Question
What is the oxidation number of manganese in KMnO4, and how does it function in redox reactions?
Solution
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To determine the oxidation number of manganese (Mn) in potassium permanganate (KMnO_4), we must consider the rules for assigning oxidation numbers and apply them systematically.
Step 1: Assign oxidation numbers to the known elements.
Potassium (K) is an alkali metal and has an oxidation number of +1 in all its compounds. Oxygen (O) typically has an oxidation number of -2, except in peroxides or when bonded to fluorine.
Step 1: Assign oxidation numbers to the known elements.
Potassium (K) is an alkali metal and has an oxidation number of +1 in all its compounds. Oxygen (O) typically has an oxidation number of -2, except in peroxides or when bonded to fluorine.
Potassium (K) is an alkali metal and has an oxidation number of +1 in all its compounds. Oxygen (O) typically has an oxidation number of -2, except in peroxides or when bonded to fluorine.
Step 2: Use the formula of the compound to set up an equation.
The formula for potassium permanganate is KMnO_4. We can set up an equation based on the sum of the oxidation numbers in the compound, which must equal zero because the compound is neutral.
Let the oxidation number of Mn be x. Then, we have:
$$
(+1) + (x) + 4(-2) = 0
$$
Step 3: Solve for the unknown oxidation number.
Solving for x gives us:
$$
1 + x - 8 = 0 \\
x - 7 = 0 \\
x = +7
$$
Therefore, the oxidation number of manganese in KMnO_4 is +7.
In redox reactions, KMnO_4 functions as an oxidizing agent. The manganese in KMnO_4 has a high oxidation state of +7, which means it has a strong tendency to gain electrons and be reduced. When KMnO_4 acts as an oxidizing agent, it will typically be reduced to Mn^2+ (with an oxidation state of +2) or MnO_2 (with manganese having an oxidation state of +4), depending on the pH of the solution.
During this reduction process, the Mn atom in KMnO_4 gains electrons, and the species that donates electrons to Mn is oxidized. This electron transfer is the essence of a redox reaction. The change in the oxidation state of manganese from +7 to a lower value is indicative of the reduction process, while the corresponding oxidation of the other species involved completes the redox cycle.
Step 2: Use the formula of the compound to set up an equation.
The formula for potassium permanganate is KMnO_4. We can set up an equation based on the sum of the oxidation numbers in the compound, which must equal zero because the compound is neutral.
Let the oxidation number of Mn be x. Then, we have:
$$
(+1) + (x) + 4(-2) = 0
$$
Step 3: Solve for the unknown oxidation number.
Solving for x gives us:
$$
1 + x - 8 = 0 \\
x - 7 = 0 \\
x = +7
$$
Therefore, the oxidation number of manganese in KMnO_4 is +7.
In redox reactions, KMnO_4 functions as an oxidizing agent. The manganese in KMnO_4 has a high oxidation state of +7, which means it has a strong tendency to gain electrons and be reduced. When KMnO_4 acts as an oxidizing agent, it will typically be reduced to Mn^2+ (with an oxidation state of +2) or MnO_2 (with manganese having an oxidation state of +4), depending on the pH of the solution.
During this reduction process, the Mn atom in KMnO_4 gains electrons, and the species that donates electrons to Mn is oxidized. This electron transfer is the essence of a redox reaction. The change in the oxidation state of manganese from +7 to a lower value is indicative of the reduction process, while the corresponding oxidation of the other species involved completes the redox cycle.
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