Chemistry
Half Equations
Half equations are chemical equations that represent either the reduction or oxidation process in a redox reaction. They show the transfer of electrons by focusing on the change in oxidation states of the elements involved. In a half equation, the reactant that loses electrons is the reducing agent, while the reactant that gains electrons is the oxidizing agent.
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9 Key excerpts on "Half Equations"
- Henry Freiser, Monika Freiser(Authors)
- 1992(Publication Date)
- CRC Press(Publisher)
In a redox halfcell, electrons are the particles transferring between the oxidized and reduced forms of a redox couple. Conjugate acid-base pair: B + H + ' BH + Redox Couple: Ox + ne-' Red In the generalized half-cell equation, n represents the number of electrons necessary to transform a species to the next stable, lower oxidat ion state. In contrast to proton transfer reactions which occur in steps of one proton at a time, many redox reactions involve the simultaneous transfer of several electrons. REDOX STOICHIOMETRY Half-cell equations are of great value in stoichiometric problems as well as in equilibrium situations. It is not necessary, in relating amounts involved in redox reactions, to use the complete overall equation since all of the necessary information (includi ng that for balancing the overall equa-tion) is contained in the half-cell equations. For example, in the reaction Chapter 1 Oxidation Reduction Equilibria 129 between Fe 3 + and Sn 2 + mentioned in the section above, the half-cell equations show that the two electrons lost by the Sn 2 + are transferred to two Fe 3 + , each of which can accept only one each. Thus, balancing redox equa-tions consists in matching the numbers of electrons lost with those gained. Balancing half-ce l l equations can be accomplished by one of several techniques, but the ion-electron method seems the most convenient and generally useful. It is worth noting that all of the uses of balanced equations, namely, stoich iometr ic and equilibrium calcu lations, are wel l served by the use of halfce l l equations. The fundamental factor in redox stoichiometry (description of combini ng power) is the number of electrons transferred (gained or lost) per species.
eBook - PDF- Ramesh Chandra, Snigdha Singh, Aarushi Singh(Authors)
- 2020(Publication Date)
- Arcler Press(Publisher)
It is to be known that from the above example, an oxidation reaction is one in which a species loses electrons and a reduction reaction is one in which a species gains electron. Both the equation must occur simultaneously as the electrons lost by one species are accepted by another. Therefore, the Eq. (1) is an oxidation half reaction (where electrons are lost by Mg) and the Eq. (2) is a reduction half reaction (where the electrons are gained by H + ). As the oxidation reaction is always accompanied by a reduction reaction, the term redox reaction is used. In the Redox reaction, the overall reaction equation is obtained by combining the separate Half Equations in such a way that the number of electrons lost is exactly equal to the number of electrons gained. The final equation thus obtained is called a redox equation. In this example, the balanced overall equation results by simply adding the two Half Equations. Mg(s) + 2H + (aq) → Mg 2 + (aq) + H 2 (g) It should be noted that, in the previous equation where 2 electrons were present on both the sides of the equation has now cancelled out. All the equations regarded as the oxidation reaction on the basis of gain of an oxygen carries the same characteristics as in the magnesium example Organic Synthesis and Organic Reagents 184 explained above, i.e., the transfer of electrons from the species oxidized to the species reduced. Therefore, the most general definitions of the oxidation-reduction reactions are: Oxidation is the loss of electron. Reduction is the gain of electrons The species which causes the oxidation to occur as H + in the above example is called the oxidizing agent or the oxidant. O the other hand, the species which is oxidized is called the reducing agent or reductant. The equations written in the manner of (1) and (2) above are called ion-electron Half Equations. 6.4. ELECTROLYSIS The concepts of compounds and of chemical change are illustrated by the electrolysis of water.
eBook - PDF- David A. Ucko(Author)
- 2013(Publication Date)
- Academic Press(Publisher)
In summary, the steps in the half-reaction method are: 1 Write a net ionic equation. 2 Identify atoms that change in oxidation number. 3 Write half-equations for the oxidation and reduction steps. 4 Balance the numbers of atoms in each half-equation. 5 Balance the charges in each half-equation. 6 Multiply the half-equations so that the number of electrons lost and the number of electrons gained are equal. 7 Add the half-equations, canceling substances that appear on both sides in equal quantities. 8 Balance the spectator ions. Both the oxidation number method and the half-reaction method lead to balanced redox equations. The oxidation number method is useful for reactions that do not take place in aqueous solution or do not involve the formation of ions. The half-reaction method is most useful for reactions in aqueous solution, since it starts with the net ionic equation. This method clearly emphasizes the actual oxidation and reduction half-reactions. We will see in the next section that these half-reactions can be physically separated. Balance the following equations by the half-reaction method. a C r 2 0 7 2 + H + + S 0 3 2 ~ C r 3 + + S 0 4 2 + H 2 0 b Zn + H N 0 3 -* Z n ( N 0 3 ) 2 + N H 4 N 0 3 + H 2 0 ANSWERS a C r 2 0 7 2 + 8 H + + 3 S 0 3 2 -> 2 C r 3+ + 3 S 0 4 2 + 4 H 2 0 b 4Zn + 10HNO 3 4 Z n ( N 0 3 ) 2 + N H 4 N 0 3 + 3 H 2 0 As we will see in this section, a redox reaction may be used to convert energy from the form of chemical bonds into the form of electricity. This type of redox reaction takes place in batteries. On the other hand, electrical energy can be used to make a chemical change take place through a redox reaction. This process is used in industry to manu-facture certain materials such as aluminum. Consider what happens if we put a piece of zinc metal into a blue aqueous solution of copper sulfate.
eBook - ePubEnvironmental Process Analysis
Principles and Modeling
- Henry V. Mott(Author)
- 2013(Publication Date)
- Wiley(Publisher)
We begin by identifying the element that gains or loses electrons as a consequence of the targeted electron transfer reaction we wish to write. Most often we see that half reactions are written as reductions, with the oxidized specie on the LHS and the reduced specie on the RHS. Since this is the chemists’ way, we will adopt their convention. Then, the specie containing the reduced element is situated on the RHS of the reaction and the specie containing the oxidized element is situated on the LHS.1. We determine the oxidation states of the target element as present with the reduced and oxidized species. 2. We balance the target element between the LHS and RHS to ensure that we have an atomic balance. 3. Once the target element is balanced, we add the requisite number of electrons on the LHS to balance the change in the overall charge of the target element from the LHS to the RHS. 4. We balance elements other than oxygen and hydrogen, taking care to ensure that the most probable species are shown on the LHS and RHS of the half reaction. 5. We then balance the excess oxygen from the LHS with water on the RHS. 6. Lastly, we balance excess hydrogen from the RHS with protons on the LHS.When we are finished, we will have both atomic and electron balances between the LHS and RHS. Most often the resultant reaction is normalized to a single electron transferred by dividing each stoichiometric coefficient by the number of electrons transferred. Sometimes this is most convenient, and other times using the half reaction as written may be the most convenient. User preference governs as long as the reaction is correctly employed. Let us get some practice with this algorithm.Example 12.2 Develop the half reactions for the following redox couples and, where possible, compare your results with published half reactions: bicarbonate–methane; nitrate–nitrogen gas; sulfate–bisulfide; glucose–carbon dioxide; glucose–methane; VBS–methane (applicable in anaerobic digestion of wastewater biosolids); and VBS/carbon dioxide.For the bicarbonate–methane couple, the oxidation state of carbon is +4 in bicarbonate and −4 in methane. We have one carbon in each specie, thus we begin with bicarbonate on the LHS and methane on the RHS and require eight electrons to balance the oxidation states of carbon in the two species:
- Peter V. Hobbs(Author)
- 2000(Publication Date)
- Cambridge University Press(Publisher)
6 Oxidation-reduction reactions 6.1 Some definitions In Chapter 5 we saw that, in terms of the Br0nsted-Lowry theory, acid-base reactions involve proton transfer. Another large and impor-tant group of chemical reactions, particularly in aqueous solutions, involves electron transfer; these are referred to as oxidation-reduction (or redox) reactions. Redox reactions are involved (1) in photosyn-thesis, which releases oxygen into the Earth's atmosphere; (2) in the combustion of fuels, which is responsible for rising concentrations of atmospheric carbon dioxide; (3) in the formation of acid precipitation; and (4) in many chemical reactions in Earth sediments. Oxidation refers to a loss of electrons, and reduction to a gain of electrons. For example, an oxidation reaction is Cu(s)->Cu 2+ (aq) + 2e-(6.1) where the symbol e~ indicates one free electron (which carries one unit of negative charge). A reduction reaction is 2Ag + (aq) + 2e-^2Ag(s) (6.2) Since electrons cannot be lost or gained overall, oxidation must always be accompanied by reduction. Thus, Eqs. (6.1) and (6.2) together form a redox reaction Cu(s) + 2Ag + (aq) -> Cu 2+ (aq) + 2Ag(s) (6.3) Equation (6.1) is called the oxidation half-reaction and Eq. (6.2) the reduction half-reaction for the overall reaction Eq. (6.3). If substance A causes the oxidation of substance B, substance A is called the oxidizing agent or oxidant. Thus, in Eq. (6.3), Ag + (aq) is the oxidant, because it causes Cu(s) to lose electrons (note that the oxidant 104 Oxidation numbers 105 is reduced, that is, it gains electrons). Similarly, if a substance A causes the reduction of substance B, substance A is called the reducing agent or reductant. In Eq. (6.3) Cu(s) is the reductant, because it causes Ag + (aq) to gain electrons (note that the reductant is oxidized, that is, it loses electrons).
eBook - PDF- Morris Hein, Susan Arena, Cary Willard(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
The main difference between balancing ionic redox equations and molecular redox equations is in how we handle the ions. In the ionic redox equations, besides having the same number of atoms of each element on both sides of the final equation, we must also have equal net charges. In assigning oxidation numbers, we must therefore remember to consider the ionic charge. Several methods are used to balance ionic redox equations, including, with slight modification, the oxidation-number method just shown for molecular equations. But the most popular method is probably the ion–electron method. The ion–electron method uses ionic charges and electrons to balance ionic redox equations. Oxidation numbers are not formally used, but it is necessary to determine what is being oxidized and what is being reduced. PROBLEM-SOLVING STRATEGY Ion–Electron Strategy for Balancing Oxidation–Reduction Reactions 1. Write the two half-reactions that contain the elements being oxidized and reduced using the entire formula of the ion or molecule. 2. Balance the elements other than oxygen and hydrogen. 3. Balance oxygen and hydrogen. Acidic solution: For reactions in acidic solution, use H + and H 2 O to balance oxygen and hydrogen. For each oxygen needed, use one H 2 O. Then add H + as needed to bal- ance the hydrogen atoms. Basic solution: For reactions in alkaline solutions, first balance as though the reaction were in an acidic solution, using Steps 1–3. Then add as many OH − ions to each side of the equation as there are H + ions in the equation. Now combine the H + and OH − ions into water (for example, 4 H + and 4 OH − give 4 H 2 O). Rewrite the equation, can- celing equal numbers of water molecules that appear on opposite sides of the equation. 4. Add electrons (e − ) to each half-reaction to bring them into electrical balance.
- Morris Hein, Scott Pattison, Susan Arena, Leo R. Best(Authors)
- 2014(Publication Date)
- Wiley(Publisher)
Oxidation numbers are not formally used, but it is necessary to determine what is being oxidized and what is being reduced. PROBLEM-SOLVING STRATEGY: Ion–Electron Strategy for Balancing Oxidation–Reduction Reactions 1. Write the two half-reactions that contain the elements being oxidized and reduced using the entire formula of the ion or molecule. 2. Balance the elements other than oxygen and hydrogen. 3. Balance oxygen and hydrogen. Acidic solution: For reactions in acidic solution, use H + and H 2 O to balance oxygen and hydrogen. For each oxygen needed, use one H 2 O. Then add H + as needed to balance the hydrogen atoms. Basic solution: For reactions in alkaline solutions, first balance as though the reaction were in an acidic solution, using Steps 1–3. Then add as many OH - ions to each side of the equation as there are H + ions in the equation. Now combine the H + and OH - ions into water (for example, 4 H + and 4 OH - give 4 H 2 O). Rewrite the equation, canceling equal numbers of water molecules that appear on opposite sides of the equation. 4. Add electrons (e - ) to each half-reaction to bring them into electrical balance. 5. Since the loss and gain of electrons must be equal, multiply each half-reaction by the appropriate number to make the number of electrons the same in each half-reaction. 6. Add the two half-reactions together, canceling electrons and any other identical substances that appear on opposite sides of the equation. LEARNING OBJECTIVE 17.3 • Balancing Ionic Redox Equations 399 E X A M P L E 1 7 . 1 0 Balance this equation using the ion–electron method: MnO 4 - + S 2- ¡ Mn 2+ + S 0 (acidic solution) SOLUTION 1. Write two half-reactions, one containing the element being oxidized and the other the element being reduced (use the entire molecule or ion): S 2- ¡ S 0 (oxidation) MnO - 4 ¡ Mn 2+ (reduction) 2. Balance elements other than oxygen and hydrogen (accomplished in Step 1: 1 S and 1 Mn on each side).
eBook - PDF- Morris Hein, Susan Arena, Cary Willard(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
The main difference between balancing ionic redox equations and molecular redox equations is in how we handle the ions. In the ionic redox equations, besides having the same number of atoms of each element on both sides of the final equation, we must also have equal net charges. In assigning oxidation numbers, we must therefore remember to consider the ionic charge. Several methods are used to balance ionic redox equations, including, with slight modifi- cation, the oxidation-number method just shown for molecular equations. But the most popu- lar method is probably the ion–electron method. The ion–electron method uses ionic charges and electrons to balance ionic redox equa- tions. Oxidation numbers are not formally used, but it is necessary to determine what is being oxidized and what is being reduced. Problem-Solving Strategy Ion–Electron Strategy for Balancing Oxidation–Reduction Reactions 1. Write the two half-reactions that contain the elements being oxidized and reduced using the entire formula of the ion or molecule. 2. Balance the elements other than oxygen and hydrogen. 3. Balance oxygen and hydrogen. Acidic solution: For reactions in acidic solution, use H + and H 2 O to balance oxygen and hydrogen. For each oxygen needed, use one H 2 O. Then add H + as needed to balance the hydrogen atoms. Basic solution: For reactions in alkaline solutions, first balance as though the reaction were in an acidic solution, using Steps 1–3. Then add as many OH − ions to each side of the equation as there are H + ions in the equation. Now combine the H + and OH − ions into water (for example, 4 H + and 4 OH − give 4 H 2 O). Rewrite the equation, canceling equal numbers of water molecules that appear on opposite sides of the equation. 4. Add electrons (e − ) to each half‐reaction to bring them into electrical balance.
eBook - ePub- Eckhard Worch(Author)
- 2015(Publication Date)
- De Gruyter(Publisher)
Section 10.3 , free electrons do not exist in aqueous systems over longer periods of time. Therefore, a complete redox reaction consists of two redox couples between which electrons are exchanged. Accordingly, no more free electrons occur in the complete redox reaction equation:(10.77)(10.78)(10.79)The law of mass action for the complete redox reaction reads:(10.80)The equilibrium constant of the complete redox reaction is related to the standard redox intensities of the half-reactions. This relationship can be derived from the equilibrium condition and the redox intensity equations for both half reactions. If the equilibrium is established in the considered water, then this water is characterized by only one single value of pe (remember that the same is true for the other master variable, the pH value). Accordingly, the equilibrium condition can be written as:(10.81)Substituting the redox intensities by the respective equations for the half-reactions gives:(10.82)Here, it is assumed that the half-reactions are written in such a manner that the same number of electrons, n e , is transferred. Rearranging Equation (10.82) leads to:(10.83)The quotient on the right-hand side equals the equilibrium constant (see Equation (10.80) ). Thus, we receive the following relationship that links the equilibrium constant of the complete redox reaction with the standard redox intensities of the half-reactions:(10.84)If we want to find the equilibrium constant from Equation (10.84) , we have to know which of the standard redox intensities is and which is , because the second is subtracted from the first and to interchange them would change the value of log K*. The decision which standard redox intensity belongs to which half-reaction is easy if we consider the addition of the half-reactions to the complete redox reaction. As we can see from Equations (10.77) -(10.79), one of the half-reactions has to be written in the reverse direction to allow addition of both half-reactions to form an overall reaction. This reverse reaction is that reaction whose standard redox intensity, pe0 , has the negative sign in Equation (10.84) (i.e. ). This is in accordance with the general relationship log that can be derived from Equation (5.13) (Chapter 5
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