Question

balance the reaction by both (oxidation number and half reaction eqn) methods
Cr(OH)3 + HO2 ----- Cr2O7+OH (basic medium)​

Answer 1

going by the data the probability of my promotion is 0.4

Answer 2

Answer:

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Explanation:

ndication of oxidation number of each element. Thus, we find that Cr in Cr(OH)3 and iodine in IO3- undergo change in oxidation number. Step 2. Writing oxidation and reduction half reactions. + 3 C r ( O H ) 3 → + 6 C r O 2 − 4 Cr+3(OH)3→Cr+6O42− (Oxidation half reaction) + 5 I O 3 → I − I+5O3→I− (Reduction half reaction) Step 3. Addition of e- to make up the difference in O.N. + 3 C r ( O H ) 3 → + 6 C r O 2 − 4 + 3 e − Cr+3(OH)3→Cr+6O42−+3e− + 5 I O − 3 + 6 e − → I − IO3−+5+6e−→I− Step 4. Balance O atoms by adding H2O molecules to the side deficient in ‘O’ atoms. Cr(OH)3 + H2O → CrO42- + 3e- IO3 + 6e- → I- + 3H2O Step 5. Balance H atoms. Since the medium is basic, therefore add proper number of H2O molecules to the falling short of H atoms and equal number of OH- ions to the other side. Cr(OH)3 + H2O + 5OH- → CrO42- + 3e- + 5H2O IO3- + 6e- + 6H2O → I- + 3H2O + 6OH- Step 6. Equalize the electrons lost and gained by multiplying the electron and gained by multiplying the oxidation half reaction with 2. Read more on Sarthaks.com - https://www.sarthaks.com/1003784/balance-the-following-redox-equations-by-half-reaction-method-cr2o7-fe-h2o-acidic-medium