What is the oxidation number change for the manganese atom in the following unbalanced reduction half reaction: MnO4 - (aq) + H + (aq) → Mn2 + (aq) + H2O (l) ? Mn goes from + 4 to + 2 Mn goes from + 3 to + 2 Mn goes from + 7 to + 2 Mn goes from + 8 to + 2

Question

Terrence Ross

Chemistry

6 December, 03:03

What is the oxidation number change for the manganese atom in the following unbalanced reduction half reaction: MnO4 - (aq) + H + (aq) → Mn2 + (aq) + H2O (l) ? Mn goes from + 4 to + 2 Mn goes from + 3 to + 2 Mn goes from + 7 to + 2 Mn goes from + 8 to + 2

+4

Answers (1)

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Dillon · Answer 1

The correct answer is Mn goes from + 7 to + 2.

Explanation:

The permanganate ion (Mn04-) forming by Oxygen (-2 its oxidation state) and the Mn (+7), throwing the count: + 7 + (-2 x4) = - 1 (the charge of the ion). Said permanganate ion is reduced in the reaction to Mn2 +.