Half-reaction equation basic form?

CN(-) + CrO4(2-) ---> CNO(-) + Cr(OH)4(-)

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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