Balance the following reaction under acidic condit...

Question

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)

Jules Bruno · Accepted Answer

We are being asked to balance the given oxidation-reduction reaction. The reaction is under acidic conditions. When balancing redox reactions under acidic conditions, we will follow the following steps.Step 1: Separate the whole reaction into two half-reactionsStep 2: Balance the non-hydrogen and non-oxygen elements firstStep 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+)[readmore]Step 5: Balance the charges: add electrons to the more positive side (or less negative side)Step 6: Balance electrons on the two half-reactionsStep 7: Get the overall reaction by adding the two reactions.Balance the redox reaction under acidic conditions:Cr2O2−7(aq) + Fe2+(aq) → Cr3+(aq)  + Fe3+(aq) *Before balancing the redox reaction:• ignore H+, OH-, and H2O in the starting reactionStep 1: Separate the whole reaction into two half-reactionsCr2O2−7(aq) → Cr3+(aq)                                                  Fe2+(aq) →  Fe3+(aq) Step 2: Balance the non-hydrogen and non-oxygen elements firstCr2O2−7(aq) → 2 Cr3+(aq)                                                  Fe2+(aq) →  Fe3+(aq)                                                                             (Fe: balanced)Step 3: Balance oxygen by adding H2O to the side that needs oxygen (1 O: 1 H2O)Cr2O2−7(aq)→ 2 Cr3+(aq) + 7 H2O(l)                                Fe2+(aq) →  Fe3+(aq)                                                                               (No Oxygen)Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen (1 H: 1 H+)Cr2O2−7(aq) + 14 H+(aq) → 2 Cr3+(aq) + 7 H2O(l)              Fe2+(aq) →  Fe3+(aq)                                                                                    (No hydrogen)Step 5: Balance the charges: add electrons to the more positive side (or less negative side)▪ For species that are neutral: oxidation state = 0▪ For ions (or species with charges): oxidation state = charge of the ion▪ The reactant side and the product side should have an equal overall charge (after adding electrons)Cr2O2−7(aq) + 14 H+(aq) → 2 Cr3+(aq) + 7 H2O(l)                      Fe2+(aq) →  Fe3+(aq) Reactants                            Products                                               Reactants             Products                               -2                                               2(+3)                                                   +2                          +3+14                                                                                                        +6 e-                                                                                                                                     +1 e------------------------                 ------------------------                                  --------------               ----------------overall = +6                             overall = +6                                         overall = +2            overall = +2                    Step 6: Balance electrons on the two half-reactions▪ multiply the reactions so both half-reactions have the same number of electrons▪ electrons should cancel in the overall reaction            Cr2O2−7(aq) + 14 H+(aq) + 6 e-→ 2 Cr3+(aq) + 7 H2O(l) (Fe2+(aq) →  Fe3+(aq) + 1 e-) x 66 Fe2+(aq) →  6 Fe3+(aq) + 6 e-▪ Oxidation half-reaction: lost electrons (electrons are on the product side)▪ Reduction half-reaction: gained electrons (electrons are on the reactant side)*Electrons are balancedOxidation half-reaction:       6 Fe2+(aq) →  6 Fe3+(aq) + 6 e-Reduction half-reaction:      Cr2O2−7(aq) + 14 H+(aq) + 6 e-→ 2 Cr3+(aq) + 7 H2O(l) Step 7: Get the overall reaction by adding the two reaction.▪ species that appear in both reactions on opposite sides are subtracted                             6 Fe2+(aq) →  6 Fe3+(aq) + 6 e-Cr2O2−7(aq) + 14 H+(aq) + 6 e-→ 2 Cr3+(aq) + 7 H2O(l) --------------------------------------------------------6 Fe2+(aq) + Cr2O2−7(aq) + 14 H+(aq) → 6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l) The Balanced Redox Reaction is: 6 Fe2+(aq) + Cr2O2−7(aq) + 14 H+(aq) → 6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)

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