Question 15.8: Balancing a Net Ionic Redox Equation Using the Oxidation Num......
Balancing a Net Ionic Redox Equation Using the Oxidation Number Method
Balance the following net ionic redox equation for the reaction between Cu and NO {_3}^- in acidic solution using the oxidation number method for balancing redox equations.
Cu + NO {_3}^- \longrightarrow Cu ^{2+} + NO _2
Step 1 The elements undergoing oxidation and reduction are identified by assigning oxidation numbers.
Cu + NO {_3}^- \longrightarrow Cu ^{2+} + NO _2
0 +5 -2 +2 +4 -2
The two elements undergoing oxidation number change are Cu and N.
Step 2 The change in oxidation number per atom is determined.
Step 3 For both Cu and N the oxidation number change per formula unit is the same as per atom. Both Cu and Cu ^{2+} contain only one Cu atom; similarly, both NO {_3}^- and NO _2 contain only one N atom.
Step 4 By multiplying the N oxidation number decrease by 2, we make the oxidation number increase and decrease per formula unit the same—two units.
Step 5 The bracket notation indicates that two N atoms and one Cu atom undergo an oxidation number change. Translating this information into coefficients, we get
1 Cu + 2 NO {_3}^- \longrightarrow 1 Cu ^{2+} + 2 NO _2
Step 6 The only atoms left to balance are those of O.
Step 7 Since this is a net ionic equation, the charges must balance; that is, the sum of the ionic charges of all species on each side of the equation must be equal. (They do not have to add up to zero; they just have to be equal.) In an acidic solution, which is the case in this example, charge balance is accomplished by adding H ^+ ion.
As the equation now stands, we have a charge of -2 on the left side (two nitrate ions each with a -1 charge) and a charge of +2 on the right side (one copper ion). By adding four H ^+ ions to the left side, we balance the charge at +2.
-2 + (+4) = +2
The equation at this point becomes
1 Cu + 2 NO {_3}^- + 4 H ^+ \longrightarrow 1 Cu ^{2+} + 2 NO _2
Step 8 The hydrogen atoms are balanced through the addition of H _2 O molecules. There are four H atoms on the left side (four H ^+ ions) and none on the right side. Addition of two H _2 O molecules to the right side will balance the H atoms at four per side.
1 Cu + 2 NO {_3}^- + 4 H ^+ \longrightarrow 1 Cu ^{2+} + 2 NO _2 + 2 H _2 O
Step 9 The O atoms automatically balance at six atoms on each side. This is our double-check that previous steps have been correctly carried out. The balanced net ionic equation is thus
Cu + 2 NO {_3}^- + 4 H ^+ \longrightarrow Cu ^{2+} + 2 NO _2 + 2 H _2 O
Answer Double Check: Is the final equation balanced relative to the charge associated with the ions present? Yes. The charge balance is at +2. On the reactant side of the equation, two -1 ions (NO {_3}^- ) and four +1 ions (H ^+ ) give a net charge of +2. On the product side of the equation, the one Cu ^{2+} ion gives a charge of +2.