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

What is the value of n in the following half equation, Cr(OH)_(4)^(-)+

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