Tollen reagent is used for the detection of aldehydes. When a solution of `AgNO_(3)` is added to glucose with `NH_(4)OH`, then gluconic acid is formed.
`Ag^(o+)+e^(-) rarr Ag," "E^(c-)._(red)=0.8V`
`C_(6)H_(12)O_(6)rarr underset(Gluconic aci d)(C_(6)H_(12)O_(7)+)2H^(o+)+2e^(-) , " "E^(c-)._(o x i d ) =-0.05V`
`[Ag(NH_(3))_(2)]^(o+)+e^(-) rarr Ag(s)+2NH_(3), " "E^(c-)._(red)=0.337V`
`[Use2.303xx(RT)/(F)=0.0592` and `(F)/(RT)=38.92at 298 K ]`
`2Ag^(o+)+C_(6)H^(12)O_(6)+H_(2)O rarr 2Ag^(s)+C_(6)H_(12)O_(7)+2H^(o+)` Find `lnK` of this reaction.

Options

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

Explanation

To find the value of \( \ln K \) for the given reaction, we will follow these steps:

### Step 1: Identify the Half-Reactions
We have the following half-reactions and their standard reduction potentials:
1. Reduction of silver ion:
\[ Ag^{+} + e^{-} \rightarrow Ag(s) \quad E^{\circ}_{\text{red}} = 0.8 \, \text{V} \]
2. Oxidation of glucose to gluconic acid:
\[ C_6H_{12}O_6 \rightarrow C_6H_{12}O_7 + 2H^{+} + 2e^{-} \quad E^{\circ}_{\text{ox}} = -0.05 \, \text{V} \]

### Step 2: Balance the Half-Reactions
To balance the number of electrons transferred, we multiply the oxidation half-reaction by 1 (it already has 2 electrons) and the reduction half-reaction by 2:
\[ 2Ag^{+} + 2e^{-} \rightarrow 2Ag(s) \quad E^{\circ}_{\text{red}} = 0.8 \, \text{V} \]
\[ C_6H_{12}O_6 \rightarrow C_6H_{12}O_7 + 2H^{+} + 2e^{-} \quad E^{\circ}_{\text{ox}} = -0.05 \, \text{V} \]

### Step 3: Calculate the Standard Cell Potential
The standard cell potential \( E^{\circ}_{\text{cell}} \) is calculated as follows:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.8 \, \text{V} - (-0.05 \, \text{V}) = 0.8 + 0.05 = 0.85 \, \text{V} \]

### Step 4: Use the Nernst Equation
The Nernst equation relates the cell potential to the equilibrium constant \( K \):
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log K \]
Where \( n = 2 \) (the number of electrons transferred). At equilibrium, \( E_{\text{cell}} = 0 \):
\[ 0 = 0.85 - \frac{0.0591}{2} \log K \]

### Step 5: Solve for \( \log K \)
Rearranging the equation gives:
\[ \frac{0.0591}{2} \log K = 0.85 \]
\[ \log K = \frac{0.85 \times 2}{0.0591} = \frac{1.7}{0.0591} \approx 28.8 \]

### Step 6: Convert to Natural Logarithm
To convert from base 10 logarithm to natural logarithm:
\[ \ln K = \log K \times 2.303 \approx 28.8 \times 2.303 \approx 66.24 \]

### Final Value of \( \ln K \)
Thus, the final value of \( \ln K \) is approximately:
\[ \ln K \approx 66.24 \]