What is the value of $n$ in the following equation?
$C r {(O H)}_{4}^{-} + O H^{-} \to C r O_{4}^{2 -} + H_{2} O + n e^{-}$

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Solution

The correct option is D 6
${C r (O H)}_{4}^{-} + (-) O H ⟶ {C r O}_{4}^{2 -} + H_{2} O + {n e}^{(-)}$
oxidation $\to$ $2 + 3 C r {(- 2 O + 1 H)}_{4}^{-} + 8 (-) O H ⟶ 2 6 + C r {- 2 O}_{4}^{2 -} + {6 e}^{(-)} + 8 H_{2} O$
Reduction $\to$ $3 {+ 1 H}_{2} {- 1 O}_{2} + 6 e^{(-)} + 6 H_{2} O ⟶ 6 {+ 1 H}_{2} - 2 O + 6 (-) O H$
So, Value of $n = 6$

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