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Solution balancing redox reactions

C04-0039. Predict whether or not a reaction will occur, and if a reaction does take place, write the half-reactions and the balanced redox reaction (a) a strip of nickel wire is dipped in 6.0 M HCl (b) aluminum foil is dipped in aqueous CaCl2 (c) a lead rod is dipped in a beaker of water (d) an iron wire is immersed in a solution of silver nitrate. [Pg.260]

Balancing Redox Reactions in Acidic and Basic Solutions... [Pg.488]

Electrolytes in Aqueous Solution 4.9 Balancing Redox Reactions ... [Pg.115]

An alternative to the oxidation-number method for balancing redox reactions is the half-reaction method. The key to this method is to realize that the overall reaction can be broken into two parts, or half-reactions. One half-reaction describes the oxidation part of the process, and the other half-reaction describes the reduction part. Each half is balanced separately, and the two halves are then added to obtain the final equation. Let s look at the reaction of aqueous potassium dichromate (K2Cr2C>7) with aqueous NaCl to see how the method works. The reaction occurs in acidic solution according to the unbalanced net ionic equation... [Pg.138]

Balancing oxidation-reduction reactions depends on whether the solution is acidic or basic. The method for balancing redox reactions in an acidic solution is as follows ... [Pg.114]

The method for balancing redox reactions in a basic solution is as follows ... [Pg.115]

Balancing redox reactions can be tricky. When you have trouble, follow the steps below to balance a redox reaction that occurs in acidic solution. [Pg.114]

Balancing Redox Reactions in Acidic Solution When a redox reaction occurs in acidic solution, H2O molecules and ions are available for balancing. Even though we ve usually used H30 to indicate the proton in water, we use H" in this chapter because it makes the balanced equations less complex. [Pg.683]

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... [Pg.684]

A redox reaction consists of two parts or half-reactions. These are the oxidation reaction in which a substance loses or donates electrons and the reduction reaction in which a substance gains or accepts electrons. An oxidation reaction and a reduction reaction must always be coupled because "free" electrons cannot exist in solution and electrons must be conserved. The coupling between the two half-reactions is by the electrons that are either generated (by oxidation) or consumed (by reduction). We will use this fact in our technique for balancing redox reactions, which basically is a stepwise stoichiometric (mass) balancing of each constituent followed by a balancing of charge (electroneutrality). [Pg.317]

What is being oxidized and what is being reduced in this reaction (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metaL Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced ... [Pg.872]

When balancing redox reactions occurring in basic solutions, the text instructs you to first use the half-reaction method as specified for acidic solutions. What if you started by adding OH first instead of H+ What potential problem could there be with this approach ... [Pg.838]

A general procedure for balancing redox reactions is given in the following procedure box. Since aqueous solutions are often acidic or basic, the procedme must accoxmt for the presence of H ions or OH ions. We cover acidic solutions in the examples that follow and demonstrate how to balance redox reactions in basic solutions in Example 16.8. [Pg.585]

EXAMPLE 16.8 Balancing Redox Reactions Occurring in Basic Solution... [Pg.588]

To balance redox reactions in aqueous acidic solutions, follow this procedure (brief version). [Pg.602]

Oxidation-reduction (redox) reactions involve the movement of electrons from one reactant to another. The half-reaction method of balancing redox reactions separates the overall reaction into two half-reactions, which mimics the actual separation of an electrochemical cell into two half-cells. Two types of electrochemical cells are distinguished by whether they generate electrical energy (voltaic) or use it (electrolytic). In both types of cell, electrodes dip into an electrolyte solution, the oxidation half-reaction occurs at the anode, and the reduction half-reaction occurs at the cathode. (Section 21.1)... [Pg.687]

To balance redox reactions occurring in basic solution, follow the half-reaction method outlined in Examples 18.1 and 18.2, but add an extra step to neutralize the acid with OH as shown in step 3. [Pg.864]

We can balance redox reactions using the half-reaction method, in which the oxidation and reduction reactions are balanced separately and then added. This method differs slightly for redox reactions in acidic and in basic solutions. [Pg.902]

Balancing Redox Reactions Occurring In Basic Solution (18.2) Example 18.3 For Practice 18.3 Exercises 41, 42... [Pg.903]

Balance the following redox reactions, and calculate the standard-state potential and the equilibrium constant for each. Assume that the [H3O+] is 1 M for acidic solutions, and that the [OH ] is 1 M for basic solutions. [Pg.177]

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. [Pg.604]

The first step In balancing a redox reaction is to divide the unbalanced equation into half-reactions. Identify the participants in each half-reaction by noting that each half-reaction must be balanced. That Is, each element In each half-reaction must be conserved. Consequently, any element that appears as a reactant In a half-reaction must also appear among the products. Hydrogen and oxygen frequently appear in both half-reactions, but other elements usually appear In just one of the half-reactions. Water, hydronium ions, and hydroxide ions often play roles In the overall stoichiometry of redox reactions occurring in aqueous solution. Chemists frequently omit these species in preliminary descriptions of such redox reactions. [Pg.1359]

C19-0083. Dichromate ions, C r2 0-j, oxidize acetaldehyde, CH3 CHO, to acetic acid, CH3 CO2 H, and are reduced to Cr . The reaction takes place in acidic solution. Balance the redox reaction and determine how many moles of electrons are required to oxidize 1.00 g of acetaldehyde. What mass of sodium dichromate would be required to deliver this many electrons ... [Pg.1421]

C19-0113. Balance the redox reactions between M11O4 and each of the following sulfur-containing species. The final products are and HSO4 and the solution is acidic (a) H2 SO3 (b) SO2 (c) H2 S ... [Pg.1424]

Redox reactions do not always take place under neutral conditions. Balancing half-reactions is more complicated for reactions that take place in acidic or basic solutions. When an acid or base is present, or OH ions must also be considered. However, the overall approach is similar. This approach involves writing the correct formulas for the reactants and products, balancing the atoms, and adding the appropriate number of electrons to one side of the half-reaction to balance the charges. [Pg.483]


See other pages where Solution balancing redox reactions is mentioned: [Pg.255]    [Pg.685]    [Pg.862]    [Pg.574]    [Pg.642]    [Pg.642]    [Pg.80]   


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