Question

What is the coefficient of H2O in the balanced equation below after balancing it in acidic medium using the smallest possible whole number coefficients? And which is the oxidizing agent? As 04 (s) + NO2 (aq) --> AsO4 (aq) + NO (g) Note that NO2 (-) is nitrite (N-O-Subscript 2 and has a negative one charge) d) 2, As204 oa) 2, NO2 (-) Ob) 4, NO2 (-) O c) 4, As204

Answer 1

Given reaction AS 9H (5) + Noz (aq) —— Aso (aq) + NO (9) ASO4 In generally 'o' oxidatiou number = -2 .. 2AS + 4(-2) = 0 = PAS = 8 = AS = +4 Na N+26-2) = -1 = N= 41+ IN = +3 ASQ3 AS + 4(2) = -3 AS = 8-] + Asate NO N+la ) = 0 5 N= +2. +5 A S O - AS 93 I oxidation oxidation number +3 increased t4 to 45 2 +2 No (11) - Noca) L. Reduction oxidizing agent ordizing agent itself undergo reduction. oxidizing agent = Nool

To balance in acidic medium separate both the half reactions. Oxidation-half-reaction Reduction half-reaction AS2O4 (5) - ASG (91) Nos (aq) No (9) stepo: Balance the atoms Except o and H. AŞ04() QASE, Cag) Ng (at) - No (9) step ©: Balance the 'o' atoms. Add to, o deficient side · AŞQ4 (5)+ 4H2O(l) 52Asa cay) No (au) - NOLI) + Hoce Step 6: Balance the Hatoms, Add ht H deficient side. AŞO4(5) +4440(8)81 Tag) +2 AS *call) Ne (aq) +24+ (24) NO(9)+ Hold step Balance the elections AS, (5) + 4+ ore) 84*(49) + 2A50"(al+ a ē a NG (99)+2 4+ (aq) + 17 Nocg) + Hollo step ① To equalise the electrons o + @x2 AS29 (5) + +14 068) 84*(44) +2 AS q? caq) + Qē 2 No (aq) + 4H+ (aq) + aé 2 NO(g) + 2 Holes AS Y (5) + 2 Hole) +2 Ng (2) -> 4H4C2BD+2 A sau cal) + 2NO 1. This is balanced equation Answer: coefficient Oxidizing of tho = 2 Agent - NO 0 J'ai