Complete and balance the equation for this reaction in basic solution? Redox reaction Cr is oxidized to CrO42– and Fe3 is reduced to Fe2...?
For a particular redox reaction Cr is oxidized to CrO42– and Fe3 is reduced to Fe2 . Complete and balance the equation for this reaction in basic solution. Phases are optional.
Cr+Fe{3+}--->CrO4{2-}+Fe{2+}
I understand how to solve a problem like this in an acidic solution. Add H+ and H2O to balance Oxygen and Hydrogen and finally add electrons and then add the equations together. I just don't understand how to do it in a basic solution. Adding the OH always confuses me because I'm adding oxygen and hydrogen at the same time and then when I go to add H2O I just get confused. Any help is really appreciated!
So in this method you don't use electrons to balance the charge you use either H+ or OH-?
Oh, balancing redox reactions in basic solutions can be a little tricky, but don't worry, I'll clown around and make it fun for you!
Let's start by balancing the atoms other than oxygen and hydrogen. We have:
Cr + Fe{3+} → CrO4{2-} + Fe{2+}
Now, let's balance the oxygens by adding water (H2O):
Cr + Fe{3+} → CrO4{2-} + Fe{2+} + H2O
Next, let's balance the hydrogens by adding hydroxide ions (OH-) to both sides. Remember, for every OH- you add, it will consume one H2O molecule on the other side:
Cr + Fe{3+} + OH- → CrO4{2-} + Fe{2+} + H2O
Great! Now, let's tackle the charge balance. On the left side, we have a total charge of +3 (from the Fe{3+}). To balance that, we will need three more OH- ions on the left side:
Cr + Fe{3+} + 3OH- → CrO4{2-} + Fe{2+} + H2O
Finally, let's check if everything is balanced. We have:
Chromium: 1 Cr on both sides
Oxygen: 4 on both sides
Hydrogen: 4 on both sides
Charge: +3 on both sides
And there you have it! The balanced equation in basic solution is:
Cr + Fe{3+} + 3OH- → CrO4{2-} + Fe{2+} + H2O
Hope I brought a smile to your face while balancing that equation! Let me know if you need any more clowning around or assistance.
To balance a redox reaction in basic solution, you can follow these steps:
1. Balance the atoms (except hydrogen and oxygen) in the equation as you normally would in an acidic solution. In this case, the atoms are already balanced:
Cr + Fe{3+} ---> CrO4{2-} + Fe{2+}
2. Balance the oxygen atoms by adding water (H2O) molecules to the appropriate side of the equation. In this case, there are 4 oxygen atoms on the right side, so you need to add 4 water molecules:
Cr + Fe{3+} ---> CrO4{2-} + 4 H2O + Fe{2+}
3. Balance the hydrogen atoms by adding equal numbers of OH- ions to the opposite side of the equation. Each OH- ion supplies one hydrogen atom. In this case, there are 8 hydrogen atoms on the left side, so you need to add 8 OH- ions to the left side:
Cr + Fe{3+} + 8 OH- ---> CrO4{2-} + 4 H2O + Fe{2+}
4. Balance the charge on both sides of the equation. In this case, the overall charge is already balanced.
The final balanced equation in basic solution is:
Cr + 3 Fe{3+} + 8 OH- ---> CrO4{2-} + 4 H2O + 3 Fe{2+}
Remember, when balancing redox reactions in basic solution, you can add OH- ions to balance the hydrogen atoms, but you need to make sure to balance the equation by adding the same number of OH- ions to both sides.
I don't do it that way. That way is the quickest and easiest, so I'm told, and the way I do it long and the hardest way BUT I do it. Perhaps this will help you.
1. Divide into the two half equations.
2. Preliminarily balance the atoms that are changing; i.e., for Cr2O7^2- to Cr^3+ you will want it to read Cr2O7^2- --> 2Cr^3+ so you are balancing apples with apples and not apples with oranges.
3. Cr^3+ ==> CrO4^2-
Determine oxidations states. Cr^3+ is 3+ and Cr in CrO4^2- is 6+.
4. Add electrons to the appropriate side to balance the change in oxidn state.
Cr^3+ ==> CrO4^2- + 3e
5. Determine charge on both sides and
a. If acid solution add H^+ to balance charge
b. If basic solution add OH^- to balance charge.
Charge is +3 on left; -5 on right. Add 8 OH^- to left to read
8OH^- + Cr^3+ ==> CrO4^2- + 3e
6. Now add H2O to the appropriate side to balance. That will read
8OH^- + Cr^3+ ==> CrO4^2- + 3e + 4H2O
7. Check it for three things.
a. atoms balance. yes. 8O, 8H, 1Cr each side.
b. charge balance. yes -5 each side
c. Oxdn state change balances. yes. Cr goes from 3+ to 6+ and that's a loss of 3e for Cr left to right.
Happy balancing.