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(A) The unbalanced equation is: $Cr(OH)_3 + ClO^{-} + OH^{-} 	o CrO_4^{2-} + Cl^{-} + H_2O$Step $1$: Assign oxidation states. $Cr$ changes from $+3$

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$2, 3, 2, 3$

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(A) The unbalanced equation is: $Cr(OH)_3 + ClO^{-} + OH^{-} 	o CrO_4^{2-} + Cl^{-} + H_2O$Step $1$: Assign oxidation states. $Cr$ changes from $+3$ to $+6$ (oxidation,loss of $3$ electrons). $Cl$ changes from $+1$ to $-1$ (reduction,gain of $2$ electrons).Step $2$: Balance the electrons. Multiply the $Cr$ half-reaction by $2$ and the $Cl$ half-reaction by $3$: $2Cr(OH)_3 + 3ClO^{-} 	o 2CrO_4^{2-} + 3Cl^{-}$.Step $3$: Balance the charge using $OH^{-}$. The left side has a charge of $-3$ and the right side has a charge of $-4 - 3 = -7$. Adding $4OH^{-}$ to the left side balances the charge: $2Cr(OH)_3 + 3ClO^{-} + 4OH^{-} 	o 2CrO_4^{2-} + 3Cl^{-} + 5H_2O$.The coefficients are $2, 3, 2, 3$.

The values of coefficients to balance the following reaction are
$Cr(OH)_3 + ClO^{-} + OH^{-} \to CrO_4^{2-} + Cl^{-} + H_2O$
Find the coefficients for $Cr(OH)_3$,$ClO^{-}$,$CrO_4^{2-}$,and $Cl^{-}$ respectively.

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The values of coefficients to balance the following reaction are$Cr(OH)_3 + ClO^{-} + OH^{-} \to CrO_4^{2-} + Cl^{-} + H_2O$Find the coefficients for $Cr(OH)_3$,$ClO^{-}$,$CrO_4^{2-}$,and $Cl^{-}$ respectively.