What is the oxidation state of a free element? Of a monoatomic ion?

1 Answer
Sep 3, 2016

Answer:

The oxidation state of a free element is #0#. The neutral element has neither gained nor donated electrons.

Explanation:

Redox reactions are considered to involve the formal TRANSFER of electrons. Oxidation involves the #"LOSS of Electrons"#. Reduction involves the #"GAIN of Electrons"#. You know the old story #"LEO says GER"#, #"Loss of Electrons, Oxidation; Gain of electrons, Reduction"#

When carbon reacts wth dioxygen this is certainly a formal redox process:

#C(s) + O_2(g) rarr CO_2(g)#

Both reactants are ZEROVALENT, i.e. a #0# oxidation state. During the reaction, #C# loses 4 electrons to give #C^(IV+)#, and oxygen gains 2 electrons to give #O^(-II)#.

Alternatively we could represent the oxidation of elemental iodine to iodate, #IO_3^-#:

#1/2I_2 + 3H_2O rarr IO_3^(-) +5e^(-) +6H^+#

Both mass and charge are balanced as is required.

For the monoatomic ion, the charge on the ion is simply the oxidation number:

#CararrCa^(2+) + 2e^-#

#1/2I_2 +e^(-) rarr I^(-)#

Again this is rationalized in terms of electron transfer.