Question

The following skeletal oxidation-reduction reaction occurs under basic conditions. Write the balanced REDUCTION half reaction. H2B03+ Br—>B + Br2 Reactants Products

Answer 1

oxidation no. + - (HBO) + Br o o B + Bra - Í oxidation Reduction baly cell (HBO) + 3 é - B _(1480) + 3e > B+31,0 Balance 3 using H20 (4,139) + 4H+ +3é - 3 + 3H0 Balance 4 using it ( HBO) + 44*+hori +3 é > 0 +380 + 40loa bom sides vaata y simplify. that of Ho. 6,803) + H2O +36 => B+ 11 +601 Add Equal rood _Oxidation bely cell oor Br + se r ie Balance two half -cello by balancing ce (14,179,) + 4,0 + 3e > Budow) x 2 LaB B Y +26)_X3 2419)+ 6.BC_+24,0 -> 23. + 3Br, + 801 BALANCED

oxidation ne LOT S + c -2 -> Sat +5 CC10,) -1 oxidation - reduction Balancing reduction-baly cell reactionis -2 Ist aé sao +5 Balancing oxidation chaly cell reactioa's cil - (Clogs + bé - Balance cl + 340 (log] the ö using to CI-+3ho (co) + 640 the balance. t using HT Guitw i a Add equal 10.4 cl +360+6014-Y (105+64 +604 té lorem J on both side simpligy 161 +604 -) (C103) + 3140_+ben Balance two half cell by balancing e L. (stré cl_+601 | 35 +41 +60 se x3. (co) + 31,0 +62 (010) 352-+ 34,0 BALANCED 1 35+cit Goh - 38 +(6103) + 34,0

+4 ta _ _ te _ 4_ Oxidation_00- Mno, +_Ni COH),_--> MOCOM), +_ Mio I reduction 1 Oxidation Reduction half cell Follow same Motae Ancona Procedure Mao, +@+sé -> MOCOH), - Mao, +2,8+20 +2 – na(OH)+ 2OH®, Maqe +240 +20=s Mocow), +2049 LOxidation haly cell NiCOH) Nio, + 20 Nicol) =Mio +2+®+24 NiCOH) +2011Nio, + 2H+ 204 +26 Nicol) + 2014-> Nio, + 2100 +21 Balance two half celle Moog +240 +4C - Mcom), +2014 NicOH) +2012 Mic +240 + Moo+Ni Colt), Mn.com), +120 1