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NEET CHEMISTRYMedium

For the disproportionation of copper: 2Cu+→Cu2++Cu2Cu^+ \rightarrow Cu^{2+} + Cu, E∘E^{\circ} is: (Given E∘E^{\circ} for Cu2+/CuCu^{2+}/Cu is 0.34 V0.34\text{ V} and E∘E^{\circ} for Cu2+/Cu+Cu^{2+}/Cu^+ is 0.15 V0.15\text{ V})

A

0.49 V

B

– 0.19 V

C

0.38 V

D

– 0.38 V

Step-by-Step Solution

Given: ECu2+/Cu∘=0.34 VE^{\circ}_{Cu^{2+}/Cu} = 0.34\text{ V} ECu2+/Cu+∘=0.15 VE^{\circ}_{Cu^{2+}/Cu^+} = 0.15\text{ V}

The corresponding half-reactions and their standard Gibbs free energy changes are:

  1. Cu2++2e−→Cu ; ΔG1∘=−nFE∘=−2×F×0.34=−0.68FCu^{2+} + 2e^- \rightarrow Cu \ ; \ \Delta G^{\circ}_1 = -nFE^{\circ} = -2 \times F \times 0.34 = -0.68 F
  2. Cu2++e−→Cu+ ; ΔG2∘=−1×F×0.15=−0.15FCu^{2+} + e^- \rightarrow Cu^+ \ ; \ \Delta G^{\circ}_2 = -1 \times F \times 0.15 = -0.15 F

To find the standard reduction potential for Cu+/CuCu^+/Cu, subtract reaction 2 from reaction 1: Cu++e−→Cu ; ΔG3∘=ΔG1∘−ΔG2∘=−0.68F−(−0.15F)=−0.53FCu^+ + e^- \rightarrow Cu \ ; \ \Delta G^{\circ}_3 = \Delta G^{\circ}_1 - \Delta G^{\circ}_2 = -0.68 F - (-0.15 F) = -0.53 F Since ΔG3∘=−nFECu+/Cu∘=−1×F×ECu+/Cu∘\Delta G^{\circ}_3 = -nFE^{\circ}_{Cu^+/Cu} = -1 \times F \times E^{\circ}_{Cu^+/Cu}, we get ECu+/Cu∘=0.53 VE^{\circ}_{Cu^+/Cu} = 0.53\text{ V}.

The disproportionation reaction is 2Cu+→Cu2++Cu2Cu^+ \rightarrow Cu^{2+} + Cu. The cell potential for this reaction is: Ecell∘=Ecathode∘−Eanode∘=ECu+/Cu∘−ECu2+/Cu+∘=0.53 V−0.15 V=0.38 VE^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}_{Cu^+/Cu} - E^{\circ}_{Cu^{2+}/Cu^+} = 0.53\text{ V} - 0.15\text{ V} = 0.38\text{ V}.

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