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+) + Cr_2 O_(7) 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 change in oxidation number: `overset(+2)(Fe^(2+)) + overset(+6 -2)(Cr_2O_(7)^(2-) ) to overset(+3)(Fe^(3+)) + overset(+3)( 2Cr^(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)(6 Fe^(2+)) + overset(+6+2)(Cr_2 O_(7)^(2-)) to overset(+3)(6 Fe^(3+)) + (2Cr^(3+))`
Step 4: As the reaction occurs in acidic medium and the ionic charges are not equal on both sides, `14 H^(+)` are added on the left to make the ionic charges equal.
` 6 Fe^(2+) + Cr_(2)O_(7)^(2-) + 14H^(+) ro 6 Fe^(3+) + 2Cr^(3+)`
Step 5: Making the number of hydrogen atoms equal on both sides by adding `7H_2 O ` on the right hand side to compensate for `14H^(+)` on the left:
` 6 Fe^(2+) + Cr_2 O_(7)^(2-) + 14 H^(+) to 6 Fe^(3+) + 2 Cr^(3+) + 2Cr^(3+) + 7 H_2 O`
The number of oxygen atoms in the reactants and products are the same, therefore, the equation represents the balanced redox reaction.