Except for the electrons you have balanced the ClO^- half equation.
1. Cl goes from +1 to -1 so add electrons.
ClO^- + 2e ==> Cl^-
2. Count the charge. -3 on left; -1 on right. Add OH^-
ClO^- + 2e ==> Cl^- + 2OH^-
3. Add water
ClO^- + 2e + H2O ==> Cl^- + 2OH^-
Check it. It's balanced.
For the Cr you are worrying about something that is a non-issue. One reason for following the procedure I gave you is that it takes care of the O and H for us.
1. Cr(OH)4^- ==> CrO4^2-
Cr changes from +3 on the left to +6 on the right. Add electrons.
Cr(OH)4^- ==> CrO4^2- + 3e
2. Count the charge. Charge on left is -1 and on the right is -5. Add OH^-
Cr(OH)4^- + 4OH^- ==> CrO4^2- + 3e
3. Add H2O
Cr(OH)4^- + 4OH^- ==> CrO4^2- + 3e + 4H2O
Balance the redox reactin in basic.
ClO- + Cr(OH)4- --> CrO42- + Cl-
H2O + ClO- --> Cl- + 2OH
ClO- has a +1 O# and Cl- has a -1 O# how do I balance that. And is what Ive done so far correct?
Cr(OH)4- --> CrO42-
Im confused on how to balance the oxygen and hydrogens since one side has both and the other just has Oxygen.
4 answers
O# of Cr(OH)4-
How did you get +3
How did you get +3
Each OH is -1. 4 of them gives -4. Cr must be +3 to leave a -1 charge on the ion. OR, if you want to do the hard way.
O = -2 x 4 -8.
H = +1 x 4 +4
-8+4+? = -1
? = -1+8-4 = +3
O = -2 x 4 -8.
H = +1 x 4 +4
-8+4+? = -1
? = -1+8-4 = +3
-7
1 answer