Question

expand using identities mn +3 p)²​

Answer 1

Answer:

Well, chromate, C(+VI), is reduced to Cr(+III):

CrO2−4+8H++3e−→Cr3++4H2O(l)

We add EIGHT hydroxides to each side (clearly the reaction is conducted under basic conditions…)

CrO2−4+4H2O+3e−→Cr3++8HO−

And ferrous ion is oxidized to ferric ion…

Fe2+→Fe3++e−

We add three of the latter to the latter…

3Fe2++CrO2−4+4H2O+3e−→3Fe3++3e−+Cr3++8HO−

And cancel away…

3Fe2++CrO2−4+4H2O→3Fe3++Cr3++8HO−