Hola friends: Which of the following is a stronger oxidising agent: MnO₄⁻ S₂O₈²⁻ and how? No Spamming
Answers
We know that In KMnO4 the Mn charge =
+7,
On writing the reaction in acidic , neutral and basic medium.
(i)So reaction occurs in acidic medium :
MnO4- + 8 H* + 5 e ----- Mn?+ + 4 H;0
As you see here , Mn gives up 5 e. (Making this as a strong oxidizing agent)
(ii) So reaction occurs in Neutral medium :
Mno,- + 4H* +3 e MnO, + 2H2O
As you can see here, In this neutral solution, it gives up 3 e in formation of MnO2 & also some OH ions.
(ii) So reaction occurs in Basic medium :
MnO4- + e - MnO42-
As you can see here, it only gives up 1 e, So it makes it weaker oxidizing agent.
Hence we can simply observe here :-
MnO4- ion + Acidic solution > Neutral solution > Basic solution.
Hence it is strong oxidising agent in
acidic medium.
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Answer:
Explanation:
We know that In KMnO₄ the Mn charge = +7,
On writing the reaction in acidic , neutral and basic medium.
(i)So reaction occurs in acidic medium :-
_______________________
MnO₄⁻ + 8 H⁺ + 5 e⁻ ----→ Mn²⁺ + 4H₂O
_______________________
As you see here , Mn gives up 5 e⁻ . (Making this as a strong oxidizing agent)
(ii) So reaction occurs in Neutral medium :-
_____________________
MnO₄⁻ + 4H⁺ + 3 e⁻ ------→ MnO₂ + 2H₂O
_____________________
As you can see here , In this neutral solution, it gives up 3 e⁻ in formation of MnO₂ & also some OH ions.
(ii) So reaction occurs in Basic medium :-
_____________________
MnO₄⁻ + e⁻ -----→ MnO₄²⁻
_____________________
As you can see here, it only gives up 1 e⁻ , So it makes it weaker oxidizing agent.
Hence we can simply observe here :-
MnO₄⁻ ion → Acidic solution > Neutral solution > Basic solution.
Hence it is strong oxidising agent in acidic medium