What is the value of $n$ in the following equation?

$C r {(O H)}_{4}^{-} + O H^{-} ⟶ C r O_{4}^{2 -} + H_{2} O + n e^{-}$

3

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6

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5

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2

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Solution

The correct option is B $6$
The given reaction is
$[C r (O H)_{4}]^{-} + O H^{-} \to [C r O_{4}]^{- 2} + H_{2} O + n e^{-}$
Now balancing the equation, $2 [C r (O H)_{4}]^{-} + 8 O H^{-} \to 2 [C r O_{4}]^{- 2} + 8 H_{2} O + 6 e^{-}$ . So the value of n is 6. Correct option is B

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