Complete balancing the equation for the oxidation of toluene with potassium permanganate in an acidic medium.

Question

Complete balancing the equation for the oxidation of toluene with potassium permanganate in an acidic medium.

(Notes: Use the half-cell method to balance the redox equation. The potassium ion is a spectator ion. The permanganate ion, MnO₄ ^- , is reduced to Mn²⁺.)

 

+ MnO₄^- -> + Mn²⁺

1. Start by substituting R for the portion of the organic molecules that do not change. Toluene becomes R-H₃ and benzoic acid HO-R=O.  

2. Then assign redox numbers to each of the atoms. H = +1, O = -2 The overall molecule is neutral, so the value of R is calculated.

     R-H₃       H = +1 x 3 = +3, so R +3 = 0 or R = -3

     HO-R=O    H = +1, O = -2 x 2 = -4, so R +1 + 2 (-2) = 0 or R = +3

   A single atom ion's redox number = valence electron charge. (e.g., Mn²⁺ = +2)

   A multiple atom ions overall value = valence electron charge. MnO4^-   Mn + 4 (-2) = -1 or Mn = +7

3. Then apply the half cell method. Two applicable mnemonics are OIL RIG (oxidation is losing electrons; reduction is gaining electrons) or LEO says GER (losing electrons is oxidation; gaining electrons is reduction).

Reduction half step:  (Gaining electrons = Mn +7-> +2) Balance the oxygens with water and the hydrogens with protons. (Determine the numbers that go in the blanks. If zero, record as the number zero "0".) Then balance the oxidation numbers using electrons.

   _____ MnO₄ ^- + _____H⁺ + _____e^- --> _____Mn²⁺ + _____H₂O 

Answer 1

MnO₄^- → Mn²⁺ (Balance 0s with water) (Actual first step is to balance the Mn - this is already fine)

MnO₄^- → Mn²⁺ + 4H₂O (Balance the H with H⁺)

MnO₄^- + 8H⁺ → Mn²⁺ + 4H₂O (Now balance charges with e^-)

5e^- + MnO₄^- + 8H⁺ → Mn²⁺ + 4H₂O (Tada!)

This is not the method that I prefer because it is not so great for basic media. I use the oxidation state method in which the steps are

1) supply electrons required to change oxidation state (In this case Mn in 7 state to 2 state requires 5 electrons)

2) balance charge with the appropriate charged ion: H⁺ for acidic and OH^- for basic

3) use H₂O to balance H or O and the other balance can be used as a check that you've done things correctly.

Please consider a tutor. Take care.

Answer 2

The permanganate oxidizes the aliphatic methyl group to a carboxylic acid while the permanganate is reduced to Mn^2+ in acidic solution

Oxidation RH3 -----> O=R-OH + 6 e- the R groups here is the C6H5-C- group

1 balance O with H₂O

RH₃ + 2 H₂O -----> O=R-OH + 6 e-

2 balance H with H⁺

RH₃ + 2 H₂O -----> O=R-OH + 6 H⁺ + 6 e- this is the balanced oxidation half rxn

Reduction

MnO₄^- + 5 e- ------> Mn²⁺

1 balance O with H2O

MnO₄^- + 5 e- -------> Mn²⁺ + 4 H₂O

2 balance H with H⁺

MnO₄^- + 5 e- + 8 H⁺ -------> Mn²⁺ + 4 H₂O this is the balanced reduction half rxn

Now you need to balance the number of electrons. since you have 6 e- in the oxidation and 5 e- in the reduction the smallest common number is 30. So multiply the oxidation by 5 to get 30 e- and multiply the reduction by 6 to get 30 e-, then add those two reaction together (the 30 e- are on both sides and can be dropped from the final equation).

5 RH₃ + 10 H₂O+ 6 MnO₄^- + 48 H⁺ -----> 5 O=R-OH + 30 H⁺ + 6 Mn²⁺ + 24 H₂O

Now we eliminate 10 H₂O and 30 H⁺ from both sides of the reaction to get the final answer.

5 RH₃ + 6 MnO₄^- + 18 H⁺ -----> 5 O=R-OH + 6 Mn²⁺ + 14 H₂O for your final overall balanced redox equation

It's always a good idea at the end to check to be sure that the overall charge of the reactants (-6 + 18 = +12) is equal to the overall charge of the products (6 * +2 = +12). Jerome