Question

0.0257 in instead of In: 0.0592 logo MnO- + 1 . $-+ H2O → MO; + S8 + OH Oxidation Half Reaction: Reduction Half Reaction: Oxidizing agent: Reducing agent 12. HS + NO:- + H → NO2 + S8 + H2O Oxidation Half Reaction: Reduction Half Reaction: Oxidizing agent: Reducing agent 13. Zn + NO3- + H → Zn2+ + N2 + H2O Oxidation Half Reaction: Reduction Half Reaction: Oxidizing agent: Reducing agent 14.__MnO4 + SO2 + H2O → SO42- + Mn2+ + + Oxidation Half Reaction:

Answer 1

11) Oxidation is the process which involves the increase in the oxidation number. Substance which undergoes oxidation is known as the reducing agent. Reduction is the process which involves the decrease in the oxidation number. Substance which undergoes reduction is known as the oxidizing agent. ition Mn07(aq) + $a- (aq) —— MnO,(s) + Se (aq) +7 -2 +4 reduction Substance being oxidized Reducing agent Substance being reduced Oxidizing agent Mno ----- Mnog - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step: 1 Oxi dati on half reaction: sa- -→S Reduction half reaction: MnO — MnO S Step: 2 Balancing oxidation half reaction: $2- Balancing S: 852- -→S Balancing charge: 852- →S,+16e- Step: 3 balancing reduction half reaction: MnO 2 MnO, Balancing O: Mnon —Mno, + 2H,0 Balancing H: MnO2 + 4H* — Mno, + 2H,0 MnO2 + 4H+ + 40H —— MnO, + 2H,0+ 4OH- Mn07 + 4H,0 –→ MO, + 2H,0+ 40H- Mn07 + 2H,0 –→ MnO, + 40H Balancing charge: MnO2 + 2H,0+ 3e —MnO, + 40H Step 4: Equalizing the no of electron lost in oxidation half reaction and no of electrons gainedin reduction half reaction. (8.52 —— Sp+166) x6 (MnO2 + 2H,0+32 —— MnO, + 40H-)x32 Step 5: Adding both the half reactions and cancel the common ions to get the balanced equation 48S2- -→ 6S+96e 32Mno+64 H, 0+966 —32MnO+ 128OH- 32Mn 07 +64 H,0+ 48S2- → 6S+ 32MnO, + 128 OH 16Mn0; +32H,0+ 2452- -→ 35+ 16MnO2 + 640H