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 ?
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.
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.