Question

KMnO4 + H2S + H2SO4 -------> KHSO4 + MnSO4 + S + H2O. Balance the equation by hit and trial method with explanation.. ​

Answer 1

Explanation:

H2S + KMnO4 + H2SO4 → S+ MnSO4 + KHSO4 + H2O Balance by oxidation. a) The oxidation number of various atoms involved in the reaction.

Answer 2

Answer:

8KMnO4 + 5H2S + 11H2SO4 → 8KHSO4 + 8MnSO4 + 12H2O

Explanation:

KMnO4 + H2S + H2SO4 = S + + K2SO4 + MnSO4 + H2O

Here,

The Oxidizing agent: KMnO4 or MnO4-1

The Reducing agent: H2S or, S-2

Reduction Half Reaction:

⇒ MnO4 -1 +5e + 8H+ = Mn2- + 4H2O … …. …. …. (1)

Oxidation Half Reaction:

⇒ S-2 – 2e– = S … … … … (2)

Now,

equation (1)x2 + (2)x5,

2MnO4-1 +10e– + 16H+ = 2Mn2- + 8H2O

5S-2 – 10e– = 5S

⇒ 2MnO4 -1 + 5 S-2 + 16H+ = 2Mn2- + 8H2O +5S

⇒ adding necessary ions and radicals we get,

⇒ 2KMnO4 + 5H2S + 3H2SO4 = 5S + K2SO4 + 2MnSO4 + 8H2O

“Answer”

⇒2KMnO4 + 5H2S + 3H2SO4 = 5S +K2SO4 + 2MnSO4 + 8H2O

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