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# Half equation redox help

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1. NiO2+2H2O+Fe --> Ni(OH)2+Fe(OH)2 (in basic solution)

No need to balance it, the equation is already balanced.

So then....

I got the half equations.

OXIDATION

Fe ----> Fe(OH)2 (Net change of +2)

Reduction

NiO2 +2H2O ----> Ni(OH)2 (Net change of -2)

Now time to balance the oxygen atoms?

So I thought:

Fe +H2O ----> Fe(OH)2 (Because I thought the LHS was the oxygen deficient side).

I figured that there was no need to add water for the reduction reaction because the 2H2O resulted in 4 H and 2 O on LHS, and there was 2 O and 2 H on RHS. So, the oxygens were balanced, the hydrogens were not.

Or would the fact that water was already present, make the RHS the oxygen defecient side ?
2. The oxidation will require 2H2O to balance the O. There is then a need to add H+ to the right to balance the H. Then you'll add e- to balance the charges.

Don't start with H2O in the reduction equation, The O is already balanced. Again, there is a need to add H+, this time to the left and again, adding e- to balance the charges finishes that equation.
3. (Original post by Pigster)
The oxidation will require 2H2O to balance the O. There is then a need to add H+ to the right to balance the H. Then you'll add e- to balance the charges.

Don't start with H2O in the reduction equation, The O is already balanced. Again, there is a need to add H+, this time to the left and again, adding e- to balance the charges finishes that equation.
You can't add H+ as it's done in basic medium.

4. Whoops, I didn't see that.

In that case, I've explained how to convert to alkaline half equations in another thread - perhaps the same one?

Nope. It turns out it was another one.

Oh well.

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