Complete and balance the following equations, and ...

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Complete and balance the following equations, and identify the oxidizing and reducing agents.I2(s) + OCl-(aq)  →  IO3-(aq) + Cl-(aq)  (acidic  solution)

Jules Bruno · Accepted Answer

We have to balance the given redox reaction and find out the oxidizing and reducing agent.I2(s) + OCl-(aq) → IO3-(aq) + Cl-(aq)                 (acidic  solution)A redox reaction can be balanced (in acidic solution) by these 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.Step 1: Separate the whole reaction into two half-reactions.I2(s) → IO3-(aq)                                                             OCl-(aq) → Cl-(aq) Step 2: Balance the non-hydrogen and non-oxygen elements first.I2(s) → 2 IO3-(aq)                                                         OCl-(aq) → Cl-(aq) (Needs 2 with IO3-)                                                                       (Cl: Balanced)Step 3: Balance oxygen by adding H2O to the side that needs oxygen. (1 O: 1 H2O)I2(s) + 6 H2O(l) → 2 IO3-(aq)                                       OCl-(aq) → Cl-(aq) + H2O(l)(Needs 6 O on the reactant side)                                               (Needs 1 O on the product side)Step 4: Balance hydrogen by adding H+ to the side that needs hydrogen. (1 H: 1 H+)I2(s) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq)                    OCl-(aq) + 2 H+(aq) → Cl-(aq) + H2O(l)(Needs 12 H on the product side)                                             (Needs 2 H on the reactant side)Step 5: Balance the charges: add electrons to the more positive side (or less negative side).For species that are neutral: oxidation state = 0For ions (or species with charges): oxidation state = charge of the ionThe reactant side and the product side should have an equal overall charge (after adding electrons).I2(s) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq)                    OCl-(aq) + 2 H+(aq) → Cl-(aq) + H2O(l)Reactants                  Products                                               Reactants                          ProductsI2 = 0                           2 IO3- = -2                                             1 OCl- = -1                          1 Cl- = -16 H2O = 0                   12 H+ = +12                                           2 H+ = +2                           1 H2O = 0                                   + 10 e- = -10                                          + 2 e- = -2-------------                   -----------------                                          ---------------                         ---------------                     Overall = 0                  Overall = 0                                            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.(I2(s) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq) + 10 e-)×1 → multiply to get 10 e-(OCl-(aq) + 2 H+(aq) + 2 e-→ Cl-(aq) + H2O(l))×5 → multiply to get 10 e-Step 7: Get the overall reaction by adding the two reactions.Species that appear in both reactions on opposite sides are subtractedI2(s) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq) + 10 e-5 OCl-(aq) + 10 H+(aq) + 10 e- → 5 Cl-(aq) + 5 H2O(l)------------------------------------------------------------------I2(s) + 5 OCl-(aq) + H2O(l) → 2 IO3-(aq) + 5 Cl-(aq) + 2 H+(aq)The overall balanced equation is:I2(s) + 5 OCl-(aq) + H2O(l) → 2 IO3-(aq) + 5 Cl-(aq) + 2 H+(aq)Reducing agentI2(s) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq) + 10 e-Recall the mnemonics LEO GER.Lose               GainElectron         ElectronsOxidation       Reductionlose electrons → oxidation → anodegain electrons → reduction → cathodeI2 is losing electrons in this reaction to form IO3-.The reducing agent is itself oxidized in a redox reaction.I2 is the reducing agent.Oxidizing agent5 OCl-(aq) + 10 H+(aq) + 10 e- → 5 Cl-(aq) + 5 H2O(l)OCl- is gaining electrons in this reaction to form Cl-.The oxidizing agent is itself oxidized in a redox reaction.OCl- is the oxidizing agent.

Problem: Complete and balance the following equations, and identify the oxidizing and reducing agents.I2(s) + OCl-(aq) → IO3-(aq) + Cl-(aq) (acidic solution)

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Problem: Complete and balance the following equations, and identify the oxidizing and reducing agents.I2(s) + OCl-(aq) → IO3-(aq) + Cl-(aq) (acidic solution)

FREE Expert Solution

Problem Details

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