What is the oxidation number of the metal in #MnO_4^(-)#, and how do we represent its reduction to #Mn^(2+)#?

3 Answers

Answer:

See below.

Explanation:

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The oxidation number of #"Mn"# is #+7#.

Jul 31, 2017

Answer:

#Mn(VII+)# is typically reduced to #Mn(II+)#......

Explanation:

This is formally a five electron reduction, and it is a useful one that while #MnO_4^(-)# is deep purple in colour........#Mn^(2+)#, a #d^5# system, is almost colourless. The reduction reaction is thus self-indiciating......and so we have to account for a 5 electron reduction....

#MnO_4^(-) + 8H^+ + 5e^(-) rarr Mn^(2+) + 4H_2O#

Mass and charge are balanced, as required..........

Jul 31, 2017

Answer:

The oxidation number of #"Mn"# is #+7#.

Explanation:

Manganese #("Mn")# is a multivalent element, meaning it can have multiple charges in its ionic state. We are trying to figure out the charge (oxidation number) of #"Mn"# in this case.

First, let's start with what we know. You want to see what elements have a definitive (single) charge. In this case, potassium #("K")# and oxygen #("O")# have definitive charges. Potassium is #+1#. Oxygen is #-2#.

Second, look at the charge of the overall compound. If it is charged, you should see a #n+# or #n-# (#n# can be any positive integer). If it is not charged, then the compound is neutral. This will be the case for #"KMnO"_4#.

Third, set up your equation. On one side, you will have the elements with their charges. On the other side, you will have the overall charge. In this case, the equation is

#(1)"K" + Z * "Mn" + (-2)"O" = 0#

Where I have the element symbols is where you will put how many of that element there are in the compound. I use #Z# to denote unknown oxidation numbers.

Last, solve for the element in question. In this case, #"Mn"#.

#(1) * "K" + Z * "Mn" + (-2)"O" = 0#

#1 * 1 + Z(1) + (-2) * 4 = 0#

#1 + Z(1) - 8 = 0#

#Z(1) = 7#

#Z = 7#

For #"Mn"#, the oxidation number is #+7#.