Problem Assignment:
1. Determine the oxidation state (also called the oxidation number) of the atoms in the
following species:
a. perchlorate (ClO4–), chlorate (ClO3–), chlorite (ClO2–), and hypochlorite (ClO–) ions
b. the peroxide ion (O22–)
c. chromate (CrO42–) and dichromate (Cr2O72–) ions
d. permanganate (MnO4–) and manganate (MnO42–) ions
e. carbonate (CO32–) and phosphate (PO43–) ions
f. the cyanide (CN–) ion
g. sodium nitrite (NaNO2) and sodium nitrate (NaNO3)
h. sodium sulfate (Na2SO4) and sodium sulfite (Na2SO3) ions
i. ammonium chloride (NH4Cl)
2. For each of the following reaction equations, first confirm that the equations are balanced. Then
determine the oxidation state of the elements in all compounds. Identify the element being reduced
and the element being oxidized, and explain why the stoichiometry must be that given.
Example:
Cr2O72– + 6 Cl– + 14 H+
3 Cl2 + 2 Cr3+ + 7 H2O
Balance: 2 Cr, 7 O, 6 Cl, 14 H, 6+ charge = 6 Cl, 2 Cr, 14 H, 7 O, 6+ charge CHECK
Oxidation States: Cr(+6),O(–2) Cl(–1) H(+1)
Cl(0)
Cr(+3)
H(+1),O(–2)
Cr is being reduced from +6 to +3; Cl is being oxidized from –1 to 0.
It must be six moles of Cl– for every mole of Cr2O72– because there are two moles of Cr(+6) going to
2 moles of Cr(+3), requiring 6 moles of electrons. Those electrons come from 6 moles of Cl(–1)
going to Cl(0).
a.
Cr2O72– + 3 HNO2 + 5 H+
b.
3 Cu + 8 HNO3
c.
2 MnO4– + 3 CN– + H2O
2 Cr3+ + 3 NO3– + 4 H2O
3 Cu(NO3)2 + 2 NO + 4 H2O
2 MnO2 + 2 OH– + 3 OCN–