Balance the following redox reactions in acidic solutions:
UO2+Cr2O2−7⟶UO2+2+Cr2++H2O
A
3UO2+Cr2O2−7+14H+⟶3UO2+2+2Cr2++7H2O
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B
2UO2+2Cr2O2−7+14H+⟶3UO2+2+2Cr2++5H2O
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C
3UO2+2Cr2O2−7+14H+⟶3UO2+2+2Cr2++5H2O
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D
None of these
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Solution
The correct option is A3UO2+Cr2O2−7+14H+⟶3UO2+2+2Cr2++7H2O The unbalanced chemical equation is as follows: UO2+Cr2O2−7⟶UO2+2+Cr3++H2O
Balance all atoms except H and O. UO2+Cr2O2−7⟶UO2+2+2Cr3++H2O
The oxidation number of U changes from +4 to +6. The change in the oxidation number is 2. The oxidation number of Cr changes from +6 to +3. The change in the oxidation number per Cr atom is 3. Total change in the oxidation number for 2 Cr atoms is 6
To balance the increase in the oxidation number with decrease in oxidation number, multiply UO2 and UO2+2 with 3. 3UO2+Cr2O2−7⟶3UO2+2+2Cr3++H2O T
o balance O atoms, add 6 water molecules of RHS. 3UO2+Cr2O2−7⟶3UO2+2+2Cr3++7H2O
To balance H atoms, add 14 H+ on LHS 3UO2+Cr2O2−7+14H+⟶3UO2+2+2Cr3++7H2O