Each has an oxidation number of O. Determine the oxidation number of O and C in

Question

Each has an oxidation number of O. Determine the oxidation number of O and C in CO_2. Determine the oxidation number of O and carbon in CO_2. Do the same for H_2O. Summarize your results in the statements given below. H in CH_4 has oxidation number = C in CH_4 has oxidation number = O in O_2 has oxidation number = C in CO_2 has oxidation number = O in CO_2 has oxidation number = O in H_2 O has oxidation number = H in H_2 O has oxidation number Now look at the blanks connected together by arrows and determine whether that element has been oxidized or reduced. Fill in your results. O undergoes because it changes its oxidation number from to. C undergoes because it changes its oxidation number from to. A list of rules for assigning oxidation numbers is given below. Learn these rules. O usually has an oxidation number of -2 except in peroxides like H_2O_2 where it is -1 H usually has an oxidation number of +1 except hydrides like LiH where it is -1. F always has an oxidation number of -1. Free elements, elements not combined with other elements have oxidation numbers of O. Examples of free elements Fe, O_2, H_2, Na. The sum of the oxidation numbers of elements in a neutral compound is 0. The sum of the oxidation numbers of elements in an ion equals the charge on the ion. For monatomic ions the oxidation number equals the charge on the ion. For example, a sodium ion Na^+ has an oxidation number equal to +1. A chloride ion, cl has an oxidation number equal to -1. Use these rules to assign oxidation numbers to the atoms in the following molecules or ions. Write the oxidation numbers above each of the atoms. An example, H_2SO_4 is shown.

Explanation / Answer

molecule oxidation state of C H O other element

CH4 -4 +1

H2O +1 -2

CO2 +4 -2

H2SO4 +1 -2 S (+6)

HClO3 +1 -2 Cl(+5)

HClO4 +1 -2 Cl(+7)

PO4-3   -2 P (+5)

C in CH4 is oxidised from -4 oxidation state to +4 oxidation state in CO2.

O is reduced from 0 in O2 to -2 in CO2.

The reaction is a redox reaction

4Fe(s) + 3O2 (g) ------> 2Fe2O3(s)

Fe (s) has oxidation state of 0

O2 (g) has oxidation state of 0

in Fe2O3 the oxidation state of Fe is +3 and that of O is -2.

Thus Fe is oxidised from 0 state to +3 oxidation state.

And oxygen is reduced from 0 to -2 oxidation state.

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