How do we represent the oxidation of #Cr^(3+)# ion to #CrO_4^(2-)# by hydrogen peroxide, using the method of half-equations?

Question

Wyatt Adkins · Accepted Answer

By the method of half-equations............

#2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+) # #"Chromic chloride"#, #CrCl_3#, is oxidized from #Cr(+III)# to #Cr(+VI)# in chromate: #Cr^(3+) + 4H_2O rarr stackrel(VI+)(Cr)O_4^(2-) +8H^(+)+ 3e^(-) # #(i)# It is your problem, not mine, to determine whether this is balanced in terms of mass and charge. Hydrogen peroxide, #H_2O_2#, is reduced from #O^(-I)# to #O^(-II)#. #H_2O_2 +2H^(+) + 2e^(-) rarr 2H_2O# #(ii)# Is everything balanced once more? To give the overall redox equation, we cross-multiply the individual redox equations: #2xx(i) + 3xx(ii)#: #2Cr^(3+) + cancel(8)2H_2O +3H_2O_2 +cancel(6H^(+)) + cancel(6e^(-)) rarr cancel(6H_2O)+2CrO_4^(2-) +cancel(16)10H^(+)+ cancel(6e^(-)) # Next, we DEDUCE the unnecessary reagents from both sides: #2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+) # But you wanted the reaction done under basic conditions. And so we add #10xxHO^(-)# to BOTH SIDES OF THE REACTION: #2Cr^(3+) +3H_2O_2 +10HO^(-) rarr 2CrO_4^(2-) +8H_2O # Is the mass to charge ratio of this balanced?

Naomi Aronson · Answer

undefined Here are the half-equations for the oxidation of Cr^(3+) ion to CrO_4^(2-) by hydrogen peroxide:

**Oxidation half-equation:**  
Cr^(3+) → CrO_4^(2-) + 3e^-

**Reduction half-equation:**  
H2O2 + 2H^+ + 2e^- → 2H2O