balance the redox reaction by using oxidation number method in acidic medium
cr2o7^-2 ( aq ) + so3^-2 ( aq ) -> cr3+ ( aq ) + so4^-2 ( aq )
Answers
We can solve this by ion electron method:
Step 1 : separate oxidation and reduction
Cr2o7^-2 ( aq) -> Cr^3+( aq) ( Reduction)
So3^-2 ( aq) -> So4^-2 (aq). ( Oxidation)
step 2 : Balance all the atom except Hydrogen and oxygen
Balancing: Cr2o7^-² -> 2Cr^3+
So4^ 2- -> so3^ 2-
Step 3 : Balance oxygen atom with the help of water ( H2O) ( add water H2O where oxygen atom is less.)
Cr²O7^ 2- -> 2 Cr+³ + 7H2o
H2O + So3^ 2- -> So4^ 2-
Step 4: Balance Hydrogen atom with the help of (H+) ( add H+ where hydrogen ( H) atom is less)
14H+ + Cr²o7^ 2- -> 2Cr^3+ +7H2O
H2O +So4^2- -> So4^2- + 2H+
Step 5 : balance charge by using electrons.
Reduction : Charge = +14 -2 = +6
= 12 = +6
-6e- = +6 = +6
Oxidation : (-2) =. -2+2 = O
- 2 = -2 ( By subtracting 2e-)
Total: we will multiply the oxidation equation because we have add 6 electrons in the reduction reaction.
( H2O + So3^²-. -> So4^²- + 2H+ + 2e- ) *6
(
( 6e- +14H+ +Cr2o7^2- -> 2Çr³+ + 7H2o ) *6
= 12e- + 28H+ + Cr2O7 ^-2 + 6H20 + 6So3^2- -> 4Cr³+ + 14H20 + 6SO4^2- + 12 H+ + 12e-
after that , 12 e- are cancelled out. and remaining is your answer .