Question

Puzzle #1 (1-1/2 pt) Balance the following reaction in acidic solution. MnO2 (s) + H3A5O3 (aq) --Mn+ (aq) + H3AsO4(aq) Puzzle #2 (3 pt) Balance the following reaction in basic solution Croa? (aq) + 12 (s) — Cr(OH); (s) + 103(aq)

Answer 1

Thank you very much increasing oxidation Number = Reach DATE +2 +4 +3 - MnO₂ (s) + H₃ AsOz (aq, +2 Mut caq) + H₂ ASOA decreasing Oxidation Number = 2 each each Now total decreasing oxidation Number = increasing oxidation Number ② charge Baloncing of Charge in left side o Change in right side = + 2 Adding 24 t on the left side Mno, + H ₂ Asoz + ant (5) (aq) (aq) Mm² + +H ₂ Asoy+ho cay, caq, a decrecy ONE 3 +5 Gro4 cag) + 12 (s) Cr (OH)315) + 2103 (aq) increasing O.N= 5 total o N = 5x2=10 Bolaning of oxidation Number 10 CrO2? + 312 → 106r(on), +6103 20-ve (6-ve) Balancing of charge left side = Right side se TQ I

thank you very much taking 14 on in Right side of Reaction 10 Cr ba+312 -> 10 Cr(OH)2 +6103 + 1401 20-ve 6-vet 14-ve = 2oove Tow final Balanced Squction 10 CrO & Cag) + 31200) + 52011) 10 Crcon)z + 610309) + 14 on Caq (5) PAGE 1