Question

Introduction Titration is a common analytical technique that is used to verify the concentrations of chemicals before they are sold as commercial products, in a process known as quality control. In this lab, you will be using a redox titration to find the concentration of the strong oxidizing agent hypochlorite (CIO) which is the active ingredient in bleach. Oxidizing and reducing agents are often determined using a redox titration. In a redox titration, the titrant (the substance of known concentration that is delivered by burette) usually oxidizes or reduces the analyte (the substance whose concentration is to be determined by the titration). In this experiment, however, the analyte (hypochlorite) will not be reduced directly by the titrant (thiosulfate). Direct reduction of hypochlorite by thiosulfate in basic solution would produce colorless products that would be useless for detecting an equivalence point in a titration: 4CIO +20H +8,0,2--> 4C1 + 250,- + H2O (This reaction will not be used because its products are colorless) Rather than using a simple redox titration, the concentration of hypochlorite is determined indirectly by an iodometric titration. In an Iodometric titration, the oxidizing agent to be determined is mixed with an iodide salt in the right acidic or basic conditions for a redox reaction with iodide to occur. The iodide is oxidized to iodine, which is brownish orange in aqueous solution: CIO- +21+ 2H* -> C- + 12 + H2O (1. turns the solution orange) The amount of iodine that is produced is limited by the amount of hypochlorite that was in the solution, so the iodine can be titrated away by another reducing agent (thiosulfate) to find its concentration and then calculate the concentration of hypochlorite that produced it: 1, + 28,0,- -21° +5,0.2 (Orange color disappears) Since it takes two thiosulfate ions to reduce each iodine molecule and only one iodine molecule was produced by each hypochlorite ion, the net stoichiometry in this reaction is 1 mole hypochlorite/2 mole thiosulfate.

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Answer 1

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in Balance the equation for Oxidation of Iodido los by 8c1 acidic solution to yield Iodine (12) cio respectively, stepti Producing unbalanced equation for the reachin jonic form and un 10 Sci av 22) + 12 (8) +cio Steep 2: Reparating the equation into half creacion - 1 Coxidation Oxidation halfi slate) o (oxidahan state) I (22) I2 (3) Reducation half : 8ch (24) -1 Stepz: Balancing the atom other than "O" and "H" in each half re a chen individually Oxidation half: 2 1 - I2 (gi Chere 2 multiply by 10) CO cil is already beelance) Reducation halfi CI

Step 4. The reactions occurring in acidic medium, O atoms and 110 so adding H₂O to balance to balance H atomo Here oxidation half: with Reducation Hall: gel no D and I atom fo14 balance other alom -+ 2H0 CIO + H₂O steps: Addling elections to half reachon to balance One side of the the charges in coach roachion Oxidation Holfi 12(1) +20- 2100 (two election added to delight side) Reducation Half: C16 +H₂O 801+ 2H+26 (to balance Left 2 positive charge by adding 2 electron) (Note: make both reachon elechan equal) Stops! Now add the two half reaching to get the overall reaction and canceling the elechens 279+ Ecle9+ 2Heron t6 1 17.09.17 + cper) to on each side

21, + 8c!eo + 2 I 2 +P + Hod ez) 'This is net ionic equation. B Balancing the equation for reducation of Lading (-2) by thiosulfate (S2032) to yield (10) and tetathion ate ( Sq0g 2-). Stepz:) producing unbalanced equation for the reaching ionic form. I2 I Step2 : separating + S, 032- +S40 2- the equation into half reaction, Oxidation half: S2 Oz ² SaO2- ( just predicahing it is going aidahin +2.ach's "atom 2. Seach"5"tem reochon) Reducation half! I I step 3: Oxidation half: 12 Balancing the atom other than "O" and "" in each haff reaction individually. 2 S203 → ( S203² multiply 2 to balance "s" atom t2 ( 18 multiply by 2 ) Sq O2- Roducation half: 210

and "H" Step 4.] The "o" atoms atom abready balance, so not require to adding that (Steps : Adding elections to one side of the half reaction to balance the charges in each reaction Oxidation halft 2520,2- - Sq q2- + 2e- ( balance 4 negative 4 negalue charge both side) Reducation Halft 12 +2e- - 270 (balance 2 negative charge both side) Steph! Now add the two half reactions to overall reaction and canceling the electrons each side get the on 12 + 25,0?-+26 → 210 + 54 0,70 +267 Iz + 2 S203² 210 & 84 og 2- This is the net ionic ionic equation