic solution. 

Example 11-1. The oxidizing agent permanganate ion, Mn04, on reduction in acid solution forms manganous ion, Mn" . Ferrous ion, Fe, can accomplish this reduction. Write the equation for the reaction between permanganate ion and ferrous ion in acid solution. 

Although the preceding skeleton equation is not complete, it does give the essential information about the oxidation-reduction reaction. Moreover, given the skeleton equation, you can complete and balance the equation using the half-reaction method. Let us see how to do that. We first look at balancing oxidation-reduction equations in acidic solution. To balance such equations in basic solution requires additional steps. 

Steps in Balancing Oxidation-Reduction Equations in Acidic Solution... 

To balance an oxidation—reduction equation in basic solution, you begin by balancing the equation as if it were a reaction in acidic solution. Then, you add the following steps. ... 

Balancing equations in acidic and basic solutions by the half-reaction method Given the skeleton equation for an oxidation-reduction equation, complete and balance it. (EXAMPLES 20.1,20.2)... 

Balance the following oxidation-reduction equations. The reactions occur in acidic or basic aqueous solution, as indicated. 

Oxidation-reduction equations are now balanced using half-reactions in acidic or basic solutions. 

The standard reduction potential of O ig) to 02(g) is +2.07 V. Write a half-equation for this reduction in acid solution. How does ozone compare with chlorine gas, Cl2( ), as an oxidizing agent Of what practical, beneficial, environmental importance is the oxidiang abiUty of ozone Hint Refer to Chapter 11 if need be.)... 

Write plausible half-equations and a balanced oxidation-reduction equation for the disproportionation of Xep4 to Xe and Xe03 in aqueous acidic solution. Xe and Xe03 are produced in a 2 1 mole ratio, and 02(g) is also produced. 

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. 

Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction. 

Though equations such as (16) (19) may be considered to represent the core steps or reactions of the reductant oxidation process in conventional electroless deposition, one or more reactions often need to be considered. In the case of reductants such as H2P02 and DMAB, codeposition of P and B also occurs, as shown here for hypo-phosphite in acidic solution ... 

B Since we need to refer to Table 21-1 in any event, it is perhaps a bit easier to locate the two balanced half-equations in the table. There is only one half-equation involving both Fe2+(aq) and Fe3+(aq) ions. It is reversed and written as an oxidation below. The halfequation involving Mn04 (aq) is also written below. [Actually, we need to know that in acidic solution Mn2+(aq) is the principal reduction product of Mn04 (aq).]... 

Write partial equations for the oxidation and the reduction. Then (1) Balance charges by adding in acid solutions or OH in basic solutions. (2) Balance the number of O s by adding H O s to one side. (3) Balance the number of H s by adding H s to one side. The number added is the number of equivalents of oxidant or reductant. 

N-Hydroxyindols (11) have been prepared in two ways,65 [Eqs. (24) and (25)]. Equation (25) is a variation in which the hydroxylamino group is oxidized to a nitroso group, which then adds to a C=C bond. In acidic solution the N-hydroxy group can be reductively removed. 

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