Question

Please Balanced this Equation

Answer 1

Answer:

$\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O$.

Explanation:

Consider the oxidation state on each of the element:

Left-hand side:

• O: -2 (as in most compounds);
• Cr: $\displaystyle \frac{1}{2}(\underbrace{-2}_{\text{ion}} - \underbrace{7\times (-2)}_{\text{Oxygen}}) = +6$;
• Fe: +2 (from the charge of the ion);

Right-hand side:

• Cr: +3;
• Fe: +3.

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 $\rm Cr_2O_7^{2-}$ 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 $\rm H^{+}$ ions to produce water $\rm H_2O$. Seven O atoms will make seven water molecules. That's fourteen H atoms and hence fourteen $\rm H^{+}$ ions on the product side of this equation. Hence the balanced equation. Double check to ensure that the charges on the ions also balance.

$\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O$.