Question 15.7: Balancing a Molecular Redox Equation Using the Oxidation Num......
Balancing a Molecular Redox Equation Using the Oxidation Number Method
Balance the following molecular redox equation for the reaction involving Cr, O_2 and HBr using the oxidation number method for balancing redox equations.
Cr + O_2 + HBr \longrightarrow CrBr _3 + H _2 O
Step 1 We identify the elements being oxidized and reduced by assigning oxidation numbers.
Cr + O _2 + HBr \longrightarrow CrBr _3 + H _2 O
0 0 +1-1 +3-1 +1-2
Chromium (Cr) and oxygen (O) are the elements that undergo oxidation number change.
Step 2 The change in oxidation number per atom is shown by drawing brackets connecting the oxidizing and reducing agents to their products and indicating the change at the middle of the bracket.
Step 3 For Cr the change in oxidation number per formula unit is the same as the change per atom because both Cr and CrBr _3 , the two Cr-containing species, contain only one Cr atom. For O, the change in oxidation number per formula unit will be double the change per atom since O _2 contains two atoms. The change per formula unit is indicated by multiplying the per-atom change by an appropriate numerical factor, which is 2 in this case.
Step 4 For Cr, the total increase in oxidation number per formula unit is +3. For oxygen, the total decrease in oxidation number per formula unit is -4. To make the increase equal to the decrease, we must multiply the oxidation number change for the element oxidized (Cr) by 4 and the oxidation number change for the element reduced (O) by 3- This will make the increase and decrease both numerically equal to 12.
Step 5 We are now ready to place coefficients in the equation in front of the oxidizing and reducing agents and their products. The bracket notation indicates that four Cr atoms undergo an oxidation number change. Place the coefficient 4 in front of both Cr and CrBr _3 . The bracket notation also indicates that six O atoms (3 × 2) undergo an oxidation number decrease of two units. Thus, we need six oxygen atoms on each side. Place the coefficient 3 in front of O _2 (six atoms of O), and the coefficient 6 in front of H _2 O (six atoms of O).
4 Cr + 3 O _2 + HBr \longrightarrow 4 CrBr _3 + 6 H _2 O
The equation is only partially balanced at this point; only Cr and O atoms are balanced.
Step 6 We next balance the element Br (by inspection). There are 12 Br atoms on the right side. Thus, to obtain 12 Br atoms on the left side, we place the coefficient 12 in front of HBr.
4 Cr + 3 O _2 + 12 HBr \longrightarrow 4 CrBr _3 + 6 H _2 O
Step 7 This step is not needed when the equation is a molecular equation.
Step 8 In this particular equation the H atoms are already balanced. There are 12 hydrogen atoms on each side of the equation.
Step 9 If all of the previous procedures (steps) have been carried out correctly, the O atoms should automatically balance. They do. There are six O atoms on each side of the equation. The balanced equation is thus
4 Cr + 3 O _2 + 12 HBr \longrightarrow 4 CrBr _3 + 6 H _2 O
Answer Double Check: Are there the same number of atoms of each kind on each side of the equation? Yes. The atom balance is 4 Cr, 6 O, 12 H, and 12 Br on each side of the equation.