Balance the Equation

Question

MnO₄^- (aq) + Cl^- (aq) --> Mn²⁺ (aq) + Cl₂ (aq) (acidic solution)

So far, I have managed to do the following:

MnO₄^- --> Mn²⁺

2MnO₄^- --> Mn²⁺

2MnO₄^- --> Mn²⁺ + 4H₂O

2MnO₄^- + 8H⁺ --> Mn²⁺ + 4H₂O

2MnO₄^- + 8H⁺ + 4e^- --> Mn²⁺ + 4H₂O

Cl^- --> Cl₂

2Cl^- --> Cl₂

2Cl^- --> Cl₂ + 2e^-

2MnO₄^- + 8H⁺ + 24e^- --> Mn²⁺ + 4H₂O

2Cl^- --> Cl₂ + 2e^-

2MnO₄^- + 2Cl^- + 8H⁺ + 2e^- --> Mn²⁺ + Cl₂ + 4H₂O

This is where I get stumped. I know that this isn't balanced, but I am uncertain what to do next.

Answer 1

The second step for Mn 1/2 reaction is wrong. Here is the procedure

Ion method

MnO₄^- = Mn²⁺ (Mn balanced - no 2!!!)

MnO₄^- = Mn²⁺+4H₂O (Balance O with water)

8H⁺+MnO₄^- = Mn²⁺+4H₂O (Balance H with H⁺)

5e^-+8H⁺+MnO₄^- = Mn²⁺+4H₂O (Balance charge with e^-)

Oxid. No. Method

MnO₄^- = Mn²⁺ (Mn balanced, oxidation state of Mn goes from +7 to +2)

5e^-+MnO₄^- = Mn²⁺ (Supply electrons for reduction in Ox. States)

5e^-+8H⁺+MnO₄^- = Mn²⁺ (Balance charge with H⁺, This method works better than other method for basic situations as you add OH^- at this point)

5e^-+8H⁺+MnO₄^- = Mn²⁺ +4H₂O (Use water to balance H or O. The other must also balance if you did it right)

That should help you out for the balancing (cancel the electrons for the overall equation by adding the weighted equations properly)