Half-reaction equations

If you look again at each half-reaction above, you will notice that the atoms and the charges are balanced. Like other types of balanced equations, half-reactions are balanced using the smallest possible whole-number coefficients. In the following equation, the atoms and charges are balanced, but the coefficients can all be divided by 2 to give the usual form of the half-reaction. 

Represent one example of a galvanic cell, and one example of an electrolytic cell, using chemical equations, half-reactions, and diagrams. 

Now write two new equations (half-reactions), using only the elements that change in oxidation number. Then add electrons to bring the equations into electrical balance. One equation represents the oxidation step the other represents the reduction step. Remember Oxidation produces electrons reduction uses electrons. 

Redox reactions, such as that shown in equation 6.22, can be divided into separate half-reactions that individually describe the oxidation and the reduction processes. 

Ladder diagrams can also be used to evaluate equilibrium reactions in redox systems. Figure 6.9 shows a typical ladder diagram for two half-reactions in which the scale is the electrochemical potential, E. Areas of predominance are defined by the Nernst equation. Using the Fe +/Fe + half-reaction as an example, we write... 

Before the equivalence point the titration mixture consists of appreciable quantities of both the oxidized and reduced forms of the analyte, but very little unreacted titrant. The potential, therefore, is best calculated using the Nernst equation for the analyte s half-reaction... 

Although EXo /ATcd is standard-state potential for the analyte s half-reaction, a matrix-dependent formal potential is used in its place. After the equivalence point, the potential is easiest to calculate using the Nernst equation for the titrant s half-reaction, since significant quantities of its oxidized and reduced forms are present. 

Substituting these concentrations into equation 9.17 along with the formal potential for the Fe 3-/pg2+ half-reaction from Appendix 3D, we find that the potential is... 

At the equivalence point, the moles of Fe + initially present and the moles of Ce + added are equal. Because the equilibrium constant for reaction 9.16 is large, the concentrations of Fe and Ce + are exceedingly small and difficult to calculate without resorting to a complex equilibrium problem. Consequently, we cannot calculate the potential at the equivalence point, E q, using just the Nernst equation for the analyte s half-reaction or the titrant s half-reaction. We can, however, calculate... 

Eeq by combining the two Nernst equations. To do so we recognize that the potentials for the two half-reactions are the same thus,... 

We begin by writing unbalanced equations for the oxidation and reduction half-reactions in part (a). 

The rate at which the corrosion of the 2iac proceeds depends on the rates of the two half reactions (eqs. 8 and 12). Equation 8, a necessary part of the desired battery reaction, fortunately represents a reaction that proceeds rather rapidly, whereas the reaction represented by equation 12 is slow. le, the generation of hydrogen on pure 2iac is a sluggish reaction and thus limits the overall corrosion reaction rate. 

When these half-reactions are summed, there is no net reaction. Thus the material balance of the cell is not altered by overcharge. At open circuit, equation 19 at the negative electrode is the sum of a two-step process, represented by equation 15 and... 

Table 1.7 shows typical half reactions for the oxidation of a metal M in aqueous solutions with the formation of aquo cations, solid hydroxides or aquo anions. The equilibrium potential for each half reaction can be evaluated from the chemical potentials of the species involved see Appendix 20.2) and it should be noted that there is no difference thermodynamically between equations 2(a) and 2(b) nor between 3(a) and 3(b) when account is taken of the chemical potentials of the different species involved. 

The data given in Tables 1.9 and 1.10 have been based on the assumption that metal cations are the sole species formed, but at higher pH values oxides, hydrated oxides or hydroxides may be formed, and the relevant half reactions will be of the form shown in equations 2(a) and 2(b) (Table 1.7). In these circumstances the a + will be governed by the solubility product of the solid compound and the pH of the solution. At higher pH values the solid compound may become unstable with respect to metal anions (equations 3(a) and 3(b), Table 1.7), and metals like aluminium, zinc, tin and lead, which form amphoteric oxides, corrode in alkaline solutions. It is evident, therefore, that the equilibrium between a metal and an aqueous solution is far more complex than that illustrated in Tables 1.9 and 1.10. Nevertheless, as will be discussed subsequently, a similar thermodynamic approach is possible. 

Pourbaix has classified the various equilibria that occur in aqueous solution into homogeneous and heterogeneous, and has subdivided them according to whether the equilibria involve electrons and/or hydrogen ions. The general equation for a half reaction is... 

Equations 1.83 and 1.84, or the equations derived from them (1.85 to 1.89), may be used to calculate and E an., providing the various parameters involved are known. The equations also serve to illustrate how and corr, depend upon a thermodynamic factor ( r,ceii. °r r.c and E, ) and the kinetic factors a and / o for each of the half reactions that constitute the corrosion reaction. 

Equations 20.176 and 20.179 emphasise the essentially thermodynamic nature of the standard equilibrium e.m.f. of a cell or the standard equilibrium potential of a half-reaction E, which may be evaluated directly from e.m.f. meeisurements of a reversible cell or indirectly from AG , which in turn must be evaluated from the enthalpy of the reaction and the entropies of the species involved (see equation 20.147). Thus for the equilibrium Cu -)-2e Cu, the standard electrode potential u2+/cu> hence can be determined by an e.m.f. method by harnessing the reaction... 

We add the half-equations to obtain the overall reaction and add the half-reaction voltages to obtain the overall voltage ... 

The Nernst equation can also be used to determine the effect of changes in concentration on the voltage of an individual half-cell, E or Consider, for example, the half-reaction... 

Strategy First (1) set up the Nemst equation for the reduction half-reaction and calculate red. Then (2) repeat the calculation for the oxidation half-reaction, finding Eox. Finally (3), add rcd + ra if the sum is positive, disproportionation should occur. 

Both of these half-reactions show production of electrons. But we know there must be an electron used for each produced, so one of the equations must be reversed. Experiment shows us it is the second because hydrogen gas is evolved from the solution. The first equation is correct as written. Lithium metal dissolves and is converted to ions. Thus,... 

When potassium chlorate solution, KClOi, is added to hydrochloric acid, chlorine gas is evolved. Although we can find the half-reaction, 2C1- = Ck(g) + 2e, in Appendix 3, we find no equation with CIQT ion involved. We can surmise that CIO3" is accepting electrons and changing into chlorine. Let us write a partial half-reaction in which we indicate an unknown number of electrons and in which we have conserved only chlorine atoms ... 

Now we can return to working out the equation for the reaction we observed. The least common multiple between 1 and 5 is 5. Writing the half-reactions to involve 5 electrons and adding them, we obtain... 

Of course, the oxidation number method gives the same balanced equation as the half-reaction method. 

One method of obtaining copper metal is to let a solution containing Cu+2 ions trickle over scrap iron. Write the equations for the two half-reactions involved. Assume the iron becomes Fe. Indicate in which half-reaction oxidation is taking place. 

By use of half-reactions, give a balanced equation for each of the following reactions ... 

Since the dichromate ion on the left side of the equation has been reduced to chromic ion, Cr+ on the right side, the conversion of methanol to formaldehyde must involve oxidation. To show more clearly that methanol has been oxidized, let us balance this reaction by the method of half-reactions. We have encountered the halfreaction involving dichromate and chromic ions before (Problem 20b in Chapter 12). It is... 

We consider oxidation first. To show the removal of electrons from a species that is being oxidized in a redox reaction, we write the chemical equation for an oxidation half-reaction. A half-reaction is the oxidation or reduction part of the reaction considered alone. For example, one battery that Volta built used silver and zinc plates to carry out the reaction... 

An oxidation half-reaction is a conceptual way of reporting an oxidation the electrons are never actually free. In an equation for an oxidation ha If-reaction, the electrons released always appear on the right of the arrow. Their state is not given, because they are in transit and do not have a definite physical state. The reduced and oxidized species in a half-reaction jointly form a redox couple. In this example, the redox couple consists of Zn2+ and Zn, and is denoted Zn2+/Zn. A redox couple has the form Ox/Red, where Ox is the oxidized form of the species and Red is the reduced form. 

This half-reaction, too, is conceptual the electrons are not actually free. In the equation for a reduction half-reaction, the electrons gained always appear on the left of the arrow. In this example, the redox couple is Agf/Ag. 

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. 

When balancing redox equations, we consider the gain of electrons (reduction) separately from the loss of electrons (oxidation), express each of these processes as a halfreaction, and then balance both atoms and charge in each of the two half-reactions. When we combine the halfreactions, the number of electrons released in the oxidation must equal the number used in the reduction. 

The general procedure for balancing the chemical equation for a redox reaction is first to balance the half-reactions separately. 

Step 6 Multiply all species in either one or both halfreactions by factors that result in equal numbers of electrons in the two half-reactions, and then add the two equations and include physical states. 

Finally, simplify the appearance of the equation by canceling species that appear on both sides of the arrow and check to make sure that charges as well as numbers of atoms balance. In some cases it is possible to simplify the half-reactions before they are combined. 

Step 6 Write the overall equation. In the half-reactions, 2 electrons are lost, but 5 are gained, so we need 10 in each half-reaction. 

Sei f-Test 12.2B When iodide ions react with iodate ions in basic aqueous solution, triiodide ions, I,, are formed. Write the net ionic equation for the reaction. (Note that the same product is obtained in each half-reaction.)... 

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