Write ionic equation for the reaction between `Cr_(2)O_(7)^(2-)` ions and `Fe^(2+)` ions in acidic medium?

Explanation

Step 1: Unbalanced equation for the reaction in ionic form: `Fe^(2+) + Cr_2O_7 to Fe^(3+) + Cr^(3+)`
Step 2: Separating the equation into two half-reactions, the oxidation half and the reduction half.
`underset("Oxidation half-reaction")(overset(+2)(Fe^(2+)) to overset(+3)(Fe^(3+))), underset("Reduction half-reaction")(overset(+6)(Cr_2)overset(-2)(O_7) to overset(+3)(Cr^(3+)))`
Step 3: Balancing the atoms other than oxygen and hydrogen in each half reaction individually.
`Fe^(2+) to Fe^(3+) , Cr_2O_7^(2-) to 2Cr^(3+)`
Step 4: As the reaction is taking place in acidic medium, adding `H_2O` to balance O atoms and `H^(+)` to balance H atoms.
`Fe^(2+) to Fe^(3+) , Cr_2O_7^(2-) + 14H^(+) to 2Cr^(3+) + 7H_2O`
Step 5: Adding electrons to one side of the half reactions to balance charges:
`Fe^(2+) to Fe^(3+) + e^(-) , Cr_2O_7^(2-) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_2O`
Multiplyingoxidation half-reaction by 6 to make the number of electrons in both the half-reactions equal:
`6Fe^(2+) to 6Fe^(3+) + 6e^(-), Cr_2O_7^(2-) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_2O`
Step 6: Adding the two halfreactions and cancelling electrons on both sides.
`6Fe^(2+) + Cr_2O_(7)^(2-) + 14H^(+) to 6Fe^(3+) + 2Cr^(3+) + 7H_2O`