Oxidation numbers writing redox equations

Oxidation numbers are used when writing redox equations to account for the transfer of electrons. For example, consider the simple reaction representing the rusting of iron ... 

Explains how to work out oxidation numbers and write redox equations... 

In earlier sections of this chapter, we showed how to write and balance equations for precipitation reactions (Section 4.2) and acid-base reactions (Section 4.3). In this section we will concentrate on balancing redox equations, given the identity of reactants and products. To do that, it is convenient to introduce a new concept, oxidation number. 

In this section, you will write balanced equations for redox reactions using the oxidation number method... 

Write an example of a balanced chemical equation for a redox reaction. Assign oxidation numbers to each element in the equation, then explain how you know it is a redox reaction. 

The first step is to write both compounds of the redox couple and when necessary to adjust the stoichiometric numbers, in order to ensure that the element with the variable oxidation number is preserved. The number of electrons exchanged is then determined from the difference between the oxidation numbers of the element in its oxidized and reduced states, taking into account the stoichiometry. Protons are added if necessary to ensure that the sum of the charges on both sides of the overall equation are kept the same. Finally, water molecules H2O are added to balance the oxygen (or hydrogen) element. It is then possible to check if the hydrogen (or oxygen) element is also balanced in the overall equation. 

The first step is to write the two compounds of the redox couple and to adjust their stoichiometric numbers, if necessary, in order to ensure that the element with the variable oxidation number is preserved. Water molecules, H2O, and protons, H", may then be added to ensure that the O and H elements are preserved respectively. The final step consists of adding electrons in order to balance the charge on both sides of the reaction equation. This method, which is more formal than the previous one, avoids calculating oxidation numbers, but may be tricky to carry out if the redox couples are not sufficiently well-identified. 

Fe is reduced (its oxidation state goes down from +3 to +2), and V is oxidized (its oxidation state goes up from +2 to +3). Appendix D discusses oxidation numbers and how to balance redox equations. You should be able to write a balanced redox reaction as the sum of two half-reactions, one an oxidation and the other a reduction. 

Analyze and Plan We are told that the reaction of NH2OH and Cu produces N2 and Cu. We are asked to write a balanced equation for the reaction. Because the Cu and N change oxidation numbers during the reaction, this is a redox reaction, and it can be balanced by the method of half-reactions discuss in Section 20.2. 

Oxides. Chlorine oxides in which Cl has an even oxidation number undergo self-redox. In basic solution, the products are oxyanions such as C102, CIOs", and CIO4". Write the balanced ionic equations for the self-redox of (a) CIO2, (b) CI2O3. 

For each of the five experiments described in Model 3, write the balanced chemical equation (no "e " appears in the balanced chemical equation) for the redox reaction that could occur between the metal bar and the ion in solution. Note that the same number of electrons must be lost and gained in the transfer process. In each case indicate the oxidizing agent and the reducing agent. 

Plan To determine whether a redox reaction is spontaneous under Equation 20. iO to calculate the standard emf, °, for the reaction. If a standard conditions, we first need to write its reduction and oxidation reaction is spontaneous, its standard emf must be a positive number, half-reactions. We can then use the standard reduction potentials and... 

---