24-5. chemical equations for Oxidation-Reduction Reactions ...

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24. OxidatiOn-ReductiOn ReactiOns. 914. exaMPLe 24-7: Hydrogen peroxide is used extensively as an oxidizing agent in acidic aqueous solution. For example ...
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24. Oxidation-Reduction Reactions

EXAMPLE 24-7: Hydrogen peroxide is used extensively as an oxidizing agent in acidic aqueous solution. For example, H2O2(aq) is used as a bleaching agent for human hair and to kill bacteria (antiseptic agent). Write a balanced half reaction equation for H2O2(aq) acting as an oxidizing agent in acidic aqueous solution. Solution: The oxidation state of each oxygen atom in H2O2 is –1 (see Example 24-1c). Because H2O2(aq) is acting as an oxidizing agent, the oxygen atom is reduced to an oxidation state of –2. This is the lowest possible oxidation state for an oxygen atom because the addition of two electrons yields a noble-gas electron configuration. The oxidation state of the oxygen atom in a water molecule is –2, and so we write H2O2 → H2O ​ ​ ​ ​ ​ ​(reduction) Balancing the oxygen atoms by adding another H2O molecule to the right side yields H2O2 → 2 H2O and balancing the hydrogen atoms by adding 2 H + ions to the left side yields 2 H + + H2O2 → 2 H2O We now balance the charge by adding 2e – to the left side and indicate phases to obtain 2 H + (aq) + H2O2(aq) + 2 e– → 2 H2O(l )

PR ACTICE PROBLEM 24-7: Write the balanced half reaction equation for H2O2(aq) acting as a reducing agent in acidic aqueous solution. Answer: H2O2(aq) → O2(g) + 2 H + (aq) + 2 e–

24-5. Chemical Equations for Oxidation-Reduction Reactions Occurring in Basic Solution Are Balanced Using OH – and H2O The reactions considered up to this stage have all taken place in acidic aqueous solution, where H + (aq) and H2O(l ) are readily available and thus can be used in balancing the equations for the half reactions. In basic solution, however, H + (aq) is not available at significant concentrations to use in balancing the equations for the half reactions with respect to hydrogen. Therefore, for reactions that take place in basic solution, we must change step 4 by using OH– ions instead of H + ions. If we simply add OH– ions to balance an equation with respect to hydrogen atoms, however, then each OH– ion adds an oxygen atom in addition to a hydrogen atom and the oxygen-atom balance that was attained in

24 -5. BAL ANCING EQUATIONS FOR OXIDATION-REDUCTION REACTIONS

step 3 will be lost. To overcome this problem, we balance the excess hydrogen atoms by adding an equal number of H2O molecules and OH– ions to opposite sides of the equation. The net result is an increase in the number of hydrogen atoms on the side of the equation to which the H2O molecules are added. By adding H2O molecules and OH– ions to opposite sides of the equation in pairs, we are able to maintain the oxygen atom balance that we established in step 3. Thus, step 4 for balancing oxidation-reduction reaction equations in basic solutions is as follows: 4. Balance each half reaction equation with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH– ions to the side opposite to the added H2O molecules. The procedure outlined above is best illustrated by example. Let’s balance the equation for the reduction half reaction OH–(aq)

ClO–(aq) → Cl–(aq)

(reduction)

in basic solution as indicated by OH– (aq) over the arrow in the equation. The half reaction equation is already balanced with respect to chlorine atoms. To balance the equation with respect to oxygen atoms we add one H2O molecule to the right side of the equation (step 3) to obtain: ClO – → Cl– + H2O The right side of this equation now has two extra hydrogen atoms. To balance the equation with respect to hydrogen atoms, we add two H2O molecules to the left side and two OH– ions to the right side (step 4). 2 H2O + ClO – → Cl– + H2O + 2 OH– Canceling the H2O molecule on the right side with one from the left side, we now balance the equation with respect to charge by adding 2e – to the left side to obtain: H2O + ClO – + 2 e– → Cl– + 2 OH– The balanced equation for the half reaction with phases indicated is H2O(l ) + ClO – (aq) + 2 e– → Cl– (aq) + 2 OH– (aq) Finally, in cases involving metal hydroxides in basic solution, we add OH – ions directly, skipping steps 3 and 4. This is illustrated in balancing the following half reaction equation: Cu(OH)2(s) → Cu(s) Cu(OH)2(s) → Cu(s) + 2 OH– (aq) Cu(OH)2(s) + 2 e– → Cu(s) + 2 OH– (aq)

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24. Oxidation-Reduction Reactions

The following Example and Practice Problem provide additional examples of balancing oxidation-reduction equations in basic solution.

EXAMPLE 24-8: Given that the oxidation-reduction reaction described by the following equation takes place in basic aqueous solution, balance the equation OH – (aq)

BrO –3(aq) + F2(g) → BrO –4(aq) + F– (aq) Solution: The oxidation state of the bromine atom changes from +5 in BrO3– (aq) to +7 in BrO4– (aq) and that of the fluorine atoms changes from 0 in F2(g) to –1 in F– (aq). Thus, we have the two half reaction equations

BrO –3 → BrO –4  ​ ​ ​ ​ ​

(oxidation)



F2 → 2 F– ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

(reduction)

where we have added a 2 to balance the fluorine atoms in the reduction half reaction equation. The oxidation half reaction equation involves one excess oxygen atom on the right side, so we add one H2O molecule to the left side (step 3).

H2O + BrO –3 → BrO –4 Next, to balance the hydrogen atoms, we add two H2O molecules to the right side and two OH– ions to the left side (step 4).



2 OH– + H2O + BrO –3 → BrO –4 + 2 H2O Canceling the H2O molecule on the left side with one from the right side, we now add 2e – to the right side to balance the half reaction equation with respect to charge:



2 OH– + BrO –3 → BrO –4 + H2O + 2 e– ​ ​ ​ ​ ​  ​(oxidation) To balance the reduction half reaction equation with respect to charge, we add 2 e – to the left side:



F2 + 2 e– → 2 F– ​ ​ ​ ​ ​ ​ ​ ​  ​ ​(reduction) Because the number of electrons is the same in both half reaction equations, we simply add the two half reaction equations and include the phase designations to obtain a complete balanced equation: 2 OH– (aq) + BrO –3 (aq) + F2(g) → BrO –4 (aq) + 2 F– (aq) + H2O(l )

PR ACTICE PROBLEM 24-8: Balance the following oxidation-reduction equation in basic aqueous solution: OH – (aq)

Fe(OH)2(s) + O2(g) → Fe(OH)3(s) Answer: 4 Fe(OH)2(s) + 2 H2O(l ) + O2(g) → 4 Fe(OH)3(s)

24 - 6. Oxidation-Reduction Reactions Are Used in Chemical Analysis

TABLE 24.3 Procedure for balancing the equations for oxidation-reduction reactions in acidic and basic aqueous solutions Step

Procedure

1. Separate the equation into an oxidation half reaction equation and a reduction half reaction equation. 2. Balance each half reaction equation with respect to all atoms other than oxygen and hydrogen. 3. Balance each half reaction equation with respect to oxygen atoms by adding the appropriate number of H2O molecules to the side deficient in oxygen atoms. (For metal hydroxides, directly add OH− ions and skip steps 3–4.) In Acidic Solution: 4. Balance each half reaction equation





with respect to hydrogen atoms by adding the appropriate number of H+ ions to the side deficient in hydrogen atoms.

In Basic Solution: 4. Balance each half reaction with respect to hydrogen atoms by adding a number of H2O molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of OH− ions to the side opposite to the added H2O molecules.

5. Balance each half reaction equation with respect to charge by adding the appropriate number of electrons to the side with the excess positive charge. 6. Multiply each half reaction equation by an integer that makes the number of electrons supplied by the oxidation half reaction equation equal to the number of electrons accepted by the reduction half reaction equation. 7. Obtain the complete balanced equation by adding the two half reaction equations and canceling or combining any like terms.

If you remember to use OH– ions and H2O molecules, or, in cases involving metal hydroxides, just OH– ions, then balancing equations in basic aqueous solution is straightforward. The procedure for balancing oxidation-reduction equations using the method of half reactions for acidic and basic aqueous solutions is summarized in Table 24.3.

24-6. Oxidation-Reduction Reactions Are Used in Chemical Analysis Oxidation-reduction reactions are used extensively in chemical analysis. To illustrate this application of oxidation-reduction reactions, let’s look at the analytical determination of iron(II) in a sample of iron ore. Suppose that we dissolve a 3.532-gram sample of iron ore in H2SO4(aq) and that we reduce any

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