Step 1 : Write the unbalanced formula equations

1. NaMnO4(aq) + HCl(aq) --------> Cl2(g) + MnCl2(aq) Step 2: Identify the species that are oxidized and reduced. Start by labeling the oxidation #s. +7 –2 +1 –1 0 +2 –1 +1 PROBLEM:Aqueous Sodium Permanganate plus Hydrochloric Acid yields Chlorine gas plus aqueous Mangenese(II) Chloride. Hint: There is another compound that is needed to balance the reaction but it is only made of spectator ions and you need to figure that out as you solve the problem. NaMnO4(aq) + HCl(aq) --------> Cl2(g) + MnCl2(aq) Since Mn goes from +7 to +2 it gained e– and is reduced. Since Cl goes from –1 to 0it lost e– and is oxidized. Do not be distracted by Cl– on the products side since that is a spectator ion. Yes the Cl– is both oxidized and a spectator at the same time Step 3: Write and balance the oxidation half reaction (Note: Chlorous acid is a weak acid) Cl– --------> HClO2 Step 1: Write the unbalanced formula equations Since the Cl is already balanced, balance the O by adding 2 H2O 2 H2O+ Cl– --------> HClO2 Next add 2 H+ to balance the H mass 2 H2O + Cl– --------> HClO2 + 3 H+ Next add e– to balance the charge – since the left is –1, and the right is +3, add 4e– to the right 2 H2O + Cl– --------> HClO2+ 3 H+ + 4 e–

2. Step 4: Write and balance the reduction half reaction CrO42–--------> Cr3+ First balance the mass – since Cr is already balanced, balance the O by adding 4 H2O. CrO42–--------> Cr3+ + 4 H2O Finish balancing the mass – by balancing the H by adding 8 H+ 8 H+ + CrO42–--------> Cr3++ 4H2O Next add e– to balance the charge – left side is +6 total, right side is +3, so add 3 e– 8 H+ + 3e– + CrO42– --------> Cr3++ 4H2O Step 5: Next make the oxidation and reduction half-reactions have the same # of e– 8 H+ + 3e– + CrO42– --------> Cr3++ 4 H2O (X4) 32 H+ + 12e– + 4 CrO42– --------> 4 Cr3++ 16 H2O 2 H2O + Cl– --------> HClO2 + 3 H+ + 4 e– (X3) 6 H2O + 3 Cl–--------> 3 HClO2+ 9 H+ + 12e– Step 6: Combine the two half-reactions together, to make the net ionic equation. Remember to cancel out the items on opposite sides of arrow (water, H+ ions, and e–). In this case in addition to the e–, 9H+ on each side, and 6 H2O on both sides cancel out. 23 H+ + 4 CrO42–+ 3 Cl– --------> 3 HClO2 + 4 Cr3++ 10 H2O

3. Step 7: Add back spectator ions and combine with other ions to write complete compounds and the balanced overall equation. Start with H+ and Cl–. Since there are more H+ than Cl–, need to add an extra 20 Cl– spectator ions. 23 HCl+ 4 CrO42–--------> 3 HClO2+ 4 CrCl3+ 10 H2O + 8 Cl– Next add back spectator Na+. 23 HCl+ 4 Na2CrO4--------> 3 HClO2+ 4 CrCl3+ 10 H2O + 8NaCl