Tollen’s reagent is used for the detection of aldehyde when a solution of AgNO3 is added to glucose with NH4OH, then gluconic acid is formed Ag++e−→Ag E0=0.8VC6H12O6+H2O→C6H12O7+2H++2e− Eoxi0=−0.05VAg(NH3)2++e−→Ag(s)+2NH3 E0=0.337VFor the following reaction Glucose +H2O+2Ag+⇌KeqGluconic acid+2Ag. Determine the value of lnKeq

In the given reaction. Ag+ ions reduces to Ag and Glucose is oxidizedto Gluconic acid as per the reaction Ag++e−→Ag;Ered0=+0.800V and C6H12O6+H2O→C6H12O6+2H++2e−;Eox0 =−0.05VGluconic acid Hence, Ecell0=−0.8−0.005=0.75VΔGcell0=−nFE=−2F×0.75=−RT In K ⇒In k=2FRT(0.75)=2×38.92×0.75=58.38

Tollen’s reagent is used for the detection of aldehyde when a solution of AgNO3 is added to glucose with NH4OH, then gluconic acid is formed Ag++e−→Ag E0=0.8VC6H12O6+H2O→C6H12O7+2H++2e− Eoxi0=−0.05VAg(NH3)2++e−→Ag(s)+2NH3 E0=0.337VFor the following reaction Glucose +H2O+2Ag+⇌KeqGluconic acid+2Ag. Determine the value of lnKeq

In the given reaction. Ag+ ions reduces to Ag and Glucose is oxidizedto Gluconic acid as per the reaction Ag++e−→Ag;Ered0=+0.800V and C6H12O6+H2O→C6H12O6+2H++2e−;Eox0 =−0.05VGluconic acid Hence, Ecell0=−0.8−0.005=0.75VΔGcell0=−nFE=−2F×0.75=−RT In K ⇒In k=2FRT(0.75)=2×38.92×0.75=58.38

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Tollen’s reagent is used for the detection of aldehyde when a solution of $A g N O_{3}$ is added to glucose with $N H_{4} O H,$ then gluconic acid is formed
$A g^{+} + e^{-} \to A g$ $E^{0} = 0.8 V$
$C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{7} + 2 H^{+} + 2 e^{-}$ $E_{o x i}^{0} = - 0.05 V$
$A g {(N H_{3})}_{2}^{+} + e^{-} \to A g (s) + 2 N H_{3}$ $E^{0} = 0.337 V$
For the following reaction
Glucose $+ H_{2} O + 2 A g^{+} \overset{K e q}{⇌} G l u c o n i c a c i d + 2 A g .$ Determine the value of $\ln K_{e q}$

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46.2

66.13

58.38

28.3

answer is A.

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Detailed Solution

In the given reaction. ${Ag}^{+}$ ions reduces to Ag and Glucose is oxidizedto Gluconic acid as per the reaction
${Ag}^{+} + e^{-} \to A g; E_{r e d}^{0} = + 0.800 V a n d C_{6} H_{12} O_{6} + H_{2} O \to C_{6} H_{12} O_{6} + 2 H^{+} + 2 e^{-}; E_{o x}^{0} = - 0.05 V$
Gluconic acid
Hence, $E_{c e l l}^{0} = - 0.8 - 0.005 = 0.75 V$
$Δ G_{c e l l}^{0} = - n F E = - 2 F \times 0.75 = - R T I n K \Rightarrow I n k = \frac{2 F}{R T} (0.75) = 2 \times 38.92 \times 0.75 = 58.38$