Tollen's test is used for detection of aldehydic group in organic compounds like glucose. The test is as follows: An aqueous solution of is added to an aqueous solution of glucose in a test tube. When concentrated ammonia is added, silver is deposited. T

Question

Tollen's test is used for detection of aldehydic group in organic compounds like glucose. The test is as follows: An aqueous solution of ${AgNO}_{3}$ is added to an aqueous solution of glucose in a test tube. When concentrated ammonia is added, silver is deposited. This is a redox reaction as ${Ag}^{+}$ is reduced to $Ag$ and glucose is oxidized to gluconic acid. The following half-cell reactions and the standard half-cell potentials are given:

${Ag}^{+} + e^{-} \to Ag E_{red}^{\circ} = 0.8 V$

$C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{7} + 2 H^{+} + 2 e^{-} E_{ox}^{\circ} = - 0.05 V$

${[Ag {({NH}_{3})}_{2}]}^{+} + e^{-} \to Ag + 2 {NH}_{3} E_{red}^{\circ} = 0.37 V$

Given: $2.303 \frac{R T}{F} = 0.0592 V, \frac{F}{R T} = 38.90 V^{- 1} at 298 K$

Then find the value of the logarithm of the equilibrium constant ( $\ln K$ ) of the following reaction:

$C_{6} H_{12} O_{6} + 2 {Ag}^{+} + H_{2} O ⇌ C_{6} H_{12} O_{7} + 2 H^{+} + 2 Ag ↓$

[Report your answer by rounding off to the nearest integer value.]

Smart Achievers · Accepted Answer

(58)

$Create Cell as:$

$\begin{matrix} Cathode: & {Ag}^{+} + e^{-} ⟶ Ag & E_{reduction}^{\circ} = 0.8 V \\ Anode: & \frac{C_{6} H_{12} O_{6} + H_{2} O ⟶ C_{6} H_{12} O_{7} + 2 H^{+} + 2 e^{-}}{2 {Ag}^{+} + C_{6} H_{12} O_{6} + H_{2} O ⟶ 2 Ag + C_{6} H_{12} O_{7} + 2 H^{+}} & E_{reduction}^{\circ} = 0.05 V \end{matrix}$

When the reaction reaches equilibrium, $Q_{cell} = K_{eq}$ and $E_{cell} = 0$

$Using: \ln K_{eq} = E_{cell}^{\circ} . n_{cell} . \frac{F}{R T} = 0.75 \times 2 \times 38.90 = 58.35$ $(E_{cell}^{\circ} = 0.8 - 0.05 = 0.75 V)$

Solution

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Solution

Ans. (58) Create Cell as: Cathode:Ag++e-⟶AgEreduction∘=0.8 VAnode:C6H12O6+H2O⟶C6H12O7+2H++2e-2Ag++C6H12O6+H2O⟶2Ag+C6H12O7+2H+Ereduction∘=0.05 V When the reaction reaches equilibrium, Qcell=Keq and Ecell=0 Using: lnKeq=Ecell∘ .ncell. FRT=0.75×2×38.90=58.35 (Ecell∘=0.8-0.05=0.75 V)

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