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What is the value of n in the following ...

What is the value of n in the following half equation, `Cr(OH)_(4)^(-) + OH^(-) rarrCrO_(4)^(2-)+H_(2)O+"ne"^(-)`

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The correct Answer is:
3
`Cr(OH)_(4)+OH^(-)rarrr CrO_(4)^(2-)+H_2O`
Write oxidation states, we have .
`[overset(+3)Cr(OH)_(4)]^(-)+OH^(-) rarr[overset(+6)CrO_(4)]^(2-)+H_(2)O`
The valance oxidation state of Cr on both sides , add `3e^(-)` on R.H.S .
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