Balance the following redox reaction by inserting ...

Question

Balance the following redox reaction by inserting the appropriate coefficients.H+ + CrO42- + NO2- → Cr3+ + H2O +NO3-

Jules Bruno · Accepted Answer

We’re being asked to balance the given oxidation-reduction reaction:H+ + CrO42– + NO2–→ Cr3+ + H2O + NO3–We assume that the redox is in acidic solution since H+ is present. When 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+)[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-reactions.Step 7: Get the overall reaction by adding the two reactions.When balancing under basic conditions, additional steps are required:Step 8: Balance remaining H+ by adding an equal amount of OH– ions to both sides.Step 9: H+(aq) will combine with OH–(aq) to form H2O(l).Step 10: Cancel out common species.Before balancing the redox reaction, we need to ignore H+, OH–, and H2O in the reaction since these will be added later.Step 1: Separate the whole reaction into two half-reactions:CrO42–→ Cr3+                                                           NO2–→ NO3–Step 2: Balance the non-hydrogen and non-oxygen elements first:CrO42–→ Cr3+                                                           NO2–→ NO3–(Cr: balanced)                                                                           (N: balanced)Step 3: Balance oxygen by adding H2O to the side that needs oxygen (1 O: 1 H2O):CrO42–→ Cr3+ + 4 H2O                                             NO2– + H2O → NO3–(needs 4 O on the product side)                                               (needs 1 O on the reactant side)Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen. (1 H: 1 H+):CrO42– + 8 H+ → Cr3+ + 4 H2O                                NO2– + H2O → NO3– + 2 H+(needs 8 H on the reactant side)                                              (needs 2 H on the product side)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)CrO42– + 8 H+ → Cr3+ + 4 H2O                               NO2– + H2O → NO3– + 2 H+Reactants                  Products                                             Reactants                  ProductsCrO42– = –2                 Cr3+ = +3                                             NO2– = –1                  NO3– = –18 H+ = +8                     4 H2O = 0                                            H2O = 0                     2 H+ = +2+ 3 e– = –3                                                                                                                 + 2 e– = –2––––––––––                ––––––––––                                        ––––––––––                ––––––––––Overall = +3                Overall = +3                                        Overall = –1                Overall = –1Step 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(CrO42– + 8 H+ + 3 e– → Cr3+ + 4 H2O) × 2 → multiply by 2 to get 6 e–(NO2– + H2O → NO3– + 2 H+ + 2 e–) × 3 → multiply by 3 to get 6 e–This gives us:2 CrO42– + 16 H+ + 6 e– → 2 Cr3+ + 8 H2O3 NO2– + 3 H2O → 3 NO3– + 6 H+ + 6 e–Step 7: Get the overall reaction by adding the two reactions:• Species that appear in both reactions on opposite sides are subtracted2 CrO42– + 16 H+ + 6 e– → 2 Cr3+ + 8 H2O3 NO2– + 3 H2O → 3 NO3– + 6 H+ + 6 e–––––––––––––––––––––––––––––––––––2 CrO42– + 3 NO2– + 10 H+→ 2 Cr3+ + 3 NO3– + 5 H2O The overall balanced redox reaction is:2 CrO42– + 3 NO2– + 10 H+ → 2 Cr3+ + 3 NO3– + 5 H2O

Problem: Balance the following redox reaction by inserting the appropriate coefficients.H+ + CrO42- + NO2- → Cr3+ + H2O +NO3-

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Problem: Balance the following redox reaction by inserting the appropriate coefficients.H+ + CrO42- + NO2- → Cr3+ + H2O +NO3-

FREE Expert Solution

FREE Expert Solution

Problem Details

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