What is the value of n in the following half equation,

${Cr(OH)}_{4}^{-} + {OH}^{-} \to {CrO}_{4}^{2 -} + H_{2} O + {ne}^{-} ?$

Question

What is the value of n in the following half equation, Cr(OH) 4 − + OH − → CrO 4 2 − + H 2 O + ne − ?

TestHub · Accepted Answer

C r ( O H ) 4 - + O H - → C r O 4 2 - + H 2 O To balance H and O atoms, add 3 H 2 O to R.H.S. and 30 H ⊖ to the L.H.S. we have, C r ( O H ) 4 - + 4 O H - → C r O 4 2 - + 4 H 2 O To balance charge, add 3 e - to the R.H.S. we have C r ( O H ) 4 - + 4 O H - → C r O 4 2 - + 4 H 2 O + 3 e -

Chemistry - Redox Reactions Question with Solution | TestHub

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