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

1 Answer
Apr 8, 2017

Answer:

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

#2Cr^(3+) + 2H_2O +3H_2O_2 rarr 2CrO_4^(2-) +10H^(+) #

Explanation:

#"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)#

Is this balanced with respect to mass and charge? It is your problem not mine.

Hydrogen peroxide, #H_2O_2#, is reduced from #O^(-I)# to #O^(-II)#.

#H_2O_2 +2H^(+) + 2e^(-) rarr 2H_2O# #(ii)#

Again is the thing balanced?

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^(-)) #

And then we SUBTRACT the superfluous reagents from each side:

#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 this balanced with respect to mass and charge?