Question

Again consider the following unbalanced oxidation - reduction reaction: Ba(OH)2(aq) + H2O2(aq) + CIO2(aq) + Ba(CIO2)2(s) + O2(g) Balance this reaction for basic conditions. Once both half reactions are ready to add to give the overall balanced reaction, the number of electrons lost ( = the number of electrons gained) is O 2 O 3 O 1 O none of these

Answer 1

oxidation Ba(CIO), +3 + 02 1) Ba(OH), +010, + H2O2 - +4 -1 redxtim Step:1 Oxi dati on half reaction: H,O, - Reduction half reaction: C10, — 0 Ba(Co), Step: 2 Balancing oxidation half reaction: H20 - 0 Balancing H: HO, — 0, +2H* H,9, +20H —— 0, +2H*+ 2OH- H20, +20H —— 0, + 2H,0 Balancing charge: H,9, +20H —— 0, +2H,0+ 2e Step: 3 balancing reduction half reaction: C10, — Ba(C102) Balancing Ba&C1: Ba(OH),+ 2010, — Ba (C107), Balancing 0: Ba(OH), + 2010, —> Ba(C102), + 2H,0 Balancing H: Ba(OH), + 2010,+2H* →Ba(C10,), + 2H,0 Ba(OH), + 2010,+2H+ + 20H — Ba(C10,)+ 2H,0 + 20H- Ba(OH), + 2C10,+2H,0 —— Ba(40,), + 2H,0 + 20H- Ba(OH), + 200, —— Ba(C10,), + 2OH- Balancing charge: Ba(OH), + 200, + 2€Ba(C102), + 20H- Step 4: Equalizing the no of electron lostin oxidation half reaction and no of electrons gainedin reduction half reaction. ( H,9, +20H —— 03 + 2H,0+ 2e-)x1 (Ba(OH), + 2010, +22 —Ba(C10,), + 2OH- )x1 Step 5: Adding both the half reactions H,02 + 2OH —0, + 2H,0+ 2e Ba(OH),+ 290, + 2 —Ba(C10,)+ 2OH- Ba(OH),+ 2010, +H,9, +20H-— , + 2H,0+ Ba(C102), + 2OH- Ba(OH), + 2010, +H,0,– 0, +2H,0+ Ba(C10,), No of electrons transferred in this reaction = 2