Plz Help!!! How would you balance this redox equation?

Question

Balance the following redox equations:

CrO4 2- Fe2 => Cr3 Fe3 (in acidic solution)

MnO4- ClO2- => MnO2 ClO4- (in basic solution)

Answer 1

he is right on the first one... here is the second one:

MnO4-  & 3 electrons  => MnO2  (when Mn goes from +7 --> +4)

ClO2- => ClO4- & 4 electrons     (when Cl goes from +3 to +7)

so , to balance the electrons taken & lost we :

4 MnO4-  & 12 electrons  => 4 MnO2

3 ClO2- => 3 ClO4- & 12 electrons  

now we combine them:

4 (MnO4)- & 3 (ClO2)- => 4 MnO2 &  3 (ClO4)-

we have a total of charges of -7 on the left , & a 3- on the right... we will add 4 OH- 's to balance charges:

4 (MnO4)- & 3 (ClO2)-   => 4 MnO2 &  3 (ClO4)- & & 4 (OH)-

we have 22 oxygens on the left , & we have 24 on the right, we will add 2H20's to balance the oxygens:

4 (MnO4)- & 3 (ClO2)-  & 2 H2O  => 4 MnO2 &  3 (ClO4)- & 4 OH-

that's your answer

======================

everything balances:

12 electrons taken = 12 electrons lost

7 - charges =  7- charges

24 oxygens = 24 oxygens

4 H, 4 Mn , 3 Cl  = 4 H, 4 Mn , 3 Cl

Answer 2

I will do each one step by step

CrO4 2-  ---->  Cr 3+

CrO4 2- ----> Cr3+ + 4H2O

CrO42-  + 8H+  ---->  Cr3+  +  4H2O

CrO42-  +  8H+  +  3e-  ------>   Cr3+ + 4H2O

Fe2+  ---->  Fe3+

Fe2+  -----> Fe3+  +  e-

3Fe2+  ----->  3Fe3+  +  3e-

Combine the last two equations

CrO42-  +  8H+  +  3e-  ------>   Cr3+ + 4H2O

3Fe2+  ----->  3Fe3+  +  3e-

Cross off the e- to give you the complete redox rxn.

CrO42- + 8H+ 3Fe2+ ----> Cr3+ + 4H2O + 3Fe3+

Check that all atoms balance and charges balance

Cr...check

O.....check

H.....check

Fe.....check

charge....check

That's the answer....