Oxidation Step 1: Chromate Cr3+ Cr6+ Step 2: Cr(OH) 3 ÆCrO 4 2-Step 3: Cr(OH) 3 ÆCrO 4 2-Step 4: (Balance O) Cr(OH) 3 + H 2OÆCrO 4 2-(Balance H) Cr(OH) 3 + H 2O + 5OH-ÆCrO 4 2-+ 5H 2O Step 5: Cr(OH) 3 + 5OH-ÆCrO 4 2-+ 4H 2O + 3e-Step 6: 3(ClO-+ H 2O +2e-ÆCl-+ 2OH-) 3ClO-+ 3H 2O +6e-Æ3Cl-+ 6OH-2(Cr(OH) 3 + 5OH-ÆCrO 4 2-+ 4H 2O + 3e-) 2Cr
Chromi­um in the ox­i­da­tion state of +2. Chromi­um (II) ox­ide and hy­drox­ide CrO and Cr (OH)₂ are com­pounds which dis­play typ­i­cal base prop­er­ties. The com­pounds are col­ored – chromi­um (II) ox­ide is black, and the hy­drox­ide is yel­low. Salts of chromi­um (II) have a bluish col­or.
To find the correct oxidation state of in [Cr(NH3)4Cl2] + (the ion), and each element in the ion, we use a few rules and some simple math.First, since the io
c) The total charge of the ion is 3-. while the oxidation state of O is always 2-. Because there are 4 oxygen atoms with 2- oxidation state which gives 8- charge in total from oxygen, P must have an oxidation state of 5+ \text{\textcolor{#c34632}{P must have an oxidation state of 5+}} P must have an oxidation state of 5+ to equal 3-.
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Lets assume oxidation number of C in C H X 3 O H \ce{CH3OH} CH X 3 OH is x x x. Oxidation number of H is +1 and O is -2. The sum of the oxidation numbers of all atoms in a neutral compound is zero.

Write separate equations for oxidation and reduction. Reduction: + 1 H + e − → 0 H ( i n H 2) Oxidation: 0 A l → + 3 A l + 3 e −. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation: Reduction: 3 H + + 3 e − 3 H 0 ( i n H 2) Oxidation:
In addition to CrO 3, other commonly used oxidizing agents include potassium permanganate (KMnO 4) and sodium dichromate (Na 2 Cr 2 O 7). ‍Any of these reagents can be used to oxidize secondary alcohols to form ketones and primary alcohols to form carboxylic acids. Tertiary alcohols remain unreactive to oxidation.
Click here:point_up_2:to get an answer to your question :writing_hand:croh3 h02alkali cro h2othe number of oh required to balance theabove equation1125335446
Cr has, at first, N° of oxydation +3 and becomes +6. For make this passage Cr gives 3 electrons #[Cr(OH)_4]^(-) = CrO_4^(2-) + 3e^-# for balance the charge, i put on the left 4 negative charges as #OH^-# and i obtain on the right 4 mol of water #[Cr(OH)_4]^(- ) +4OH^(-) = CrO_4^(2-) + 3e^(-) + 4 H_2O#. The Oxygen, at first,has oxydation number
1dH1L.
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  • cr oh 3 oxidation number