One uses standard solution of potassium dichromate (K2Cr2O7) to determine the percent by weight of iron (as Fe2+) in an unknown solid.
Dichromate ion reduces to two chromium(III) ions. This reaction requires 6 electrons and 14 (!) hydrogen ions:
Cr2O72-+ 14H+ + 6 e- ® 2Cr3+ + 7H2O
Only one electron is necessary to reduce Fe(III) to Fe(II)
Fe3+ + e- ® Fe2+
Therefore, 1 mole of Cr2O72-(the oxidizing agent) reacts with 6 moles of Fe2+ (the reducing agent) to form 6 moles of Fe3+ and 2 moles of Cr3+. Thus, in net ionic form:
Cr2O72-+ 6Fe2+ + 14H+ ® 6Fe3+ + 2Cr3+ + 7H2O
The molecular form ofthe reaction equation can be written as:
K2Cr2O7 + 6Fe(NH4)2(SO4)2 + 7H2SO4 ®
3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 6(NH4)2SO4 + 7H2O
The 1:6 mole ratio with respect to the amounts of Cr2O72-and Fe2+ consumed.