Question

Difficulty: MediumBalancing Redox Equations and Half-Reactions
Consider the following ionic equation for the redox reaction between dichromate(VI) ions and iodide ions in an acidic medium:
Cr2O72(aq)+xI(aq)+yH+(aq)2Cr3+(aq)+zI2(aq)+wH2O(l)\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + x\text{I}^-(\text{aq}) + y\text{H}^+(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + z\text{I}_2(\text{aq}) + w\text{H}_2\text{O}(\text{l})
What are the correct values of the stoichiometric coefficients xx, yy, and zz respectively when the equation is completely balanced?
  1. 6, 14, 3Answer
  2. B
    2, 14, 1
  3. C
    6, 7, 3
  4. D
    3, 14, 6

Answer

The correct stoichiometric coefficients for x, y, and z are 6, 14, and 3 respectively.
The reduction of one dichromate ion (Cr2O72\text{Cr}_2\text{O}_7^{2-}) requires 6 electrons and 14 hydrogen ions to yield two Cr3+\text{Cr}^{3+} ions and 7 water molecules. Oxidation of iodide ions (I\text{I}^-) to iodine (I2\text{I}_2) releases 2 electrons per molecule formed. To equalize electron exchange at 6 electrons, 6 moles of iodide ions (x=6x = 6) produce 3 moles of iodine molecules (z=3z = 3), requiring 14 moles of H+\text{H}^+ (y=14y = 14).

Step-by-Step Solution

1
Write and balance the reduction half-reaction for dichromate(VI) ions
Cr2O72+14H++6e2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
Chromium changes oxidation state from +6 to +3 (total 6 electrons gained for two Cr atoms), and 7 oxygen atoms require 14H+14\text{H}^+ to form 7H2O7\text{H}_2\text{O}.
2
Write and balance the oxidation half-reaction for iodide ions
2II2+2e2\text{I}^- \rightarrow \text{I}_2 + 2e^-
Iodine changes oxidation state from -1 to 0, losing 1 electron per iodide ion.
3
Equalize the number of electrons transferred in both half-reactions
Multiply the oxidation half-reaction by 3: 6I3I2+6e6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-
6 electrons lost must equal 6 electrons gained.
4
Combine the half-reactions and determine coefficients
Cr2O72+6I+14H+2Cr3++3I2+7H2O\text{Cr}_2\text{O}_7^{2-} + 6\text{I}^- + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}
Matching coefficients gives x=6x = 6, y=14y = 14, and z=3z = 3.

Key Concept

Balancing Redox Equations using Half-Reactions in Acidic Medium
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