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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2006 · Chemistry · Electrochemistry

Paragraph: Tollen's reagent is used for the detection of aldehyde...

Paragraph: Tollen's reagent is used for the detection of aldehyde when a solution of AgNO3\mathrm{AgNO}_3 is added to glucose with NH4OH\mathrm{NH}_4 \mathrm{OH} then gluconic acid is formed Ag++e−→AgEred ∘=0.8 VC6H12O6+H2O→ Gluconic acid (C6H12O7)+2H++2e−;−Ered∘=−0.05 VAg(NH3)2++e−→Ag(s)+2NH3;Ered∘=0.337 V\begin{array}{rr}\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} & E_{\text{red }}^{\circ}=0.8 \mathrm{~V} \\ \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text{ Gluconic acid }\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)+2 \mathrm{H}^{+}+2 e^{-} ; & -E_{\mathrm{red}}^{\circ}=-0.05 \mathrm{~V} \\ \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ; & E_{\mathrm{red}}^{\circ}=0.337 \mathrm{~V}\end{array} [Use 2.303×RTF=0.05922.303 \times \frac{R T}{F}=0.0592 and FRT=38.92\frac{F}{R T}=38.92 at 298 K298 \mathrm{~K} ]Question: 2Ag++C6H12O6+H2O→2Ag(s)+C6H12O4+2H+2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_4+2 \mathrm{H}^{+}. Find ln⁡K\ln \mathrm{K} of this reaction.

Options

  1. A66.1366.13
  2. B58.3858.38
  3. C28.3028.30
  4. D46.2946.29

Correct answer

B. 58.3858.38

Step-by-step solution

Cell reaction, (i) Ag++e−⟶Ag(s),Ered∘=0.8 V\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag}(\mathrm{s}), E_{\mathrm{red}}^{\circ}=0.8 \mathrm{~V} (ii) C6H12O6+H2O⟶C6H12O7+2H++2e−,Eoxid =−0.05 VHence,\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7+2 \mathrm{H}^{+}+2 e^{-}, E_{\text{oxid }}=-0.05 \mathrm{~V}Hence, for reaction 2Ag++C6H12O6+H2O⟶2Ag(s)+C6H12O7+2H+Ecell ∘=(0.8−0.05)=0.75 VEcell ∘=Eanode ∘−Ecathode ∘=[−0.05−(−0.8)]V\begin{aligned} 2\mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} & \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O} 7+2 \mathrm{H}^{+} \\ E_{\text{cell }}^{\circ} & =(0.8-0.05)=0.75 \mathrm{~V}\\ E_{\text{cell }}^{\circ} & =E_{\text{anode }}^{\circ}-E_{\text{cathode }}^{\circ}=[-0.05-(-0.8)]_{\mathrm{V}}\end{aligned} Onthebasisofoxidisedelectrodepotential=0.75 Vthe basis of oxidised electrode potential=0.75 \mathrm{~V} By Nernst equation, Ecell =Ecell ∘−2.303RTnFlog⁡10KE_{\text{cell }}=E_{\text{cell }}^{\circ}-\frac{2.303 R T}{n F} \log _{10} K At equilibrium, Ecell =0E_{\text{cell }}=0 Ecell ∘=2.303RTnFlog⁡10K0.75=0.05922log⁡10Klog⁡10K=2×0.750.0592=25.33\begin{aligned} E_{\text{cell }}^{\circ} & =\frac{2.303 R T}{n F}\log _{10} K \\ 0.75 & =\frac{0.0592}{2}\log _{10} K \\ \log _{10} K & =\frac{2 \times 0.75}{0.0592}=25.33 \end{aligned} or InK=2.303×log⁡10K=2.303×25.33=58.35InK=2.303 \times \log _{10} K=2.303 \times 25.33=58.35
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