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Step 1 : Write the unbalanced formula equations

NaMnO 4 ( aq ) + HCl ( aq ) --------> Cl 2 (g) + MnCl 2 ( aq ).

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Step 1 : Write the unbalanced formula equations

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  1. NaMnO4(aq) + HCl(aq) --------> Cl2(g) + MnCl2(aq) 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. Step 2: Identify the species that are oxidized and reduced. Start by labeling the oxidation #s. Since Mn goes from +7 to +2 it gained e– and is reduced. Since Cl goes from –1 to 0 it 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 +1 +7 –2 +1 –1 0 +2 –1 NaMnO4(aq) + HCl(aq) --------> Cl2(g) + MnCl2(aq) Step 1: Write the unbalanced formula equations Step 3: Write and balance the oxidation half reaction Cl– --------> Cl2 First balance the mass by adding a 2 in front of the Cl– 2 Cl– --------> Cl2 Next add e– to balance the charge 2Cl– --------> Cl2 + 2e–

  2. Step 4: Write and balance the reduction half reaction MnO41– --------> Mn2+ First balance the mass – since Mn is already balanced, balance the O by adding 4 H2O. MnO41– --------> Mn2+ + 4 H2O Finish balancing the mass – by balancing the H by adding 8 H+ 8 H+ + MnO41– --------> Mn2+ + 4 H2O Next add e– to balance the charge – left side is +7 total, right side is +2, so add 5 e– 5 e– + 8 H+ + MnO41– --------> Mn2+ + 4 H2O Step 5: Next make the oxidation and reduction half-reactions have the same # of e– 5 e– + 8 H+ + MnO41– --------> Mn2+ + 4 H2O (X2) 10 e– + 16 H+ + 2 MnO41– --------> 2 Mn2+ + 8 H2O 2Cl– --------> Cl2 + 2e– (X5) 10 Cl– --------> 5 Cl2 + 10e– 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 only the e– 16 H+ + 2MnO41– + 10Cl– --------> 5 Cl2 + 2 Mn2+ + 8 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 6 Cl– spectator ions. 2 MnO41–+ 16 HCl --------> 5 Cl2 + 2 MnCl2 + 8 H2O + 2 Cl– Next add back spectator Na+. 2 NaMnO4(aq) + 16 HCl(aq) --------> 5 Cl2(g) + 2 MnCl2(aq) + 8 H2O + 2NaCl (aq)

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