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 ]`
When ammonia is added to the solution, `pH` is raised to `11`. Which half cell reaction is affected by `pH` and by how much ?

Options

A. `E_(o x i d )` will increase by a factor of `0.65 ` from `E^(c-)._(o x i d)`.
B. `E_(o x i d )` will decrease by a factor of `0.65 ` from `E^(c-)._(o x i d)`.
C. `E_(red )` will increase by a factor of `0.65 ` from `E^(c-)._(red)`.
D. `E_(red )` will decrease by a factor of `0.65 ` from `E^(c-)._(red)`.

Explanation

To solve the problem, we need to analyze the half-cell reactions provided and determine which one is affected by the change in pH when ammonia is added to the solution, raising the pH to 11.

### Step 1: Identify the half-cell reactions
The half-cell reactions given are:
1. \( Ag^{+} + e^{-} \rightarrow Ag \) with \( E^{\circ}_{\text{red}} = 0.8 \, V \)
2. \( C_{6}H_{12}O_{6} \rightarrow \text{Gluconic acid} (C_{6}H_{12}O_{7}) + 2H^{+} + 2e^{-} \) with \( E^{\circ}_{\text{ox}} = -0.05 \, V \)
3. \( [Ag(NH_{3})_{2}]^{+} + e^{-} \rightarrow Ag(s) + 2NH_{3} \) with \( E^{\circ}_{\text{red}} = 0.337 \, V \)

### Step 2: Determine the effect of pH on the reactions
In electrochemistry, the pH of the solution can affect the concentration of \( H^{+} \) ions, which in turn can influence the equilibrium of reactions that involve \( H^{+} \).

Among the reactions, the second half-cell reaction involves \( H^{+} \):
\[ C_{6}H_{12}O_{6} \rightarrow C_{6}H_{12}O_{7} + 2H^{+} + 2e^{-} \]

### Step 3: Analyze the effect of pH on the second half-cell reaction
As the pH increases (which means \( [H^{+}] \) decreases), the equilibrium of the second half-cell reaction will shift to the left according to Le Chatelier's principle. This means that the formation of gluconic acid will be less favored, and thus the standard reduction potential \( E^{\circ}_{\text{ox}} \) will be affected.

### Step 4: Calculate the change in potential due to pH
The Nernst equation can be used to calculate the effect of pH on the potential:
\[ E = E^{\circ} - \frac{0.0592}{n} \log Q \]
where \( n \) is the number of moles of electrons transferred, and \( Q \) is the reaction quotient.

For the reaction:
\[ C_{6}H_{12}O_{6} \rightarrow C_{6}H_{12}O_{7} + 2H^{+} + 2e^{-} \]
Here, \( n = 2 \).

As the pH increases to 11, the concentration of \( H^{+} \) ions decreases. At pH 11, \( [H^{+}] = 10^{-11} \, M \).

### Step 5: Calculate the change in \( E \)
Using the Nernst equation, we can find the new potential:
\[ E = E^{\circ}_{\text{ox}} - \frac{0.0592}{2} \log \left( \frac{1}{[H^{+}]^2} \right) \]
Substituting \( [H^{+}] = 10^{-11} \):
\[ E = -0.05 - \frac{0.0592}{2} \log \left( \frac{1}{(10^{-11})^2} \right) \]
\[ = -0.05 - \frac{0.0592}{2} \log(10^{22}) \]
\[ = -0.05 - \frac{0.0592}{2} \times 22 \]
\[ = -0.05 - 0.0592 \times 11 \]
\[ = -0.05 - 0.6512 \]
\[ = -0.7012 \, V \]

### Conclusion
The half-cell reaction affected by pH is the oxidation of glucose to gluconic acid, and the potential decreases significantly due to the increase in pH.