J.R. S. answered 04/12/23
Ph.D. University Professor with 10+ years Tutoring Experience
Cr₂O₇²⁻(aq) → Cr³⁺(aq)
Cr2O72-(aq) ==> 2Cr3+(aq) + 7H2O(l) .. balanced for Cr and O
Cr2O72-(aq) + 14H2O(l) ==> 2Cr3+(aq) + 7H2O(l) + 14OH-(aq).. balanced for Cr, O and H (using base, OH-)
Cr2O72-(aq) + 14H2O(l) + 6e- ==> 2Cr3+(aq) + 7H2O(l) + 14OH-(aq) .. balance for mass and charge
Combine and cancel like terms to get the final balanced equation:
Cr2O72-(aq) + 7H2O(l) + 6e- ==> 2Cr3+(aq) + 14OH-(aq) .. balanced reduction half reaction