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AIIMS 2010
Exam
AIIMS
Year
2010
Subject
Chemistry
Chapter
Ionic Equilibrium

AIIMS 2010 · Chemistry · Ionic Equilibrium

The standard half-cell reduction potential for Ag^ + Ag is 0.7991...

The standard half-cell reduction potential for Ag+∣Ag\mathrm{Ag}^{+} \mid \mathrm{Ag} is 0.7991 V0.7991 \mathrm{~V} at 25∘C25^{\circ} \mathrm{C}. Given the experimental value Ksp=1.56×10−10K_{s p}=1.56 \times 10^{-10} for AgCl\mathrm{AgCl}, calculate the standard half-cell reduction potential for the Ag∣AgCl\mathrm{Ag} \mid \mathrm{AgCl} electrode.

Options

  1. A0.2192 V0.2192 \mathrm{~V}
  2. B−0.2192 V-0.2192 \mathrm{~V}
  3. C−1.2192 V-1.2192 \mathrm{~V}
  4. D1.2192 V1.2192 \mathrm{~V}

Correct answer

A. 0.2192 V0.2192 \mathrm{~V}

Step-by-step solution

For the desired reaction, AgCl+\mathrm{AgCl}+1 e^{-} →Ag+Cl−,\rightarrow \mathrm{Ag}+\mathrm{Cl}^{-}, \quadE^{{}^{\circ}}=−ΔrG∘(1)F\frac{-\Delta_r G^{{}^{\circ}}}{(1) F}The needed ΔrG∘\Delta_r G^{{}^{\circ}} can be obtained by adding the values of ΔrG∘\Delta_r G^{{}^{\circ}} for the reactions Ag++1e−→Ag;\DeltaG∘=−nFE∘AgCl→Ag++Cl−;\DeltaG∘=−RTln⁡Ksp, giving ΔrG∘=−(1 mol)(9.648×104 J mol−1 V−1)(0.7991 V)−(8.314JK−1 mol−1)(298 K)ln⁡(1.56×10−10)=−77.10 kJ+55.95 kJ=−21.15 kJ\begin{aligned}& \mathrm{Ag}^{+}+1 e^{-} \rightarrow \mathrm{Ag}; \DeltaG^{{}^{\circ}}=-n F E^{{}^{\circ}} \\ & \mathrm{AgCl} \rightarrow \mathrm{Ag}^{+}+\mathrm{Cl}^{-}; \DeltaG^{{}^{\circ}}=-R T \ln K_{s p}, \text{ giving }\\ & \Delta_r G^{{}^{\circ}}=-(1 \mathrm{~mol})\left(9.648 \times10^4 \mathrm{~J} \mathrm{~mol}{ }^{-1} \mathrm{~V}^{-1}\right)(0.7991 \mathrm{~V})\\ & \quad-\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(298 \mathrm{~K}) \ln \left(1.56 \times10^{-10}\right) \\ & =-77.10 \mathrm{~kJ}+55.95 \mathrm{~kJ}=-21.15 \mathrm{~kJ} \end{aligned}The potential is E∘=−(−21.15)(1)(96.485)=+0.2192 VE^{{}^{\circ}}=\frac{-(-21.15)}{(1)(96.485)}=+0.2192 \mathrm{~V}.
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