Aqueous half reactions

Half-reaction (i) means that Co(II) in aqueous solution cannot be oxidised to Co(III) by adding ammonia to obtain the complexes in (ii), oxidation is readily achieved by, for example, air. Similarly, by adding cyanide, the hexacyanocobaltate(II) complex becomes a sufficiently strong reducing agent to produce hydrogen from water ... 

When an element can exist in several oxidation staie.s it is sometimes convenient to display the various reduction potentials diagramaltcally. the corresponding half-reactions under standard conditioas being implied. Thus, in acidic aqueous soiultons... 

In addition to simple dissolution, ionic dissociation and solvolysis, two further classes of reaction are of pre-eminent importance in aqueous solution chemistry, namely acid-base reactions (p. 48) and oxidation-reduction reactions. In water, the oxygen atom is in its lowest oxidation state (—2). Standard reduction potentials (p. 435) of oxygen in acid and alkaline solution are listed in Table 14.10- and shown diagramatically in the scheme opposite. It is important to remember that if or OH appear in the electrode half-reaction, then the electrode potential will change markedly with the pH. Thus for the first reaction in Table 14.10 O2 -I-4H+ -I- 4e 2H2O, although E° = 1.229 V,... 

Half reactions involving the oxidation of a metal in aqueous solutions... 

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... 

Chromium metal can be electroplated from an aqueous solution of potassium dichromate. The reduction half-reaction is... 

Aluminum metal reacts with aqueous acidic solutions to liberate hydrogen gas. Write the two half-reactions and the net ionic reaction. 

In the dectrolysis of aqueous cupric bromide, CuBr2, 0.500 gram of copper is deposited at one dectrode. How many grams of bromine are formed at the other electrode Write the anode and cathode half-reactions. 

The I——12 half-reaction has many applications in aqueous solution chemistry. The use of I- as a reducing agent and I2 as an oxidizing agent, particularly for quantitative purposes, is called iodimetry. 

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. 

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.)... 

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. 

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. 

Separate the following unbalanced redox processes into half-reactions. All occur in aqueous solution ... 

The spontaneous redox reaction shown in Figure 19-7 takes place at the surfaces of metal plates, where electrons are gained and lost by metal atoms and Ions. These metal plates are examples of electrodes. At an electrode, redox reactions transfer electrons between the aqueous phase and the external circuit. An oxidation half-reaction releases electrons to the external circuit at one electrode. A reduction half-reaction withdraws electrons from the external circuit at the other electrode. The electrode where oxidation occurs is the anode, and the electrode where reduction occurs is the cathode. 

In contrast with these active electrodes, a passive electrode conducts electrons to and from the external circuit but does not participate chemically in the half-reactions. Figure 19-8 shows a redox setup that contains passive electrodes. One compartment contains an aqueous solution of iron(III) chloride in contact with a platinum electrode. Electron transfer at this electrode reduces Fe " (a q) to Fe " ((2 q) ... 

The electrical current needed to start an automobile engine is provided by a lead storage battery. This battery contains aqueous sulfuric acid in contact with two electrodes. One electrode is metallic lead, and the other is solid Pb02. Each electrode becomes coated with solid PbSOq as the battery operates. Determine the balanced half-reactions, the overall redox reaction, and the anode and cathode in this galvanic cell. 

C19-0014. The mercury battery, use of which is being discontinued because of the toxicity of mercury, contains HgO and Zn in contact with basic aqueous solution. The redox products are Hg and ZnO. Determine the oxidation and reduction half-reactions and the overall reaction for these batteries. 

The cell shown in Figure 19-16 can serve as an example for calculations using Equation. One cell contains aqueous 1.00 M iron(in) chloride in contact with an iron metal electrode, and the other cell contains 1.00 M KCl in contact with a silver-silver chloride (AgCl/Ag) electrode. The half-reactions for these electrodes follow ... 

C19-0050. What are the half-reactions for these redox processes (a) Aqueous hydrogen peroxide acts on Co, and the products are hydroxide and Co , in basic solution, (b) Methane reacts with oxygen gas and produces water and carbon dioxide, (c) To recharge a lead storage battery, lead(II) sulfate is converted to lead metal and to lead(IV) oxide, (d) Zinc metal dissolves in aqueous hydrochloric acid to give ions and hydrogen gas. 

C21-0070. In basic aqueous solution, A1 acts as a strong reducing agent, being oxidized to AIO2. Balance this half-reaction, and determine balanced net reactions for A1 reduction of the following (a) NO3 to NH3 (b) H2 O to H2 and (c) Sn03 to Sn. 

It is not possible to prepare F2 by electrolysis of an aqueous NaF solution. In electrolysis, the most easily oxidized and reduced species are the ones involved. To prepare F2, the oxidation of F would have to occur. However, water is more easily oxidized than is F, as seen by its position in the standard reduction potential chart (Appendix J and below). By inspection, H20 is a stronger reducing agent than F because the reduction half-reaction has a less positive E°. So H20 s oxidation is preferable to F s oxidation. F2 can be prepared from molten NaF, but not aqueous NaF. 

A different view of the OMT process is that the molecule, M, is fully reduced, M , or oxidized, M+, during the tunneling process [25, 26, 92-95]. In this picture a fully relaxed ion is formed in the junction. The absorption of a phonon (the creation of a vibrational excitation) then induces the ion to decay back to the neutral molecule with emission (or absorption) of an electron - which then completes tunneling through the barrier. For simplicity, the reduction case will be discussed in detail however, the oxidation arguments are similar. A transition of the type M + e —> M is conventionally described as formation of an electron affinity level. The most commonly used measure of condensed-phase electron affinity is the halfwave reduction potential measured in non-aqueous solvents, Ey2. Often these values are tabulated relative to the saturated calomel electrode (SCE). In order to correlate OMTS data with electrochemical potentials, we need them referenced to an electron in the vacuum state. That is, we need the potential for the half reaction ... 

Write balanced half-reactions from the net ionic equation for the reaction between solid aluminum and aqueous iron(in) sulfate. The sulfate ions are spectator ions, and are not included. 

Write the net ionic equation and the half-reactions for the disproportionation of mercury(I) ions in aqueous solution to give liquid mercury and aqueous mercury(II) ions. Assume that mercury(I)... 

Recall that, if you consider a redox reaction as two half-reactions, electrons are lost in the oxidation half-reaction, and electrons are gained in the reduction half-reaction. For example, you know the reaction of zinc with aqueous copper(II) sulfate. 

You know from Investigation 10-A that magnesium metal, Mg(s), displaces aluminum from an aqueous solution of one of its compounds, such as aluminum nitrate, Al(N03)3(aq). To obtain a balanced net ionic equation for this reaction, you can start by looking at tbe half-reactions. Magnesium atoms undergo oxidation to form magnesium ions, which have a 2+ charge. The oxidation half-reaction is as follows. 

A comma separates the formulas Fe and Fe for the ions involved in the reduction half-reaction. The formulas are not separated by a vertical line, because there is no phase boundary between these ions. The Fe and Fe " ions exist in the same aqueous solution. 

To predict the products of an electrolysis involving an aqueous solution, you must examine all possible half-reactions and their reduction potentials. Then, you must find the overall reaction that requires the lowest external voltage. That is, you must find the overall cell reaction with a negative cell potential that is closest to zero. The next Sample Problem shows you how to predict the products of the electrolysis of an aqueous solution. 

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