Examkin

Question

Difficulty: MediumBalancing Redox Equations and Half-Reactions
Consider the redox reaction between dichromate ions (Cr2O72−\text{Cr}_2\text{O}_7^{2-}) and iron(II) ions (Fe2+\text{Fe}^{2+}) in an acidic medium:
Cr2O72−+xFe2++yH+→2Cr3++xFe3++zH2O\text{Cr}_2\text{O}_7^{2-} + x\text{Fe}^{2+} + y\text{H}^+ \rightarrow 2\text{Cr}^{3+} + x\text{Fe}^{3+} + z\text{H}_2\text{O}
What is the stoichiometric coefficient xx of Fe2+\text{Fe}^{2+} when the ionic equation is completely balanced?

Answer: 6

Answer

The stoichiometric coefficient x of Fe²⁺ in the balanced redox reaction is 6.
In the reduction half-reaction, dichromate (Cr2O72−\text{Cr}_2\text{O}_7^{2-}) contains two Cr atoms in the +6 oxidation state converting to two Cr3+\text{Cr}^{3+} ions in the +3 state, which consumes 6 electrons. In the oxidation half-reaction, each Fe2+\text{Fe}^{2+} ion loses 1 electron to form Fe3+\text{Fe}^{3+}. To balance charge transfer, 6 Fe2+\text{Fe}^{2+} ions are needed for every 1 Cr2O72−\text{Cr}_2\text{O}_7^{2-} ion, making the stoichiometric coefficient xx equal to 6.

Step-by-Step Solution

1
Determine the oxidation state changes for Chromium and Iron.
Chromium changes from +6 in Cr2O72−\text{Cr}_2\text{O}_7^{2-} to +3 in Cr3+\text{Cr}^{3+}, requiring 3 electrons per Chromium atom (6e−6e^- total for two Cr atoms). Iron changes from +2 in Fe2+\text{Fe}^{2+} to +3 in Fe3+\text{Fe}^{3+}, releasing 1e−1e^- per Iron atom.
Identifying the number of electrons transferred in each half-reaction is required to balance the overall redox equation.
2
Balance the electron gain and loss.
The oxidation half-reaction (Fe2+→Fe3++e−\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-) must be multiplied by 6 to balance the 6 electrons required by the dichromate ion.
The total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidizing agent.
3
Combine the half-reactions and read the coefficient xx.
The balanced chemical equation is Cr2O72−+6Fe2++14H+→2Cr3++6Fe3++7H2O\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}, giving x=6x = 6.
The coefficient xx corresponds directly to the stoichiometric multiplier applied to Fe2+\text{Fe}^{2+}.

Key Concept

Balancing Redox Equations via Half-Reactions in Acidic Medium
Rate this question

Topics

Balancing Redox Equations and Half-Reactions

More questions from this topic