Problem: Balance the following reaction under acidic conditions.Cr2O2−7(aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq)A. 7H2O(l) + Cr2O2−7(aq) + Fe2+(aq) → 2Cr3+(aq) + Fe3+(aq) + 14OH−B. 14H+ + Cr2O2−7(aq) + Fe2+(aq) → 2Cr3+(aq) + Fe3+(aq) + 7H2O(l)C. 7H2O(l) + Cr2O2−7(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 14OH−D. 14H+ + Cr2O2−7(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l) 

FREE Expert Solution

We are being asked to balance the given oxidation-reduction reaction. The reaction is under acidic conditionsWhen balancing redox reactions under acidic conditions, we will follow the following steps.



Step 1: Separate the whole reaction into two half-reactions

Step 2: Balance the non-hydrogen and non-oxygen elements first

Step 3: Balance oxygen by adding H2O to the side that needs oxygen (1 O: 1 H2O)

Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen (1 H: 1 H+)


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Problem Details

Balance the following reaction under acidic conditions.

Cr2O2−7(aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq)

A. 7H2O(l) + Cr2O2−7(aq) + Fe2+(aq) → 2Cr3+(aq) + Fe3+(aq) + 14OH

B. 14H+ + Cr2O2−7(aq) + Fe2+(aq) → 2Cr3+(aq) + Fe3+(aq) + 7H2O(l)

C. 7H2O(l) + Cr2O2−7(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 14OH

D. 14H+ + Cr2O2−7(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)