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CHAP. 14]                        OXIDATION AND REDUCTION                              209


               EXAMPLE 14.21. Complete and balance the following equation in basic solution:
                                                       2−
                                           Fe(OH) 2 + CrO 4  −→ Fe(OH) 3 + Cr(OH) 3
                Ans.  Step 1:       Fe(OH) 2 −→ Fe(OH) 3                          CrO 4  2−  −→ Cr(OH) 3
                     Step 2: Already done.                                            Already done
                                                                               −
                     Step 3:        Fe(OH) 2 −→ Fe(OH) 3 + e  −              3 e + CrO 4 2−  −→ Cr(OH) 3
                                                                               −
                                                        −
                                                                            +
                     Step 4:        Fe(OH) 2 −→ Fe(OH) 3 + e + H +       5H 3 e + CrO 4 2−  −→ Cr(OH) 3
                     Step 5:   H 2 O + Fe(OH) 2 −→ Fe(OH) 3 + e + H +  5H + 3 e + CrO 4 2−  −→ Cr(OH) 3 + H 2 O
                                                                          +
                                                        −
                                                                               −
                     Step 6: 3 H 2 O + 3Fe(OH) 2 −→ 3Fe(OH) 3 + 3 e + 3H +
                                                           −
                                                +    −      2−                −     +
                            3H 2 O + 3Fe(OH) 2 + 5H + 3 e + CrO 4  −→ 3Fe(OH) 3 + 3 e + 3H + Cr(OH) 3 + H 2 O
                                                     +      2−
                                  2H 2 O + 3Fe(OH) 2 + 2H + CrO 4  −→ 3Fe(OH) 3 + Cr(OH) 3
                     To eliminate the H , which cannot exist in basic solution, add 2 OH to each side, forming 2 H 2 O on the left:
                                                                       −
                                   +
                                                            2−                          −
                                 2H 2 O + 3Fe(OH) 2 + 2H 2 O + CrO 4  −→ 3Fe(OH) 3 + Cr(OH) 3 + 2OH
                     Finally,
                                                            2−  −→ 3Fe(OH) 3 + Cr(OH) 3 + 2OH −
                                       4H 2 O + 3Fe(OH) 2 + CrO 4
               14.6. ELECTROCHEMISTRY
                   The interaction of electricity with matter was introduced in Chap. 8, where the electrical decomposition of
               a melted salt was used to prepare active elements from their compounds. An illustration of an electrolysis was
               shown in Fig. 5-2. Chemical reactions occur at the two electrodes. The electrode at which oxidation occurs is
               called the anode; the one at which reduction takes place is called the cathode. Electricity passes through a circuit
               under the influence of a potential or voltage, the driving force of the movement of charge. There are two different
               types of interaction of electricity and matter, as follows:
               Electrolysis: Electric current causes chemical reaction.
               Galvanic cell action: Chemical reaction causes electric current, as in the use of a battery.



               Electrolysis
                   The following are the requirements for electrolysis:

               1. Ions (There must be charged particles to carry current. It might not be the ions that react, however.)
               2. Liquid, either a pure liquid or a solution, so that the ions can migrate
               3. Source of potential (In a galvanic cell, the chemical reaction is the source of potential, but not in an electrolysis
                   cell.)
               4. Mobile ions, complete circuit (including wires to carry electrons), and electrodes (at which the current
                   changes from the flow of electrons to the movement of ions or vice versa)

                   If you electrolyze a solution containing a compound of a very active metal and/or a very active nonmetal, the
               water (or other solvent) might be electrolyzed instead of the ion. For example, if you electrolyze molten sodium
               chloride, you get the free elements:

                                                      electricity
                                              2 NaCl(l) −−−→ 2Na(l) + Cl 2 (g)
               However, if you electrolyze a dilute aqueous solution of NaCl, the water is decomposed. (The NaCl is necessary
                                             +
               to conduct the current, but neither Na nor Cl reacts at the electrodes.)
                                                    −
                                                       electricity
                                                 2H 2 O −−−−−→ 2H 2 + O 2
                                                      NaCl(dilute)
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