55 Balance the following redox equations by oxidation number method. The reactions occur in acidic medium a) H,02109) + Croa) -7020g) + 02(g) + Crea
Answers
For the equation:
Cr₂O + H₂O2 → Cr³+ + 0₂
Step 1: Assign oxidation number for each atom in the equation
+3 Cr³+ +0₂ → Cr³+ 0 +6 -2 +1 −1 Cr₂O + H₂O2
Step 2: Divide equation in two half - (1) reduction half where reactant being reduced (oxidation no decreases) and (2) oxidation half (oxidation no increases) And also write the transfer of electrons to make numbers of oxidized & reduced atoms equal.
Oxidation half:
+1 -1 0 H₂O2 → O₂ +2e (here oxygen is oxidized
& oxidation number increases)
Reduction half:
+6-2 Cr₂O7 +6e +3 → 2Cr³+ (here Cr reduced by
3 electrons each)
Step 3 (a): Balance the charge is acidic medium add H+ ion to the side deficient in positive charge
+1 -1
0
H₂O2 → 0₂ +2e¯ + 2H+
+6 -2 Cr₂O +6e +14H+ → 2Cr³+
(b): Balance oxygen atom if oxygen atoms are not balanced add water molecules for it
H₂O2 → O₂ +2e¯ + 2H+
+6 -2 + Cr₂O² + 6e +14H* +3 → 2Cr³+ +7H₂O
Step 4: The electron lost in oxidation half must be equal to electron gained in reduced early so we multiply oxidation half by 3 and reduction early by 1.
+1 -1 0:3H202 0 30₂ + 6e + 6H+ 2
R: Cr₂O + 6e¯ + 14H+ 7H₂O →2Cr³+ +
+1 -1 0 0:3H202 → 30₂ + 6e + 6H* +
R: Cr₂0² +6e¯ +14H+ 2Cr³+ +
7H₂O
Step $$5$: Add half-reaction by adding all the reactants on one side and product on the other side.
0 3H₂O2 + Cr₂O¾¯ +6e¯ +14H+ 6e + 6H+ + 2Cr³+ + 7H₂O 3+ → 30₂ +
+1 +6 → 3H₂O₂ + Cr₂O² + 8H+ → 30₂ + 2Cr³+ +7H₂O
Finally, always check the equation is balanced and contain substance type & the number of atoms both sides and sum of charges on both sides are equal.
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