Q30 / 120 · JEE Advanced 2006 · Chemistry
Paragraph:
Tollen's reagent is used for the detection of aldehyde when a solution of $\mathrm{AgNO}_3$ is added to glucose with $\mathrm{NH}_4 \mathrm{OH}$ then gluconic acid is formed
$\begin{array}{rr}\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} & E_{\text {red }}^{\circ}=0.8 \mathrm{~V} \\ \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text { Gluconic acid }\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)+2 \mathrm{H}^{+}+2 e^{-} ; & -E_{\mathrm{red}}^{\circ}=-0.05 \mathrm{~V} \\ \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ; & E_{\mathrm{red}}^{\circ}=0.337 \mathrm{~V}\end{array}$
[Use $2.303 \times \frac{R T}{F}=0.0592$ and $\frac{F}{R T}=38.92$ at $298 \mathrm{~K}$ ]Question:
$2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_4+2 \mathrm{H}^{+}$. Find $\ln \mathrm{K}$ of this reaction.
- A$66.13$
- Correct answerB$58.38$
- C$28.30$
- D$46.29$
Solution
Cell reaction,
(i) $\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag}(\mathrm{s}), E_{\mathrm{red}}^{\circ}=0.8 \mathrm{~V}$
(ii) $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7+2 \mathrm{H}^{+}+2 e^{-}, E_{\text {oxid }}=-0.05 \mathrm{~V}$
Hence, for reaction
$$
\begin{aligned}
2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} & \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O} 7+2 \mathrm{H}^{+} \\
E_{\text {cell }}^{\circ} & =(0.8-0.05)=0.75 \mathrm{~V} \\
E_{\text {cell }}^{\circ} & =E_{\text {anode }}^{\circ}-E_{\text {cathode }}^{\circ}=[-0.05-(-0.8)]_{\mathrm{V}}
\end{aligned}
$$
On the basis of oxidised electrode potential $=0.75 \mathrm{~V}$
By Nernst equation, $E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log _{10} K$
At equilibrium, $E_{\text {cell }}=0$
$$
\begin{aligned}
E_{\text {cell }}^{\circ} & =\frac{2.303 R T}{n F} \log _{10} K \\
0.75 & =\frac{0.0592}{2} \log _{10} K \\
\log _{10} K & =\frac{2 \times 0.75}{0.0592}=25.33
\end{aligned}
$$
or In $K=2.303 \times \log _{10} K=2.303 \times 25.33=58.35$
Hence, the correct answer is option: 'B'.