studybloomer
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.A disproportionation reaction occurs when a species M+ spontaneously undergoes simultaneous oxidation and reduction.

2M+(aq) → M2+(aq) + M(s)

The table below contains E data for copper and mercury species.

E / V
Cu2+(aq) + e− → Cu+(aq) + 0.15
Cu+(aq) + e− → Cu(s) + 0.52
Hg2+(aq) + e− → Hg+(aq) + 0.91
Hg+(aq) + e− → Hg(l) + 0.80

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

A Both Cu(I) and Hg(I) undergo disproportionation.
B Only Cu(I) undergoes disproportionation.
C Only Hg(I) undergoes disproportionation.
D Neither Cu(I) nor Hg(I) undergoes disproportionation.
Answer: B



Q7.Use the data in the table below to answer this question.

E / V
MnO (aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) + 1.52
Cr2O (aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) + 1.33
Fe3+(aq) + e− → Fe2+(aq) + 0.77
Cr3+(aq) + e− → Cr2+(aq) − 0.41
Zn2+(aq) + 2e− → Zn(s) − 0.76

Which one of the following statements is not correct?

A Fe2+(aq) can reduce acidified MnO (aq) to Mn2+(aq)
B CrO (aq) can oxidise acidified Fe2+(aq) to Fe3+(aq)
C Zn(s) can reduce acidified Cr2O (aq) to Cr2+(aq)
D Fe2+(aq) can reduce acidified Cr3+(aq) to Cr2+(aq)
Answer: B

Please help, as I am really stuck on these questions. Thank you
Last edited by studybloomer; 9 months ago
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charco
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(Original post by studybloomer)
.A disproportionation reaction occurs when a species M+ spontaneously undergoes simultaneous oxidation and reduction.

2M+(aq) → M2+(aq) + M(s)

The table below contains E data for copper and mercury species.

E / V
Cu2+(aq) + e− → Cu+(aq) + 0.15
Cu+(aq) + e− → Cu(s) + 0.52
Hg2+(aq) + e− → Hg+(aq) + 0.91
Hg+(aq) + e− → Hg(l) + 0.80

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

A Both Cu(I) and Hg(I) undergo disproportionation.
B Only Cu(I) undergoes disproportionation.
C Only Hg(I) undergoes disproportionation.
D Neither Cu(I) nor Hg(I) undergoes disproportionation.
Answer: B

Please help, as I am really stuck on these questions. Thank you
For a reaction to be spontaneous the proposed (cell) reaction must be positive.
E(cell) = E(red) - E(ox)

For the copper disproprortionation:
The reduced half-equation
Cu+(aq) + e− → Cu(s) +0.52V
The oxidised half-equation:
Cu2+(aq) + e− → Cu+(aq) +0.15V (this one is driven to the LHS)

E(cell) = 0.52 - 0.15 = +0.37V
In other words the disproportionation is thermodynamically spontaneous.

For the mercury disproportionation:
The reduction half-equation
Hg+(aq) + e− → Hg(l) +0.80V
The oxidation half-equation
Hg2+(aq) + e− → Hg+(aq) +0.91V (this one driven to the LHS)

E(cell) = 0.80 - 0.91 = -0.11V
Hence, the disproportionation is NOT thermodynamically spontaneous.
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