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

Question

What is the value of n in the following half equation? Cr ( OH )4-+ OH arrow CrO 42-+ H 2 O +n e- (A) 3 (B) 6 (C) 5 (D) 2

Tardigrade Admin · Accepted Answer

Cr ( OH )4-+ OH - arrow CrO 42-+ H 2 O Writing oxidation states, we have [ Cr ( OH )4]-+ OH - arrow[ CrO 4]2-+ H 2 O To balance oxidation state of Cr on both sides, add 3 e- on R. H.S.

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

3

6

5

2

$C r (O H)_{4}^{-} + O H^{-} \to C r O_{4}^{2 -} + H_{2} O$

Writing oxidation states, we have $[C r (O H)_{4}]^{-} + O H^{-} \to [C r O_{4}]^{2 -} + H_{2} O$

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

Q. What is the value of n in the following half equation? Cr(OH)4−​+OH→CrO42−​+H2​O+ne−

3

6

5

2

Cr(OH)4−​+OH−→CrO42−​+H2​O Writing oxidation states, we have [Cr(OH)4​]−+OH−→[CrO4​]2−+H2​O To balance oxidation state of Cr on both sides, add 3e− on R.H.S.

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

$C r (O H)_{4}^{-} + O H^{-} \to C r O_{4}^{2 -} + H_{2} O$

Writing oxidation states, we have $[C r (O H)_{4}]^{-} + O H^{-} \to [C r O_{4}]^{2 -} + H_{2} O$

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