Question

Practice Exercise 8.12 Hydroperoxide ions, HO, react with permanganate ions, MnO producing MnO: and O, gas. Write a balanced net ionic equation for this reaction in a basid solution

Answer 1

8.12 Aus- The reaction can lau written as Hoa cao) + mm (aq) → nO2 (s) + 02 (9) Oxidation state change :- O change from -2 to 0 and my change from tt to ta. Thus, half reactions are? Oxidation : Hoa (aq) → O2(g) + 20 Reduction :- MnO,- (aq) +bé mm Oz (s). Now in order to balance the number of electrons, we should multiply the egn. by hast common multiple. In this case the value is 6.50 and add up the two half-reactions. Oxidation : (HO₂ (aq) → 029) + 2 e) x 3 Redulion:- (mmogłegt he – MnO2 (s)) x2. 3 HO2 (aq) + 2 mda (aq) + be → 302(g) + 2 MnO, (s) + bé Here the factor 2 and 3 multiply the egn e To balance Orygen, add 40 to the deficient sido .. 3HO (aq) + a Mmoa (aq) + 30(9)+2 MnO2(3) +4 406) to balance hydrogen, add Ht rons. a (as) + amma (ag) + 5Hat > 30279) + 2 MQ,(s) + 4B01)

Since the reaction occurs in basin medium so for SH+ ions, add 5oh ions to both sides of the equation. 3H0,- (aq) +2mm04 (aq) + S Hitapolog > 30,(g) + 2 Mn0,(s) +46011) +50H%) Replaing 4t and oft ions with 120, we get 3H0,- (aq) + 2 MnO2 (aq) + 5 HOCA) 30 (9) +2000, (+415 011) +501(aq) The esalanced equation is 3H07 (aq) + 2m nos (aq) + H2O(l) → 30) + 2 Mm2, (+ 50H" (ay) The net ronie equation is BH0z (ag) + ammoa (aq) → 501" (ay)