Question

Tollen's reagent is used for the detection of aldehyde when a solution of $AgNO _{3}$ is added to glucose with $NH _{4} OH$ then gluconic acid is formed
$ Ag ^{+}+e^{-} \rightarrow Ag ; E_{ red }^{+}=0.8 \,V$
$ C _{6} H _{12} O _{6}+ H _{2} O \rightarrow $ Gluconic acid $\left( C _{6} H _{12} O _{7}\right)+2 H ^{+}+2 e^{-} $
$E_{ red }{ }^{\circ}=-0.05 \,V $
$ Ag \left( NH _{3}\right)_{2}{ }^{+}+e^{-} \rightarrow Ag ( s )+2 NH _{3} ; E_{ red }{ }^{\circ}=0.337 \,V$
[Use $2.303 \times \frac{R T}{F}=0.0592$ and $ \frac{F}{R T}=38.92$ at $398\, K ]$
$2 Ag ^{+}+ C _{6} H _{12} O _{6}+ H _{2} O \rightarrow 2 Ag ( s )+ C _{6} H _{12} O _{7}+2 H ^{+}$ Find In $K$ of this reaction.

• A. 66.13
• B. 58.38
• C. 28.30
• D. 46.29

Answer 1

Answer: (B)

We know that
$E_{\text {cell }}{ }^{\circ}=\frac{R T \ln K}{n F}$
We know that $E_{\text {cell }}{ }^{\circ}=E_{\text {anode }}{ }^{\circ}-E_{\text {cathode }}{ }^{\circ}$
and at equilibrium $E_{\text {cell }}=0 .$ Substituting given values:
$0.8-0.05=\left(\frac{1}{2}\right) \times\left(\frac{0.0592}{2.303}\right) \ln K $
$\ln L=(0.8-0.05) \times 2 \times\left(\frac{2.303}{0.0592}\right)=58.38$