Balance the given reaction by oxidation number method in acidic medium.
Cr 2 O 7(aq)2− +I (aq)− →Cr (aq)3+ +I 2
Answers
Answer:
For the equation:
Cr
2
O
7
2−
+H
2
O
2
⟶Cr
3+
+O
2
Step 1: Assign oxidation number for each atom in the equation
Cr
2
+6
O
7
2−
−2
+
H
2
+1
O
2
−1
⟶
Cr
3+
+3
+
O
2
0
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:
H
2
+1
O
2
−1
⟶
O
2
0
+2e
−
(here oxygen is oxidized & oxidation number increases)
Reduction half:
Cr
2
+6
O
7
−2
+6e
−
⟶
2Cr
3+
+3
(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
H
2
+1
O
2
−1
⟶
O
2
0
+2e
−
+2H
+
Cr
2
+6
O
7
2−
−2
+6e
−
+14H
+
⟶2Cr
3+
(b): Balance oxygen atom if oxygen atoms are not balanced add water molecules for it
H
2
O
2
⟶O
2
+2e
−
+2H
+
Cr
2
+6
O
7
2−
−2
+6e
−
+14H
+
⟶
2Cr
3+
+3
+7H
2
O
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.
O:3
H
+1
2
O
−1
2
⟶3
O
0
2
+6e
−
+6H
+
R:Cr
2
O
7
2−
+6e
−
+14H
+
⟶2Cr
3+
+7H
2
O
Step $$5$: Add half-reaction by adding all the reactants on one side and product on the other side.
3
H
0
2
O
2
+Cr
2
O
7
2−
+6e
−
+14H
+
⟶3O
2
+6e
−
+6H
+
+2Cr
3+
+7H
2
O
⇒3
H
+1
2
O
−1
2
+
Cr
2
+6
O
7
2−
+8H
+
⟶
3O
0
2
+2Cr
3+
+7H
2
O
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.