JEE Challenger
More from Electrochemistry
ChemEasyAdv 2007 • Paper 2

Identification of Spontaneous Redox Reaction from Potentials

Comprehension Passage

Redox reactions play a pivotal role in chemistry and biology. The values of standard redox potential (E∘E^\circ) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zinc goes into solution and copper gets deposited.

Given below are a set of half-cell reactions (acidic medium) along with their E∘E^\circ (V\text{V} with respect to normal hydrogen electrode) values:

I2+2e−→2I−E∘=0.54 VFe3++e−→Fe2+E∘=0.77 VO2(g)+4H++4e−→2H2OE∘=1.23 VCl2+2e−→2Cl−E∘=1.36 VMn3++e−→Mn2+E∘=1.50 V\begin{aligned} \text{I}_2 + 2e^- &\rightarrow 2\text{I}^- && E^\circ = 0.54\text{ V} \\ \text{Fe}^{3+} + e^- &\rightarrow \text{Fe}^{2+} && E^\circ = 0.77\text{ V} \\ \text{O}_2(g) + 4\text{H}^+ + 4e^- &\rightarrow 2\text{H}_2\text{O} && E^\circ = 1.23\text{ V} \\ \text{Cl}_2 + 2e^- &\rightarrow 2\text{Cl}^- && E^\circ = 1.36\text{ V} \\ \text{Mn}^{3+} + e^- &\rightarrow \text{Mn}^{2+} && E^\circ = 1.50\text{ V} \end{aligned}

Among the following, identify the correct statement.

Options

A

Chloride ion is oxidised by O2\text{O}_2

B

Fe2+\text{Fe}^{2+} is oxidised by iodine

C

Iodide ion is oxidised by chlorine

Correct
D

Mn2+\text{Mn}^{2+} is oxidised by chlorine

Topics & Concepts

ElectrochemistryStandard Reduction PotentialRedox SpontaneityHalogens

Step-by-Step Solution

A species with a higher reduction potential (E∘E^\circ) will oxidize a species with a lower reduction potential:

  • E∘(Cl2/Cl−)=1.36 VE^\circ(\text{Cl}_2/\text{Cl}^-) = 1.36\text{ V}
  • E∘(I2/I−)=0.54 VE^\circ(\text{I}_2/\text{I}^-) = 0.54\text{ V}

For the reaction Cl2+2I−→2Cl−+I2\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2:
Ecell∘=E∘(Cl2/Cl−)−E∘(I2/I−)=1.36−0.54=+0.82 V>0E^\circ_{\text{cell}} = E^\circ(\text{Cl}_2/\text{Cl}^-) - E^\circ(\text{I}_2/\text{I}^-) = 1.36 - 0.54 = +0.82\text{ V} > 0
Because Ecell∘>0E^\circ_{\text{cell}} > 0, chlorine spontaneously oxidizes iodide ion to iodine.

Let us verify the other options:

  • A: E∘(O2/H2O)=1.23 V<E∘(Cl2/Cl−)=1.36 VE^\circ(\text{O}_2/\text{H}_2\text{O}) = 1.23\text{ V} < E^\circ(\text{Cl}_2/\text{Cl}^-) = 1.36\text{ V} (non-spontaneous).
  • B: E∘(I2/I−)=0.54 V<E∘(Fe3+/Fe2+)=0.77 VE^\circ(\text{I}_2/\text{I}^-) = 0.54\text{ V} < E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 0.77\text{ V} (non-spontaneous).
  • D: E∘(Cl2/Cl−)=1.36 V<E∘(Mn3+/Mn2+)=1.50 VE^\circ(\text{Cl}_2/\text{Cl}^-) = 1.36\text{ V} < E^\circ(\text{Mn}^{3+}/\text{Mn}^{2+}) = 1.50\text{ V} (non-spontaneous).

Correct Answer: C

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