Answer:
[tex]\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O[/tex].
Explanation:
Consider the oxidation state on each of the element:
Left-hand side:
- O: -2 (as in most compounds);
- Cr: [tex]\displaystyle \frac{1}{2}(\underbrace{-2}_{\text{ion}} - \underbrace{7\times (-2)}_{\text{Oxygen}}) = +6[/tex];
- Fe: +2 (from the charge of the ion);
Right-hand side:
Change in oxidation state:
- Each Cr atom: decreases by 3 (reduction).
- Each Fe atom: increases by 1 (oxidation).
Changes in oxidation states shall balance each other in redox reactions. Thus, for each Cr atom on the left-hand side, there need to be three Fe atoms.
Assume that the coefficient of the most complex species [tex]\rm Cr_2O_7^{2-}[/tex] is 1. There will be two Cr atoms and hence six Fe atoms on the left-hand side. Additionally, there are going to be seven O atoms.
Atoms are conserved in chemical reactions. As a result, the right-hand side of this equation will contain
- two Cr atoms,
- six Fe atoms, and
- seven O atoms.
O atoms seldom appear among the products in acidic environments; they rapidly combine with [tex]\rm H^{+}[/tex] ions to produce water [tex]\rm H_2O[/tex]. Seven O atoms will make seven water molecules. That's fourteen H atoms and hence fourteen [tex]\rm H^{+}[/tex] ions on the product side of this equation. Hence the balanced equation. Double check to ensure that the charges on the ions also balance.
[tex]\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O[/tex].
