Redox reaction : Br2 —> Br- + BrO3- in basic solution?

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Br2 ——-> Br- + BrO3-
separating this reaction into oxidation and reduction half reations :
reduction half reaction: Br2 ——> Br-…..(1)
oxidation half reaction: Br2——–> BrO3-……(2)
now balancing reduction half first – balance all atoms other than O and H ….
Br2 ——-> 2Br-…..(4)
now balancing oxidation number by adding electrons….oxidation no. of Br in Br2 is 0 while in Br- is -1 …thus each Br accpets 1 electron.since there are 2 Br atoms on right side of eqn (4) therefore adding 2 e- to left side of eqn (4)
Br2 + 2e- ——–> 2Br- ….(5)
balancing charge – not needed as charge on either side is already balanced …
thus eqn(5) represents balanced reduction half reaction
now balancing oxidation half reaction…
first balancing all atoms other than O and H….
Br2 ——> 2BrO3- ……..(6)
now balancing oxidation no. by adding electrons….oxidation of Br in Br2 is 0 and in BrO3- is +5 . thus each Br atom looses 5 electrons since there are two atoms of Br in right side of (6) therefore adding 10 e- to right side of (6)..
Br2 ——–> 2BrO3- + 10e-…….(7)
now balacing charge by adding OH- (as we are under basic condition) …..the total charge on right side of (7) is -12 and zero on left side ….so adding 12 OH- ions to left side of (7)
Br2 + 12OH- ——-> 2BrO3- + 10e-…..(8)
now balancing O atoms on both sides by adding H2O
Br2 + 12OH- ——–> 2BrO3- + 10e- + 6H2O….(9)
H atoms are already balanced……therefore (9) represents a balanced oxidation half reaction…
now multiplying (5) by 5 and adding in (9)
5Br2 + 10e- ——–> 10Br-
+
Br2 + 12OH- ——–> 2BrO3- + 6H2O + 10e-
= 6Br2 + 12OH- ——–> 10Br- + 2BrO3- + 6H2O
dividing by 2…
3Br2 + 6OH- ——> 5Br- + BrO3- + 3H2O
this represents the final balanced redox reaction

Also Read :   What is the empirical formula for the gas formed?

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