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Example 4.7: Balancing Redox Reactions in Acidic Solution

Balancing Redox Reactions in Acidic Solution

Write a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution.

Solution

Write the two half-reactions.
Each half-reaction will contain one reactant and one product with one element in common.



Balance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.



Balance oxygen atoms by adding \(\ce{H2O}\) molecules. The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side.



Balance hydrogen atoms by adding H^ +ions. The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side.



Balance charge by adding electrons. The iron half-reaction shows a total charge of 2 + on the left side ( \(\ce{1 Fe^2+}\) ion) and 3 + on the right side ( \(\ce{1 Fe^3+}\) ion). Adding one electron to the right side bring that side’s total charge to (3+) + (1-) \( = \) \(\ce{2+}\), and charge balance is achieved.
The chromium half-reaction shows a total charge of (1 × \(\ce{2-}\)) + (14 × 1 +) \( = 12 + \mathrm{on}\) the left side ( \(\ce{1 Cr2O7^2-}\) ion and \(\ce{14 H+}\) ions). The total charge on the right side is (2 × 3 +) \( = 6\) + ( \(\ce{2 Cr^3+}\) ions). Adding six electrons to the left side will bring that side’s total charge to (12 + + \(\ce{6-}\)) \( = 6\) +, and charge balance is achieved.



Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction’s coefficient must be multiplied by 6.



Add the balanced half-reactions and cancel species that appear on both sides of the equation.


Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:


A final check of atom and charge balance confirms the equation is balanced.