Basic solutions balancing redox equations

The following redox reactions occur in basic solution. Balance the equations using the oxidation number method. 

Compare and contrast balancing redox equations in acidic and basic solutions. 

EXERCISE 20.3 Balancing Redox Equations in Basic Solution... 

To balance redox equations for reactions in basic solution ... 

In this case, the hydrogen ion and the water molecule are eliminated because neither is oxidized nor reduced. The only additional information needed is that the reaction takes place in acid solution. In acid solution, hydrogen ions (H+) and water molecules are abundant and free to participate in redox reactions as either reactants or products. Some redox reactions can occur only in basic solution. When you balance equations for these reactions, you may add hydroxide ions (OH ) and water molecules to either side of the equation. Basic solutions have an abundance of OH ions instead of H30" ions. 

Before you attempt to balance the equation for a redox reaction, why do you need to know whether the reaction takes place in acidic or basic solution (20.3)... 

Use the half-reaction method to balance these equations for redox reactions. Add water molecules and hydrogen ions (in acid solutions) or hydroxide ions (in basic solutions) as needed. 

An alternative procedure for basic solutions is to balance the redox equation for acid solutions first. Then, add... 

Even though there is only a minor difference between the balance of redox reactions in acid and basic solutions the principles differ in that the equation are balanced with OH" ions in stead of H. this we will look into in the following example ... 

Balancing Redox Reactions in Basic Solution As you just saw, in acidic solution, H2O molecules and H ions are available for balancing. As Sample Problem 21.1 shows, in basic solution, H2O molecules and 0H ions are available. Only one additional step is needed to balance a redox equation that takes place... 

Some redox reactions can occur only in basic solution. When you balance equations for these reactions, you can add hydroxide ions (OH ) and water molecules to either side of the equation. 

If a redox reaction occurs in basic solution, the equation must be balanced by using OH and H2O rather than and H2O. One approach is to first balance the halfreactions as if they occurred in acidic solution and then count the number of in each... 

In the half-reaction method, we usually begin with a skeleton ionic equation showing only the substances undergoing oxidation and reduction. In such cases, we assign oxidation numbers only when we are unsure whether the reaction involves oxidation-reduction. We will find that (for acidic solutions), OH (for basic solutions), and H2O are often involved as reactants or products in redox reactions. Unless H, OH , or H2O is being oxidized or reduced, these species do not appear in the skeleton equation. Their presence, however, can be deduced as we balance the equation. 

If a redox reaction occurs in basic solution, the equation must be balanced by using OH and H2O rather than and H2O. Because the water molecule and the hydroxide ion both contain hydrogen, this approach can take more moving back and forth from one side of the equation to the other to arrive at the appropriate half-reaction. An alternate approach is to first balance the half-reactions as if they occurred in acidic solution, count the number of in each half-reaction, and then add the same number of OH to each side of the half-reaction. This way, the reaction is mass-balanced because you are adding the same thing to both sides. In essence, what you are doing is neutralizing the protons... 

Complete and balance this equation for a redox reaction that takes place in basic solution CN (aq) + Mn04 (oq) --------> CNO (oq) + Mn02(s) (basic solution)... 

Balance these ionic redox equations using the ion-electron method. These reactions occur in basic solution. 

Problem Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction between NaMn04 and Na2C204 in basic solution ... 

A few redox reactions have more than one oxidation half-reaction or more than one reduction half-reaction. Balancing the equations for these reactions is more complicated. However, the multiple half-reactions are often stoichiometricaUy hnked. Maintain the correct ratio of the elements and balance the electron transfer by multiplying them both by the same integer. Balance the following net ionic equation for a reaction in basic solution. OH" or HjO (but not H" ") may be added as necessary. 

Balance the following redox equations. AU occur in basic solution. (See Sections 11-4 and 11-5.)... 

We need only one additional step to balance a redox reaction that takes place in basic solution. It appears after we balance the half-reactions as if they occurred in acidic solution and were combined (step 4). At this point, we add one 0H to both sides of the equation for every //+ present. (This step is labeled 4 Basic in Sample Problem 21.1.) The OR- ions added on the side with H" ions combine with them to form H2O molecules, while the OH" on the other side remain in the equation, and then excess H2O is canceled. 

The following unbalanced equation represents a redox reaction that takes place in a basic solution containing KOH. Balance the redox equation. 

Redox equations can be balanced via the half-reaction method, which allows for the addition of H2O to balance O, to balance H, and OH for reactions taking place in basic solution. 

Oxides. Chlorine oxides in which Cl has an even oxidation number undergo self-redox. In basic solution, the products are oxyanions such as C102, CIOs", and CIO4". Write the balanced ionic equations for the self-redox of (a) CIO2, (b) CI2O3. 

To balance equations for redox reactions in a basic solution, a step or two must be added to the procedure used in Example 5-6. In basic solution, OH , not H, must appear in the final balanced equation. (Recall that, in basic solutions, OH ions are present in excess.) Because both OH and H2O contain H and O atoms, at times it is hard to decide on which side of the halfequations to put each one. One simple approach is to treat the reaction as though it were occurring in an acidic solution, and balance it as in Example 5-6. Then, add to each side of the overall redox equation a number of OH ions equal to the number of H ions. Where H and OH appear on the same side of the equation, combine them to produce H2O molecules. If H2O now appears on both sides of the equation, subtract the same number of H2O molecules from each side, leaving a remainder of H2O on just one side. This method is illustrated in Example 5-7. For ready reference, the procedure is summarized in Table 5.6. 

An effective way to balance a redox equation is to break down the reaction into separate half-reactions, write and balance half-equations for these half-reactions, and recombine the balanced half-equations into an overall balanced equation (Table 5.5). A slight variation of this method is used for a reaction that occurs in a basic aqueous solution (Table 5.6). A redox reaction in which the same substance is both oxidized and reduced is called a disproportionation reaction. 

In solving this problem the major effort was to balance a redox equation for a reaction under basic conditions. This allowed us to find the molar relationship between dithionite and chromate ions. The remainder of the problem was a stoichiometry calculation for a reaction in solution, much like Example 4-10 (page 127). A quick check of the final result involves (1) ensuring that the redox equation is balanced, and (2) noting that the number of moles of Cr04 is about 1.5 (i.e., 100 X 0.0148), that the number of moles of 8204 is about 2.25 (i.e., 1.5 X 3/2), and that the mass of Na2S204 is somewhat more than 350 (i.e., 2.25 X 175). 

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