Question

What is the oxidizing agent in the following reaction? H2C2O4(aq) + 2KMnO4(aq) + 6HCl → 2MnCl2(aq) + 2KCl(aq) + 10CO2(g) + 8H2O(l)

Answer 1

H2C2O4(aq) + 2KMnO4(aq) + 6HCl → 2MnCl2(aq) + 2KCl(aq) + 10CO2(g) + 8H2O(l)

oxidizing agents and reducing agents are always on the reactant section so

H2C2O4(aq) + 2KMnO4(aq) + 6HCl →

now,

identify the oxidation numbers

H2C2O4

(2)(+1) + (2)(X) + 4(-2) = 0

x = +3; that is the charge of C

for KMnO4

KMnO4

(1)(+1) + (1)(M) + 4*(-2) =

M = +7

now... in the products

note that MnCl2 is formed

(1)M + (2)(-1) = +2

Manganese goes from +7 to +2; it is being reduced, therefore; it is the OXIDIZING AGENT

and makes sense since KMnO4 has plenty of oxygens