`Fe^(+2)` in acidic medium is convert `Cr_(2)O_(7)^(-2)` ion into `Cr^(+3)` ion by reduction `Fe^(+3)` is obtained balance these redox reaction with equation.

Explanation

Step-1 : Produce unbalanced equation for the reaction in ionic form :
`Fe_((aq))^(+2)+Cr_(2)O_(7(aq))^(-2)toFe_((aq))^(+3)+Cr_((aq))^(+3)`
Step-2 : Separate the equation into half reactions.
Oxidation half : `Fe_((aq))^(+2)toFe_((aq))^(+3)`
Reduction half : `Cr_(2)O_(7(aq))^(-2)toCr_((aq))^(+3)`
Step-3 : Balance the atoms other than O and H in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms. For the reduction half reaction, we multiply the `Cr^(3+)` by 2 to balance Cr atoms.
`Cr_(2)O_(7(aq))^(-2)to2Cr_((aq))^(+3)`
Step-4 : For reactions occurring in acidic medium, add `H_(2)O` to balance O atoms and `H^(+)` to balance H atoms.
`Cr_(2)O_(7(aq))^(-2)+14H_((aq))^(+)to2Cr_((aq))^(+3)+7H_(2)O_((l))`
Step-5 : Add electrons to one side of the half Oxidation half reaction to balance the charges. If need be, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate number.
`OHR:Fe_((aq))^(+2)toFe_((aq))^(+3)+e^(-)`
`RHR:Cr_(2)O_(7(aq))^(-2)+14H_((aq))^(+)+6^(-)to2Cr_((aq))^(+3)+7H_(2)O_((l))`
To equalise the number of electrons in both the half reaction, we add two water half reactions, we multiply the oxidation half reaction by 6 and write as : `6Fe_((aq))^(+2)to6Fe_((aq))^(+3)+6e^(-)`
Step-6: We add the two half reactions to achieve the overall reaction and cancel the electrons on each side.
`6Fe_((aq))^(+2)+Cr_(2)O_(7(aq))^(-2)+14H_((aq))^(+)to6Fe_((aq))^(+3)2Cr_((aq))^(+3)+7H_(2)O_((l))`