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.
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
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
-7
To balance the redox reaction in basic solution, you need to follow these steps:
Step 1: Separate the reaction into half-reactions:
The given reaction can be split into two half-reactions:
1. Oxidation half-reaction: ClO- --> Cl- (where Cl goes from +1 to -1)
2. Reduction half-reaction: Cr(OH)4- --> CrO42- (where Cr goes from +6 to +3)
Step 2: Balance the non-hydrogen and non-oxygen atoms:
Starting with the oxidation half-reaction, we see that there is only one Cl on each side, so the non-hydrogen and non-oxygen atoms are already balanced in this half-reaction.
ClO- --> Cl-
Step 3: Balance the oxygen atoms:
In the oxidation half-reaction, the number of oxygen atoms changes from 1 on the left side to 0 on the right side. To balance the oxygen atoms, we need to add water (H2O) to the right side of the equation. Since there is one oxygen atom in ClO-, we need to add one water molecule (H2O):
ClO- + H2O --> Cl-
Step 4: Balance the hydrogen atoms:
In the oxidation half-reaction, there are no hydrogen atoms. Therefore, we do not need to balance hydrogen in this half-reaction.
Step 5: Balance the charge:
In the oxidation half-reaction, the total charge on the left side is -1 (from ClO-), and the total charge on the right side is -1 (from Cl-). The charges are already balanced in this half-reaction.
Now, let's move on to the reduction half-reaction:
Step 6: Balance the non-hydrogen and non-oxygen atoms:
Starting with the reduction half-reaction, we see that there is one Cr on each side, so the non-hydrogen and non-oxygen atoms are already balanced in this half-reaction.
Cr(OH)4- --> CrO42-
Step 7: Balance the oxygen atoms:
In the reduction half-reaction, the number of oxygen atoms changes from 4 on the left side to 2 on the right side. To balance the oxygen atoms, we need to add water (H2O) to the right side of the equation. Since there are 4 oxygen atoms in Cr(OH)4-, we need to add 2 water molecules (H2O):
Cr(OH)4- --> CrO42- + 2H2O
Step 8: Balance the hydrogen atoms:
In the reduction half-reaction, there are 4 hydrogen atoms on the left side (from the hydroxide ions in Cr(OH)4-). To balance the hydrogen atoms, we need to add hydroxide ions (OH-) to the left side of the equation. Since there are 4 hydrogen atoms, we need to add 4 hydroxide ions:
Cr(OH)4- + 4OH- --> CrO42- + 2H2O
Step 9: Balance the charge:
In the reduction half-reaction, the total charge on the left side is -4 (from Cr(OH)4- and OH-), and the total charge on the right side is -2 (from CrO42-). To balance the charges, we need to balance the electrons. Since the net charge decreases by 2, we need to add 2 electrons (e-) to the left side of the equation:
Cr(OH)4- + 4OH- + 2e- --> CrO42- + 2H2O
Step 10: Combine the half-reactions:
To combine the two half-reactions, we need to make the number of electrons in both reactions equal. In the oxidation half-reaction, there are no electrons. Therefore, we need to multiply the reduction half-reaction by 2 to equalize the number of electrons:
2[Cr(OH)4- + 4OH- + 2e- --> CrO42- + 2H2O]
Now, let's add the two half-reactions together:
ClO- + H2O --> Cl- + 2OH-
2[Cr(OH)4- + 4OH- + 2e- --> CrO42- + 2H2O]
After combining, cancel out the common species on both sides:
ClO- + 3H2O + 2[Cr(OH)4- + 4OH- + 2e- --> CrO42- + 2H2O]
ClO- + 3H2O + 2Cr(OH)4- + 8OH- + 4e- --> Cl- + 2OH- + 2CrO42- + 4H2O
Finally, simplify the equation:
ClO- + 3H2O + 2Cr(OH)4- --> Cl- + 2OH- + 2CrO42- + 4H2O
So the balanced redox reaction in basic solution is:
ClO- + 3H2O + 2Cr(OH)4- --> Cl- + 2OH- + 2CrO42- + 4H2O
O# of Cr(OH)4-
How did you get +3