Balance the following equation in acid medium by Ion-electron method :
`Fe_("(aq)")^(+2) + Cr_(2) O_(7 "(aq)")^(2-) to Fe_("(aq") ^(3+)j + Cr_("(aq))^(+3)`

Explanation

1) write the reduction half reaction and oxidation half reaction separately
Oxidation `Fe^(2+) to Fe^(3+)`
Reduction `Cr_(2)O_(7(aq))^(2-)to Cr_((aq))^(3+)`
2) Balance the oxidation half reaction
`Fe^(2+)-e to Fe^(3+) ..... (1)`
3) Balance the reduction half reaction
`Cr_(2)O_(7)^(2-) to Cr^(3+)`
Balance Cr
`Cr_(2)O_(7)^(2-) to 2Cr^(3+)`
Balance O by adding `H_(2)O`
`Cr_(2)O_(7)^(2-) to 2Cr^(3+)+7H_(2)O`
Balance .H. by using `H^(+)` as the reaction is taking place in acid medium
`Cr_(2)O_(7)^(2-)+14H^(+) to 2Cr^(3+)+7H_(2)O`
Balance changes by adding .e.
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O ....... (2)`
Multiply equation (1) by `6` and add to (2)
`6Fe^(2+)-6e^(-) to 6Fe^(3+)`
`(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O)/(Cr_(2)O_(7)^(2-)+6Fe^(2+)+14H^(+) to 2Cr^(3+)+7H_(2)O+6Fe^(3+))`