Balance the following redox reaction by half reaction method.
`Fe^(2+)(aq)+Cr_(2)O_(7)^(2-)(aq)toFe^(3+)(aq)+Cr^(3+)(aq)` (In acid medium)

Explanation

Step 1: (LEO= Loss of Electron Oxidation)

(GER= Gain of Electron Reduction)
Step 2:
`Fe^(2+)+Fe^(3+).....(1)(2to3=1 unitON)`
`Cr_2O_7^(2-)to2Cr^(3+)....(2)(+6to+3=3 units 2 sets implies 6 units ON)`
Step 3: Cross Multiplication i.e., (1) `times`6 and (2) `times` 1
`(1)times6to6Fe^(2+)to6Fe^(3+)......(3)`
`(2)times1toCr_2O_7^2to2Cr^(3+).....(4)`
Step 4: Add (3)+(4) `to6Fe^(2+)+CrO_7^(2-)to 6Fe^(3+)+2Cr^(3+)`
Step 5: Balance oxygen and then hydrogen
`6Fe^(2+)+CrO_7^(2-)+14H^(+) to 6Fe^(3+)+2Cr^(3+)+7H_2O`