Balance the following equation by ion electron method.
`Cr_(2) O_(7)^(2-) + Fe^(2+) rarr Cr^(3+) + Fe^(2+)` in acidic medium.

Explanation

`Cr_(2) O_(7)^(-2) rarr Cr^(+3)` ( reduction )( The O.N. of Cr decreases from `+6` to `+3`)
`Fe^(+2) rarr Fe^(+3)` ( oxidation ) ( The O.N. of Fe increases from `+2` to `+3` )
Balancing oxidation and reduction half reactions separately.
`Cr_(2) O_(7)^(_2) rarr Cr^(+3) ` `Fe^(+2) rarr Fe^(+3)`
Step-1 `:` Balance all atoms, other than hydrogen and oxygen.
`Cr_(2) o_(7)^(-2) rarr 2Cr^(+3) ` `Fe^(+2) rarr Fe^(+3)`
Step - 2 `:` Balance oxygen atom by adding equal number of `H_(2) O` molecules on the side deficient of oxygen.
`Cr_(2) O_(7)^(-2) rarr 2Cr^(+3) + 7H_(2) O`
`Fe^(+2) rarr Fe^(+3) `
Step -3 `:` To balance , hydrogen atom in acidic medium, add equal number of `H^(+)` ions on the side deficient of hydrogen atom.
`14H^(+) + Cr_(2) O_(7)^(-2) rarr 2Cr^(+3) + 7H_(2) O`
`Fe^(+2) rarr Fe^(+3)`
Step-4 `:` Balance the charges in such a way the charges on either side be the same,
`6e+ Cr_(2) O_(7)^(_2) + 14 H^(+) rarr 2Cr^(+3) + 7H_(2) O`
`Fe^(+2) rarr Fe^(+3)`
Add both equations in such a way the number of electrons taen up by the oxidising agent and the number of electron given by the reducing agent be the same,
`Cr_(2) O_(7)^(-2) + 14H^(+) + 6e rarr 2Cr^(+3) + 7H_(2) O`
`6e^(+2) rarr Fe^(+3) + e xx 6`
`bar( Cr_(2)O+(7)^(-2) + 14H^(+) + 6Fe^(+2) rarr 2r^(+3) + 6Fe^(+3) + 7H_(2) O)`