Half-reaction equation basic form?
Answer:
You need to identify the oxidation and reduction.
For instance, Cr+6 is changing to Cr+3. This is the reduction:
3e- + 4H+(aq) + CrO4^(-2)(aq) ---> Cr(OH)4^(-)(aq)
The oxidation is a little trickier:
H2O(l) + CN-(aq) ---> CNO-(aq) + 2e- + 2H+(aq)
You must multiply by a common factor to get 6e-:
2H+(aq) + 2CrO4^(-2)(aq) + 3H2O(l) + 3CN-(aq) --->
3CNO-(aq) +2Cr(OH)4^-(aq)
I guarantee that this reaction is "in base" in which we would add 2OH- to both sides to give 5H2O on the left and 2OH- on the right.
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