Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Tollens’s reagent is used for the detection of aldehyde when a solution of $\text{ AgN}{{\text{O}}_{\text{3}}}\text{ }$ added to glucose with $\text{ N}{{\text{H}}_{\text{4}}}\text{OH }$ the gluconic acid is formed
$\begin{align}
  & \text{ A}{{\text{g}}^{\text{+}}}\text{ + }{{\text{e}}^{-}}\text{ }\to \text{ Ag ; E}_{\text{red}}^{\text{0}}=0.8\text{ V } \\
 & {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{12}}}{{\text{O}}_{\text{6}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\to \text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{12}}}{{\text{O}}_{\text{7}}}\text{(Gluconic acid) + 2}{{\text{H}}^{\text{+}}}\text{ + 2}{{\text{e}}^{-}}\text{ ; E}_{\text{red}}^{\text{o}}\text{ = }-\text{0}\text{.05 V } \\
 & \text{Ag(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{ + }{{\text{e}}^{-}}\text{ }\to \text{ Ag(s) + 2N}{{\text{H}}_{\text{3}}}\text{ } \\
\end{align}$
(Use $\text{ 2}\text{.302 }\!\!\times\!\!\text{ }\dfrac{\text{RT}}{\text{F}}\text{ = 0}\text{.0592 }$ and $\text{ }\dfrac{\text{F}}{\text{RT}}\text{=38}\text{.92 }$ at $\text{ 298 K }$ )
$\text{ 2A}{{\text{g}}^{\text{+}}}\text{ + }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{12}}}{{\text{O}}_{\text{6}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\to \text{ 2Ag(s) + }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{12}}}{{\text{O}}_{\text{7}}}\text{ + 2}{{\text{H}}^{\text{+}}}\text{ }$
Find$\text{ ln K }$ of this reaction?


seo-qna
Last updated date: 13th Jun 2024
Total views: 392.4k
Views today: 9.92k
Answer
VerifiedVerified
392.4k+ views
Hint: For a reversible reaction shown by the general reaction,
$\text{ aA + bB }\rightleftharpoons \text{ cC + dD }$
The Nernst equation is written in terms of the equilibrium constant is written below,
$\text{ }{{\text{E}}_{\text{cell}}}\text{= }\dfrac{\text{RT}}{\text{nF}}\text{ln}\dfrac{{{\left[ \text{C} \right]}^{\text{c}}}{{\left[ \text{D} \right]}^{\text{d}}}}{{{\left[ \text{A} \right]}^{\text{a}}}{{\left[ \text{B} \right]}^{\text{b}}}}\text{ }$
Where $\text{ K = }\dfrac{{{\left[ \text{C} \right]}^{\text{c}}}{{\left[ \text{D} \right]}^{\text{d}}}}{{{\left[ \text{A} \right]}^{\text{a}}}{{\left[ \text{B} \right]}^{\text{b}}}}\text{ }$ and K is an equilibrium constant for a reaction, R is gas constant, T is absolute temperature, n is the number of electrons in redox reaction and F is faraday's constant.

Complete step by step solution:
The nearest equation can be written in terms of equilibrium constant .It is given as follows,
$\text{ E}_{\text{cell}}^{\text{0}}\text{ = }\dfrac{\text{RT}}{\text{nF}}\text{ln K }$ (1)
Where $\text{ E}_{\text{cell}}^{\text{0}}\text{ }$ is a cell constant, R is gas constant, T is absolute temperature, n is the number of electrons involved in a redox reaction, F is faraday's constant and k is the equilibrium constant of the reaction.
We have given that silver from silver nitrate undergoes a reduction reaction. The reduction potential for the reaction is,
$\text{ A}{{\text{g}}^{\text{+}}}\text{ + }{{\text{e}}^{-}}\text{ }\to \text{ Ag E}_{\text{red}}^{\text{0}}=0.8\text{ V }$
Similarly, glucose (aldehyde) forms gluconic acid. the reduction potential for the oxidation reaction of glucose to gluconic acid is as shown below,
$\text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{12}}}{{\text{O}}_{\text{6}}}\text{ + }{{\text{H}}_{\text{2}}}\text{O }\to \text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{12}}}{{\text{O}}_{\text{7}}}\text{(Gluconic acid) + 2}{{\text{H}}^{\text{+}}}\text{ + 2}{{\text{e}}^{-}}\text{ ; E}_{\text{red}}^{\text{o}}\text{ = }-\text{0}\text{.05 V }$
Thus cell potential $\text{ E}_{\text{Cell}}^{\text{0}}\text{ }$ is the difference in reduction potential of reduction reaction and oxidation reaction. Thus standard electrode potential is determined as,
$\text{ E}_{\text{Cell}}^{\text{0}}\text{ }=\text{ }{{\left( \text{E}_{\text{red}}^{\text{0}} \right)}_{\text{Red}}}+{{\left( \text{E}_{\text{red}}^{\text{0}} \right)}_{\text{Ox}}}\text{ = 0}\text{.8 }-0.05\text{ = 0}\text{.75 V }$
Therefore electrode standard potential is equal to $\text{ 0}\text{.75 V }$ .
Substitute values in the equation (1) we have,
$\text{ 0}\text{.75 = }\dfrac{\text{RT}}{\text{nF}}\text{ln K = 2}\text{.302 }\dfrac{\text{RT}}{\text{nF}}\text{log K }$ (2)
We are requested to use $\text{ 2}\text{.302 }\!\!\times\!\!\text{ }\dfrac{\text{RT}}{\text{F}}\text{ = 0}\text{.0592 }$ and $\text{ }\dfrac{\text{F}}{\text{RT}}\text{=38}\text{.92 }$ . Let's substitute all value given in the equation (2) we get the following relation,
$\text{ 0}\text{.75 = }\dfrac{\text{RT}}{\text{2}\times \text{F}}\text{ ln K = }\dfrac{1}{2}\times \dfrac{0.0592}{2.303}\ln \text{ K }$
Rearrange above equation with respect to the natural logarithmic value of equilibrium constant K is given as,
$\text{ ln K = }\dfrac{\left( 0.75 \right)\left( 2\times 2.303 \right)}{0.0592}\text{ = 58}\text{.38 }$
Thus the correct answer of the ln K value is $\text{ 58}\text{.38 }$ .

Hence, (B) is the correct option.

Note: Note that, for electrochemical reaction value of equilibrium constant K is very large.it indicates that electrochemical reactions are more favoured towards the product. Remember that the negative value of $\text{ }{{\text{E}}_{\text{cell}}}\text{ }$ means that the cell under study is not feasible or not possible. Thus the value of $\text{ }{{\text{E}}_{\text{cell}}}\text{ }$is used to determine whether the cell is spontaneous or not.