Question

Tollen's reagent is used for the detection of aldehyde when a solution of $AgN{O}_3$ is added to glucose with $N{H}_{4}OH$ then gluconic acid is formed
$A{g}^{+}+e^{-}\to Ag;E_{red}^{+}=0.8V$
$C_6{H}_{12}{O}_6+H_{2}O\to$ Gluconic acid $\left(C_6{H}_{12}{O}_7\right)+2H^{+}+2e^{-}$
$E_{red}{}^{\circ }=-0.05V$
$Ag{\left(N{H}_3\right)}_2{}^{+}+e^{-}\to Ag(s)+2N{H}_3;E_{red}{}^{\circ }=0.337V$
[Use $2.303\times \frac{RT}{F}=0.0592$ and $\frac{F}{RT}=38.92$ at $398K]$
$2A{g}^{+}+C_6{H}_{12}{O}_6+H_{2}O\to 2Ag(s)+C_6{H}_{12}{O}_7+2H^{+}$ 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
$E_{\text{cell }}{}^{\circ }=\frac{RT\ln K}{nF}$
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:
$0.8-0.05=\left(\frac{1}{2}\right)\times \left(\frac{0.0592}{2.303}\right)\ln K$
$\ln L=(0.8-0.05)\times 2\times \left(\frac{2.303}{0.0592}\right)=58.38$