Complete and balance the following equations, and ...

Question

Complete and balance the following equations, and identify the oxidizing and reducing agents.

Part A

Cr₂O₇²⁻(aq) + I ⁻(aq) → Cr ³⁺(aq) + IO₃⁻(aq) (acidic solution)

Express your answer as a chemical equation. Identify all of the phases in your answer.

Part B

oxidizing agent, reducing agent

Express your answers as chemical expressions separated by a comma.

Jules Bruno · Accepted Answer

Balance the redox reaction and determine the oxidizing/reducing agent in each reaction.Recall that the oxidizing agent is the one being reduced (GER-OA = Gain electrons, reduced- Oxidizing agent) while reducing agent is the one being oxidized (LEO-RA = Lose electrons, oxidized – Reducing agent)PART AWe are being asked to balance the given oxidation-reduction reaction. The reaction is under acidic conditions. Balance the redox reaction under acidic conditions:Cr2O72−(aq) + I− (aq) → Cr 3+(aq) + IO3−(aq)[readmore]Step 1: Separate the whole reaction into two half-reactionsCr2O72−(aq)  → Cr 3+(aq)    I− (aq)  → IO3−(aq)Step 2: Balance the non-hydrogen and non-oxygen elements firstCr2O72−(aq)  → 2Cr 3+(aq)                       I− (aq)  → IO3−(aq)                                                                            (I: balanced)Step 3: Balance oxygen by adding H2O(l) to the side that needs oxygen (1 O : 1 H2O)Cr2O72−(aq)  → 2Cr 3+(aq) + 7H2O (l)                   3H2O(l)+ I− (aq)  → IO3−(aq)            Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen (1 H: 1 H+)14H+(aq) +Cr2O72−(aq)  → 2Cr 3+(aq) + 7H2O(l)3H2O(l)+ I− (aq)  → IO3−(aq) + 6H+(aq)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)14H+(aq) +Cr2O72−(aq)  → 2Cr 3+(aq) + 7H2O(l)          3H2O(l)+ I− (aq)  → IO3−(aq) + 6H+(aq)Reactants                             Products                               Reactants                            Products                               H+ = +14                                    2Cr2+ = +6                            3H2O = 0                                   IO3- = -1     Cr2O72- = -2                                                                                 I- = -1                                       6H+ = +6 =+12                                       overall = +6                            overall = -1                                          =+5       +6e-                                                                                                                                                 +6e-         +6                                                                                                                                                    -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***both reactions have 6 electrons in it            ▪ Oxidation half-reaction: lost electrons (electrons are on the product side) ▪ Reduction half-reaction: gained electrons (electrons are on the reactant side)*Electrons are balancedReduction half-reaction:       6e- + 14H+(aq) +Cr2O72−(aq)  → 2Cr 3+(aq) + 7H2O(l)Oxidation half-reaction:      3H2O(l)+ I− (aq)  → IO3−(aq) + 6H+(aq) + 6e-Step 7: Get the overall reaction by adding the two reaction.▪ species that appear in both reactions on opposite sides are subtracted6e- + 14H+(aq) +Cr2O72−(aq)  → 2Cr 3+(aq) + 7H2O(l)3H2O(l) + I− (aq)  → IO3−(aq)+ 6H+(aq) + 6e-8H+(aq) + Cr2O72−(aq) + I- (aq)→ 2Cr 3+(aq) + 4H2O(l) + IO3- (aq)Balanced Redox Reaction:         8H+(aq) + Cr2O72−(aq) + I- (aq)→ 2Cr 3+(aq) + 4H2O(l) + IO3- (aq)PART BDetermine the OA and RA by examining the oxidation numbers of Cr and IStep 1. Determine the ON of Cr in the reactant sideThe oxidation number of Cr in Cr2O72- is:Recall: Oxygen has an oxidation number of -2, except peroxides and superoxides.2Cr + 7(-2) = -22Cr - 14 = -22Cr = -2+14Cr = 12/2Cr = +6Step 2. Determine ON of Cr in the product sideFor Cr in the product side:The oxidation number of Cr in Cr2+ is:Recall: The oxidation number of charged species is equal to its chargeON Of Cr2+ is +2Oxidation number of Cr decreased from +6 to +2; therefore, it is reduced. Cr2O72- is the oxidizing agent.Step 3. Determine the ON of I in the reactant sideThe oxidation number of I in I- is:Recall: The oxidation number of charged species is equal to its chargeON Of I- is -1Step 4. Determine ON of I in the product sideFor I in the product side:The oxidation number of I in IO3- is:Recall: Oxygen has an oxidation number of -2, except peroxides and superoxides.I + 3(-2) = -1I - 6 = -1I = -1+6I = +5Oxidation number of I increased from -1 to +5; therefore, it is oxidized. I- is the reducing agent.Summary:Balanced redox reaction: 8H+(aq) + Cr2O72−(aq) + I- (aq)→ 2Cr 3+(aq) + 4H2O(l) + IO3- (aq)Oxidizing agent – Cr2O72-Reducing agent – I-

FREE Expert Solution

Problem Details

FREE Expert Solution

Problem Details

Sign up for free to watch this video!