Answered

A disproportionation reaction occurs when a species $M^{+}$ spontaneously undergoes simultaneous oxidation and reduction.



$2M^{+}(aq) \rightarrow M^{2+}(aq) + M(s)$



The table below contains $E^{\ominus}$ data for copper and mercury species.




E \Theta / V

$Cu^{2+}(aq) + e^{-} \rightarrow Cu^{+}(aq)$		+0.15

$Cu^{+}(aq) + e^{-} \rightarrow Cu(s)$		+0.52

$Hg^{2+}(aq) + e^{-} \rightarrow Hg^{+}(aq)$		+0.91

$Hg^{+}(aq) + e^{-} \rightarrow Hg(l)$		+0.80






Using these data, which one of the following can be predicted?


ABoth $Cu(I)$ and $Hg(I)$ undergo disproportionation.
BOnly $Cu(I)$ undergoes disproportionation.
COnly $Hg(I)$ undergoes disproportionation.
DNeither $Cu(I)$ nor $Hg(I)$ undergoes disproportionation.

Verified Answer :

  • The disproportionation reaction \(2M^+ \rightarrow M^{2+} + M\) is spontaneous if the cell potential \(E^\ominus_{cell} = E^\ominus_{red}(M^+/M) - E^\ominus_{red}(M^{2+}/M^+) > 0\).
  • For copper(I), \(E^\ominus_{cell} = 0.52\,V - 0.15\,V = +0.37\,V\), which is positive, so it disproportionates.
  • For mercury(I), \(E^\ominus_{cell} = 0.80\,V - 0.91\,V = -0.11\,V\), which is negative, so it does not disproportionate.
  • Only copper(I) undergoes spontaneous disproportionation, making the final answer \(\boxed{B}\).

Explanation

  1. Analysis of Disproportionation and Cell Potential

A disproportionation reaction is defined as a process where a species undergoes both oxidation and reduction simultaneously. The reaction provided is:
\[2M^+(aq) \rightarrow M^{2+}(aq) + M(s)\]
To analyze whether this reaction is spontaneous, we split it into two half-reactions:

  1. Reduction (at the cathode): \(M^+(aq) + e^- \rightarrow M(s)\) with potential \(E^\ominus_{red}(M^+/M)\)
  2. Oxidation (at the anode): \(M^+(aq) \rightarrow M^{2+}(aq) + e^-\) with potential \(E^\ominus_{ox}(M^+/M^{2+}) = -E^\ominus_{red}(M^{2+}/M^+)\)

The overall standard cell potential for the disproportionation is given by:
\[E^\ominus_{cell} = E^\ominus_{red}(M^+/M) - E^\ominus_{red}(M^{2+}/M^+)\]
A reaction is spontaneous if the standard cell potential \(E^\ominus_{cell}\) is greater than \(0\).

  1. Calculations for Copper and Mercury Species

Now we apply the derived formula \(E^\ominus_{cell} = E^\ominus_{red}(M^+/M) - E^\ominus_{red}(M^{2+}/M^+)\) to both copper(I) and mercury(I).

For Copper(I):

  • Given \(E^\ominus(Cu^+/Cu) = +0.52 \, V\)
  • Given \(E^\ominus(Cu^{2+}/Cu^+) = +0.15 \, V\)
  • Calculation: \(E^\ominus_{cell}(Cu^+) = 0.52 \, V - 0.15 \, V = +0.37 \, V\)

Since \(+0.37 \, V > 0\), the disproportionation of \(Cu^+\) is spontaneous.

For Mercury(I):

  • Given \(E^\ominus(Hg^+/Hg) = +0.80 \, V\)
  • Given \(E^\ominus(Hg^{2+}/Hg^+) = +0.91 \, V\)
  • Calculation: \(E^\ominus_{cell}(Hg^+) = 0.80 \, V - 0.91 \, V = -0.11 \, V\)

Since \(-0.11 \, V < 0\), the disproportionation of \(Hg^+\) is not spontaneous.

  1. Conclusion

Based on our calculations, the standard cell potential for the disproportionation reaction of \(Cu(I)\) is positive, indicating that it proceeds spontaneously. Conversely, the potential for \(Hg(I)\) is negative, indicating that it does not undergo spontaneous disproportionation under standard conditions. Therefore, only \(Cu(I)\) can be predicted to undergo disproportionation, which corresponds to option B. The final answer is \(\boxed{B}\).

Examples

In chemistry, the spontaneity of reactions is crucial for understanding how substances interact and decompose. The study of disproportionation, as seen with copper, helps explain why copper(I) salts are often unstable in aqueous solutions, readily breaking down into copper(II) and solid copper. This is fundamental in electrochemistry, for example, when designing battery electrolytes where preventing unwanted self-reaction (disproportionation) ensures the battery remains stable over time. By calculating standard cell potentials, chemists can predict the stability of metal ions in various oxidation states, which is essential for synthesis and analysis in the laboratory.

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