# Chemistry

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!

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1. 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.

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2. So in this method you don't use electrons to balance the charge you use either H+ or OH-?

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