Balancing Oxidation-Reduction Reactions Quiz

PDF page 858: Electrochemistry studies oxidation-reduction reactions, which were first discussed in an earlier chapter, where we learned that oxidation was the loss of electrons and reduction was the gain of electrons. The reactions discussed tended to be rather simple, and conservation of mass (atom counting by type) and deriving a correctly balanced

PDF page 858: Electrochemistry studies oxidation-reduction reactions, which were first discussed in an earlier chapter, where we learned that oxidation was the loss of electrons and reduction was the gain of electrons. The reactions discussed tended to be rather simple, and conservation of mass (atom counting by type) and deriving a correctly balanced

PDF page 859: chemical equation were relatively simple. In this section, we will concentrate on the half-reaction method for balancing oxidation-reduction reactions. The use of half-reactions is important partly for balancing more complicated reactions and partly because many aspects of electrochemistry are easier to discuss in terms of half-reactions. There are alternate methods of balancing these reactions; however, there are no good alternatives to half-reactions for discussing what is occurring in many systems. The half-reaction method splits oxidation-reduction reactions into their oxidation “half” and reduction “half” to make finding the overall equation easier. Electrochemical reactions frequently occur in solutions, which could be acidic, basic, or neutral. When balancing oxidation-reduction reactions, the nature of the solution may be important. It helps to see this in an actual problem. Consider the following unbalanced oxidation-reduction reaction in acidic solution: 3+ 2+ 2+ − MnO (aq) + Fe (aq) ⟶ Mn (aq) + Fe (aq) 4 We can start by collecting the species we have so far into an unbalanced oxidation half-reaction and an unbalanced reduction half-reaction. Each of these half-reactions contain the same element in two different oxidation states. The

PDF page 859: If the atoms and charges balance, the half-reaction is balanced. In oxidation half-reactions, electrons appear as products (on the right). As discussed in the earlier chapter, since iron underwent oxidation, iron is the reducing agent. Now return to the reduction half-reaction equation: 2+ − + reduction (unbalanced): MnO (aq) + 8H (aq) ⟶ Mn (aq) + 4H O(l) 4 2 The atoms are balanced (mass balance), so it is now necessary to check for charge balance. The total charge on the left of the reaction arrow is [(−1) × (1) + (8) × (+1)], or +7, while the total charge on the right side is [(1) × (+2) +

PDF page 860: MnO was reduced to Mn . In all reduction half-reactions, electrons appear as reactants (on the left side). As 4 − discussed in the earlier chapter, the species that was reduced, MnO in this case, is also called the oxidizing agent. 4 We now have two balanced half-reactions. 2+ 3+ − oxidation: Fe (aq) ⟶ Fe (aq) + e 2+ − + − reduction: MnO (aq) + 8H (aq) + 5e ⟶ Mn (aq) + 4H O(l) 4 2 It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives

PDF page 860: MnO was reduced to Mn . In all reduction half-reactions, electrons appear as reactants (on the left side). As 4 − discussed in the earlier chapter, the species that was reduced, MnO in this case, is also called the oxidizing agent. 4 We now have two balanced half-reactions. 2+ 3+ − oxidation: Fe (aq) ⟶ Fe (aq) + e 2+ − + − reduction: MnO (aq) + 8H (aq) + 5e ⟶ Mn (aq) + 4H O(l) 4 2 It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives

PDF page 859: If the atoms and charges balance, the half-reaction is balanced. In oxidation half-reactions, electrons appear as products (on the right). As discussed in the earlier chapter, since iron underwent oxidation, iron is the reducing agent. Now return to the reduction half-reaction equation: 2+ − + reduction (unbalanced): MnO (aq) + 8H (aq) ⟶ Mn (aq) + 4H O(l) 4 2 The atoms are balanced (mass balance), so it is now necessary to check for charge balance. The total charge on the left of the reaction arrow is [(−1) × (1) + (8) × (+1)], or +7, while the total charge on the right side is [(1) × (+2) +

PDF page 860: MnO was reduced to Mn . In all reduction half-reactions, electrons appear as reactants (on the left side). As 4 − discussed in the earlier chapter, the species that was reduced, MnO in this case, is also called the oxidizing agent. 4 We now have two balanced half-reactions. 2+ 3+ − oxidation: Fe (aq) ⟶ Fe (aq) + e 2+ − + − reduction: MnO (aq) + 8H (aq) + 5e ⟶ Mn (aq) + 4H O(l) 4 2 It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives

PDF page 859: (4) × (0)], or +2. The difference between +7 and +2 is five; therefore, it is necessary to add five electrons to the left

PDF page 860: MnO was reduced to Mn . In all reduction half-reactions, electrons appear as reactants (on the left side). As 4 − discussed in the earlier chapter, the species that was reduced, MnO in this case, is also called the oxidizing agent. 4 We now have two balanced half-reactions. 2+ 3+ − oxidation: Fe (aq) ⟶ Fe (aq) + e 2+ − + − reduction: MnO (aq) + 8H (aq) + 5e ⟶ Mn (aq) + 4H O(l) 4 2 It is now necessary to combine the two halves to produce a whole reaction. The key to combining the half-reactions is the electrons. The electrons lost during oxidation must go somewhere. These electrons go to cause reduction. The number of electrons transferred from the oxidation half-reaction to the reduction half-reaction must be equal. There can be no missing or excess electrons. In this example, the oxidation half-reaction generates one electron, while the reduction half-reaction requires five. The lowest common multiple of one and five is five; therefore, it is necessary to multiply every term in the oxidation half-reaction by five and every term in the reduction half-reaction by one. (In this case, the multiplication of the reduction half-reaction generates no change; however, this will not always be the case.) The multiplication of the two half-reactions by the appropriate factor followed by addition of the two halves gives

PDF page 861: of hydroxide ions to each side of the equation so that all the hydrogen ions (H ) are removed and mass balance is

PDF page 861: maintained. Hydrogen ion combines with hydroxide ion (OH ) to produce water. Let us now try a basic equation. We will start with the following basic reaction: − − −

PDF page 858: Electricity refers to a number of phenomena associated with the presence and flow of electric charge. Electricity includes such diverse things as lightning, static electricity, the current generated by a battery as it discharges, and many other influences on our daily lives. The flow or movement of charge is an electric current (Figure 16.2). Electrons or ions may carry the charge. The elementary unit of charge is the charge of a proton, which is equal in magnitude to the charge of an electron. The SI unit of charge is the coulomb (C) and the charge of a proton is 1.602 ×