Thank you!! Please show each step as you do it!!
Ce(Io3)4+H2C2O4= Ce2(C204)3+I2+H2O+Co2
2007-01-08 05:21:01 · 1 answers · asked by laur664664 1 in Science & Mathematics ➔ Chemistry
The equation is balanced by using oxidation and reduction techniques.
2Ce(IO3)4+24H2C2O4= Ce2(C204)3+4I2+24H2O+42CO2
Reduction reactions
2Ce+4 + 2e ===> 2Ce+3
8 I+5 = 40e ===> 4I2
(Total of 42 electrons needed for reduction)
Oxidation reaction
{2C+3 (from oxalate) ===>2C+4 + 2e}
Multiply above rxn by 21 to obtain an equal number of electrons in oxidation as in reduction.
21x{2C+3 (from oxalate) ===>2C+4 + 2e}
Add the 2 half reactions together
2Ce+4 + 8I+5 + 42e + 21(2C+3) ===> 2Ce+3 + 4I2 +42C+4 + 42e
cancel the 42 e on each side and then balance the rest by inspection. Notice that you will have to add 3 to the number of oxalates on the left (21 +3) to include those that do not change and are part of Cerium oxalate in the products.
Be sure to recount all atoms.
2007-01-08 06:53:35 · answer #1 · answered by docrider28 4 · 0⤊ 0⤋