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
\(\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} E_{\text {red }}^{\circ}=0.8 \mathrm{~V} \)
\( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text { Gluconic acid }\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)\) \(+~2 \mathrm{H}^{+}+2 e^{-} ; -E_{\mathrm{red}}^{\circ}=-0.05 \mathrm{~V} \)
\( \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ; E_{\mathrm{red}}^{\circ}\) \(=0.337 \mathrm{~V}\)
[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.

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
\(2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{Ag}(s)~+\) \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O} 7+2 \mathrm{H}^{+} \)
\( E_{\text {cell }}^{\circ} =(0.8-0.05)=0.75 \mathrm{~V} \)
\( E_{\text {cell }}^{\circ} =E_{\text {anode }}^{\circ}-E_{\text {cathode }}^{\circ}=[-0.05\) \(-(-0.8)]_{\mathrm{V}}\)
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, \(E_{\text {cell }}=0\)
\(E_{\text {cell }}^{\circ} =\frac{2.303 R T}{n F} \log _{10} K \)
\( 0.75 =\frac{0.0592}{2} \log _{10} K \)
\( \log _{10} K =\frac{2 \times 0.75}{0.0592}=25.33\)
or In \(K=2.303 \times \log _{10} K=2.303 \times 25.33=58.35\)