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3 years ago

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What is the oxidation number of manganese (Mn) in potassium permanganate (KMnO4)?

2 answers:

irina [24]3 years ago

8 0

Answer:

+7

Explanation:

Within every molecule, there is always an oxidation number assigned to every individual atom within the molecule. Hydrogen will always have an oxidation number of +1 (because it only has one electron to give, and when it gives that electron, it has a charge of +1).

Elements in group 1 always have an oxidation number of +1, and elements in group 16 always have an oxidation number of -2. Using this information we can calculate the oxidation number of Mn in the molecule KMnO₄.

Overall the molecule is neutral, so all the oxidation numbers would add up to zero:

K + Mn + O₄ = 0

(+1) + Mn + 4(-2) = 0

(+1) + Mn + (-8) = 0

Mn + (-7) = 0

Mn = +7

andrezito [222]3 years ago

7 0

What is the oxidation number of manganese (Mn) in potassium permanganate (KMnO4)?

+3
+5
+6
+7

The oxidation number of manganese (Mn) in potassium permanganate (KMnO4) is +5.

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Read 2 more answers

Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. (Assume standard conditions).

babunello [35]

Answer:

-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

Oxidation half equation;

Sn(s) ------> Sn^2+(aq) + 2e

Reduction half equation:

Mn^2+(aq) + 2e ----> Mn(s)

Cell voltage= E°cathode - E°anode

E°cathode= -1.19V

E°anode= -0.14 V

Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V

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3 years ago