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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Chemistry
Chapter
Electrochemistry

BITSAT 2016 · Chemistry · Electrochemistry

The equilibrium constant for the disproportionation reaction 2 Cu^...

The equilibrium constant for the disproportionation reaction 2Cu+(aq)⟶Cu(s)+Cu2+(ag)2 \mathrm{Cu}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{ag}) at 25∘C(E∘Cu+/Cu=0.52V,E∘Cu2+/Cu=0.16V)25^{{}^{\circ}} \mathrm{C}\left(\mathrm{E}^{{}^{\circ}} \mathrm{Cu}^{+} / \mathrm{Cu}=0.52 \mathrm{ V}, \mathrm{E}^{{}^{\circ}} \mathrm{Cu}^{2+} / \mathrm{Cu}=0.16 \mathrm{ V}\right) is

Options

  1. A6×1046 \times 10^{4}
  2. B6×1066 \times 10^{6}
  3. C1.2×1061.2 \times 10^{6}
  4. D1.2×10−61.2 \times 10^{-6}

Correct answer

C. 1.2×1061.2 \times 10^{6}

Step-by-step solution

The reaction \begin{array}{l}2 \mathrm{Cu}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq})\\ \mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}^{{}^{\circ}}-\frac{0.0592}{1} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Cu}^{+}\right]^{2}} \end{array} At equilibrium Ecell=0∴Ecell ∘=0.0592log⁡Kc\mathrm{E}_{\mathrm{cell}}=0 \therefore \quad \mathrm{E}_{\text{cell }}^{{}^{\circ}}=0.0592 \log \mathrm{K}_{\mathrm{c}} or, log⁡Kc=0.52−0.160.0592∴Kc=1.2×106\quad \log \mathrm{K}_{\mathrm{c}}=\frac{0.52-0.16}{0.0592}\therefore \mathrm{K}_{\mathrm{c}}=1.2 \times 10^{6}
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