Figure 2-112. Balancing the equation for the redox reaction between Al and O2. In step 1
, we identify the elements that change their oxidations states in the reaction. Both the Al and O do. Before the reaction, the aluminum was a simple substance and so was the O2
. The oxidation state of any simple substance is always 0. In the reaction, the oxidation state of the aluminum metal changes from 0 to +3 and that of the oxygen from 0 to -2. The aluminum is therefore oxidized and the oxygen is reduced. In step 2
, we write half-equations for the oxidation and reduction (Figure 2-112). One Al atom loses three electrons to get to the oxidation state of +3. One molecule of O2
gains four electrons, as each O atom gains two to attain the oxidation state of -2. In step 3
, we balance the number of electrons donated and the number of electrons accepted. For that, we multiply the oxidation half-equation by 4 and the reduction half-equation by 3. Having done that, we can see that the number of electrons in both half-equations is the same, 12, the least common multiple
for 3 and 4. In step 4
, we use the multipliers found in step 3 (4 for Al and 3 for O2
) as the correct coefficients for the reagents
. These coefficients should be used as the guidance to balance the product
. Placing the coefficient 2 before the Al2
formula on the right side concludes the balancing.
The next equation we will balance is for the reaction that is used for the recovery of refined gold from gold scraps. Watch this video
showing this fascinating process that employs aqua regia
, a mixture of concentrated hydrochloric and nitric acids. Aqua regia
, "royal water" or "king's water" in Latin, takes its name from its unique ability to dissolve gold, the "king of metals" that no other individual acid dissolves. An inexperienced person should keep away from extremely corrosive and hazardous aqua regia.
The reaction of the dissolution of gold in aqua regia is as follows.Au + HNO3 + HCl = AuCl3 + NO + H2O
To balance this equation, first we identify the elements that change their oxidation states (Figure 2-113, step 1). These elements are Au, which is oxidized (the oxidation state changes from 0 to +3), and N, which is reduced (the oxidation state changes from +5 to +2). Next, we write the half-equations for the oxidation and reduction (step 2). As the number of electrons one Au atom loses equals the number of electrons one N atom gets, we do not need to use any multipliers in this case (step 3).