Question

Acidified potassium dichromate cannot oxidize ________.

• A. Iodides to iodine
• B. Iron (II) salt to iron (III) salt
• C. Tin (II) salt to tin (IV) salt
• D. $H_{2}S$ to sulphur
• E. Fluoride to fluorine

Hint:

Nothing can chemically oxidize Fluoride because Fluorine is the ultimate oxidizing agent.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
Acidified potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+\)) is a strong oxidizing agent. An oxidizing agent causes another substance to be oxidized (lose electrons) while it is itself reduced. The ability of dichromate to oxidize a species depends on the standard electrode potentials of the two half-reactions. Dichromate can only oxidize a species if its own standard reduction potential is higher than that of the species it is trying to oxidize.
Step 2: Key Formula or Approach
The reduction half-reaction for dichromate is: \[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O}; \quad E^\circ = +1.33 \text{ V} \] For dichromate to oxidize a species 'X' to 'Y', the standard reduction potential of the Y/X couple must be less than +1.33 V. We are looking for the species that dichromate \textit{cannot} oxidize, which means its corresponding reduction potential will be higher than +1.33 V.
Step 3: Detailed Explanation
Let's check the reduction potentials for the species in the options.
(A) Iodides (\(\text{I}^-\)) to iodine (\(\text{I}_2\)): The reduction is \(\text{I}_2 + 2e^- \to 2\text{I}^-\), which has \(E^\circ = +0.54 \text{ V}\). Since \(+0.54<+1.33\), dichromate can oxidize iodide.
(B) Iron (II) (\(\text{Fe}^{2+}\)) to iron (III) (\(\text{Fe}^{3+}\)): The reduction is \(\text{Fe}^{3+} + e^- \to \text{Fe}^{2+}\), which has \(E^\circ = +0.77 \text{ V}\). Since \(+0.77<+1.33\), dichromate can oxidize Fe(II).
(C) Tin (II) (\(\text{Sn}^{2+}\)) to tin (IV) (\(\text{Sn}^{4+}\)): The reduction is \(\text{Sn}^{4+} + 2e^- \to \text{Sn}^{2+}\), which has \(E^\circ = +0.15 \text{ V}\). Since \(+0.15<+1.33\), dichromate can oxidize Sn(II).
(D) H\(_2\)S to sulphur (S): The reduction is \(\text{S} + 2\text{H}^+ + 2e^- \to \text{H}_2\text{S}\), which has \(E^\circ = +0.14 \text{ V}\). Since \(+0.14<+1.33\), dichromate can oxidize hydrogen sulfide.
(E) Fluoride (\(\text{F}^-\)) to fluorine (\(\text{F}_2\)): The reduction is \(\text{F}_2(g) + 2e^- \to 2\text{F}^-\), which has \(E^\circ = +2.87 \text{ V}\). Here, the reduction potential of fluorine is much higher than that of dichromate (\(+2.87>+1.33\)). This means F\(_2\) is a much stronger oxidizing agent than dichromate. Consequently, dichromate is not a strong enough oxidizing agent to take electrons from F\(^-\) to form F\(_2\).
Step 4: Final Answer
Acidified potassium dichromate cannot oxidize fluoride to fluorine.