Question

2. What useful information is obtained from the oxidation state of an atom? 3. Which compound is the stronger oxidizing agent, KMnO, or MnO2? 4. Distinguish between the equivalence point and the end point of a titration. 5. What advice will you give to your lab partner if he or she decides to add the starch indicator to the mixture prior to the titration?

Answer 1

Solution oxidation states It is referred to as oxidation number. At describes the clegrll of oxidation (loss of electrons) of an atom in a chemical compound. The oxidation state of an atom does not represent the real charge on the atom. So the oxidation number of an atom is the charge that an atom would have if the compound was composed of ions. It plays an important role in the systematic momenclature of chemical compound. © By comparing the standard reduction potential we can tell which compound is stronger oxidizing agent, KMnOy on MnO2 Standar Half reaction of KM Mnoucaq + shall) + se'>> Muay +47406). E=1:57 Half reaction of MnO2 MnO2 (s) + 4Hea tzé Moulav) + 2920(8) E*= 1-23V More positive reduction potentio i sitive reduction potential is the stronga oxidizing agent. This KMnOy is stronger oxidizing agent than Ecs Scanned mnoz Samséashe

Equivalence point I End point a) Number of moles of titrant | a) The colour changes to are equal to the number indicate a complete of males of analyte titration 6) There may be multiple 6) Only one end point equivalence points c) (C) occurs before the end point Occurs after the equivalence point ③ I suggest my las partner that the starch indicator added near the end of the titration and not at the beginning because starch - iodine complex at high Iz concentration is relatively stakle. s Scanned with SamScanner