Write the net ionic equation for the reaction of potassium dichromate (VI), `K_2Cr_2O_7` with ferrous sulphate `(FeSO_4)` in sulphuric acid to ferric `(Fe^(3+))` ion and chromium(III) ion.

Explanation

Step 1: The skeletal ionic equation : `Fe^(2+) to Cr_2O_7^(2-) to Fe^(3+) + Cr^(3+)`
all atoms other than oxygen and hydrogen equal on both sides,
Step 2: Assigning oxidation numbers to all elements and identifying atoms which undergo charige in oxidation number: `overset(+2)(Fe^(2+)) + overset(+6)(Cr_2)overset(-2)(O_7^(2-)) to overset(+3)(Fe^(3+)) + overset(+3)(Cr^(3+))`
Dichromate ion is the oxidant and ferrous ion is the reductant.
Step 3: Calculating the increase and decrease of oxidation number and making them.equal:
`overset(+2)(6Fe^(2+)) + overset(+6)(Cr_2)overset(-2)(O_7^(2-)) to overset(+3)(6Fe^(3+)) + overset(+3)(2Cr^(3+))`
Step 4: As the reaction occurs in acidic medium and the ionic charges are not equal on both sides, `14H^(+)` are added on the left to make the ionic charges equal.
`6Fe^(2+) + Cr_2O_(7)^(2-) + 14H^(+) to 6Fe^(3+) + 2Cr^(3+)`
Step 5: Making the number of hydrogen atoms equal on both sides by adding `7H_2O` on the right hand side to compensate for `14H^(+)` on the left:
`6Fe^(2+) +Cr_2O_7^(2-) + 14H^(+) to 6Fe^(3+) + 2Cr^(3+) + 7H_2O`
The number of oxygen atoms in the reactants and products are the same, therefore, the equation represents the balanced redox reaction.