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  4. What is the value of n in the following half equation? Cr ( OH )4-+ OH arrow CrO 42-+ H 2 O +n e-

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Q. What is the value of $n$ in the following half equation? $Cr ( OH )_{4}^{-}+ OH \rightarrow CrO _{4}^{2-}+ H _{2} O +n e^{-}$

Redox Reactions

Solution:

$ Cr ( OH )_{4}^{-}+ OH ^{-} \rightarrow CrO _{4}^{2-}+ H _{2} O$

Writing oxidation states, we have $\left[ Cr ( OH )_{4}\right]^{-}+ OH ^{-} \rightarrow\left[ CrO _{4}\right]^{2-}+ H _{2} O$

To balance oxidation state of $Cr$ on both sides, add $3 e^{-}$ on $R. H.S$.