Balancing a Redox Reaction in Basic Solution

• 1). Determine the charge of the molecules that are being oxidized and reduced. In our example equation, we see that sulfur (S) and nitrogen (N) are the two molecules participating in the reaction. On the reactant side of the equation, [S(s) + NO3-(aq)] we see that S has a neutral charge. The charge of the N molecule can be determined by looking at the other molecule of the compound and the overall charge of the compound. NO3- has total charge of -1. There are three oxygen molecules which will each have a -2 charge as indicated by its location in the periodic table resulting in a total charge of -6 for the O3 term. So, the charge of N (x) plus the charge of the 3 O (-6) molecules is equal to the overall charge of the molecule (-1), or x - 6 = -1. Solving for x gives a charge of +5 for the nitrogen molecule. Repeat the same procedure for the reactants which results in a charge of +4 for the S molecule in SO2 and a charge of +2 for the N molecule in NO. The SO2 molecule has an overall neutral charge and each molecule of oxygen has a -2 charge resulting in a -4 charge; therefore, S must have a +4 charge to balance this out and give an overall neutral charge for the SO2 molecule. Likewise, the NO molecule has an overall neutral charge with the oxygen having a -2 charge. The N molecule must then have a +2 charge to make the molecule neutral.
  
  
• 2). Divide the equation into two half reactions: oxidation half reaction and reduction half reaction. This is accomplished by determining which molecule is oxidized and donates electrons (reducing agent) and which molecule accepts electrons and is reduced (oxidizing agent). Using our example equation, we would have [ S(s) -> SO2(g)] and [ NO3- -> NO(g)] . Using our previously determine charges on the S and N molecule, we see that S goes from a neutral charge (S) to a +4 charge (SO2) and N goes from a +5 charge (NO3-) to a +2 charge (NO). So, the N molecule had a reduction in charge (+5 to +2) meaning that it gained electrons (reduced) which indicates it as the reduction half reaction. The other reaction must, therefore, be the oxidation half reaction.
  
  
• 3). Balance the elements other than H and O of each half reaction. This is already accomplished for our example as the number of S of N reactants matches the number of product atoms for each half-reaction respectively.
  
  
• 4). Balance the O atoms by adding H2O.
  
  The oxidation reaction would be S + 2 H2O -> SO2
  
  The reduction reaction would be NO3- -> NO + 3 H2O
  
  
• 5). Balance the H atoms by adding H+ to each half reaction giving us
  
  S + 2 H2O -> SO2 + 4 H+ for the oxidation half reaction
  
  and
  
  4 H + NO3- -> NO + 2 H2O for the reduction half reaction.
  
  
• 6). Balance the charges between the reactants and products of each have reaction by adding electrons. This yields:
  
  S + 2 H2O -> SO2 + 4 H+ + 4e­- for the oxidation half
  
  and
  
  4 H+ + NO3 + 3e- -> NO + 2 H2O for the reduction half.
  
  
• 7). Multiply each term in the half reactions by the other half reaction's number of electrons. This results in the cancellation of the electrons. So, we would multiply the oxidation half reaction by 3e- and the reduction half reaction by 4e- giving us:
  
  3 S + 6 H2O -> 3 SO2 + 12 H+
  
  and
  
  16H+ + 4 NO3- -> 4 NO + 8 H2O
  
  
• 8). Recombine the reactants and products of the two half reactions and cancel the like terms on each side.
  
  Combining the two equations gives us:
  
  3 S + 4 NO3- +6 H2O + 16H+ -> 3 SO2 + 4 NO + 12 H+ + 8 H2O
  
  Canceling the number of H+ and H2O molecules on each side we get:
  
  3 S + 4 NO3- + 4H+ -> 3 SO2 + 4 NO + 2 H2O
  
  
• 9). Add the number OH- molecules to each side equal to the amount of H+ and combine the remaining H+ and OH- on each side as water molecules. The solution is basic so OH- ions will combine with all of the H+ ions which gives us:
  
  3 S + 4 NO3- + 4H+ + 4OH- -> 3 SO2 + 4 NO + 2 H2O + 4OH-
  
  Combining the OH- and H+ molecules and canceling the resulting water molecules on each side gives:
  
  3 S + 4 NO3- + 2H20 -> 3 SO2 + 4 NO + 4OH-
  
  
• 10
  
  Ensure that the charge of the reactants is equal to the charge of the products in the final balanced equation. In our example we have the charge of the reactants as -4 (4 NO3-) molecules and the charge of the products as 4 (4 OH- molecules). All of the elements and charges are balanced from reactant to product; thus, the equation is balanced.