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 and their standard electrode potentials.
The half-reactions provided are:
1. Reduction of silver ion:
\[ Ag^{+} + e^{-} \rightarrow Ag(s) \quad E^{\circ}_{\text{red}} = 0.80 \, V \]
2. Oxidation of glucose to gluconic acid:
\[ C_{6}H_{12}O_{6} \rightarrow C_{6}H_{12}O_{7} + 2H^{+} + 2e^{-} \quad E^{\circ}_{\text{ox}} = -0.05 \, V \]

### Step 2: Calculate the standard cell potential \( E^{\circ}_{\text{cell}} \).
The standard cell potential is calculated using the formula:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \]
Substituting the values:
\[ E^{\circ}_{\text{cell}} = 0.80 \, V - (-0.05 \, V) = 0.80 \, V + 0.05 \, V = 0.85 \, V \]

### Step 3: Use the Nernst equation to relate \( E^{\circ}_{\text{cell}} \) to \( K \).
The Nernst equation at equilibrium (where \( E_{\text{cell}} = 0 \)) is:
\[ 0 = E^{\circ}_{\text{cell}} - \frac{0.0592}{n} \log K \]
Here, \( n = 2 \) (the number of moles of electrons transferred).

Rearranging the equation gives:
\[ \log K = \frac{n \cdot E^{\circ}_{\text{cell}}}{0.0592} \]
Substituting the values:
\[ \log K = \frac{2 \cdot 0.85}{0.0592} = \frac{1.70}{0.0592} \approx 28.66 \]

### Step 4: Convert \( \log K \) to \( \ln K \).
To convert from \( \log K \) to \( \ln K \), use the relationship:
\[ \ln K = 2.303 \cdot \log K \]
Substituting the value of \( \log K \):
\[ \ln K = 2.303 \cdot 28.66 \approx 65.95 \]

### Step 5: Final answer.
Thus, the value of \( \ln K \) for the reaction is approximately:
\[ \ln K \approx 65.95 \]