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_(oxd)` will increases by a factor of `0.65` from `E_(oxd)^(@)`
B. `E_(oxd)` will decrease by a factor of `0.65` from `E_(oxd)^(@)`
C. `E_(red)` will increases by a factor of `0.65` from `E_(red)^(@)`
D. `E_(red)` will decreases by a factor of `0.65` from `E_(red)^(@)`

Explanation

To solve the problem, we need to analyze the effect of pH on the half-cell reactions provided. Let's break down the steps:

### Step 1: Understand the Reaction
We have several half-reactions, but we need to identify which one is affected by the change in pH when ammonia is added, raising the pH to 11. The relevant half-reaction that involves protons (H⁺) is:

\[ 2H^+ + 2e^- \rightarrow H_2 \]

### Step 2: Calculate the Concentration of H⁺ Ions
Given that the pH is raised to 11, we can calculate the concentration of H⁺ ions:

\[ \text{pH} = -\log[H^+] \]
\[ [H^+] = 10^{-\text{pH}} = 10^{-11} \, \text{M} \]

### Step 3: Use the Nernst Equation
The Nernst equation for the half-reaction can be expressed as:

\[ E_{\text{reduction}} = E^\circ - \frac{0.0592}{n} \log [\text{products}]/[\text{reactants}] \]

For the reaction \( 2H^+ + 2e^- \rightarrow H_2 \), we have:

- \( n = 2 \) (number of electrons transferred)
- \( E^\circ \) for the hydrogen electrode is 0 V.

Thus, we can rewrite the Nernst equation as:

\[ E_{\text{reduction}} = 0 - \frac{0.0592}{2} \log \left( \frac{1}{[H^+]^2} \right) \]

### Step 4: Substitute the Concentration of H⁺
Substituting \( [H^+] = 10^{-11} \):

\[ E_{\text{reduction}} = -\frac{0.0592}{2} \log \left( \frac{1}{(10^{-11})^2} \right) \]
\[ = -\frac{0.0592}{2} \log \left( 10^{22} \right) \]
\[ = -\frac{0.0592}{2} \times 22 \]
\[ = -0.0592 \times 11 \]
\[ = -0.6512 \, \text{V} \]

### Step 5: Calculate the Final E Reduction
Since we are looking for the increase in the reduction potential:

\[ E_{\text{reduction}} = 0 - (-0.6512) = 0.6512 \, \text{V} \]

### Conclusion
The half-cell reaction affected by pH is \( 2H^+ + 2e^- \rightarrow H_2 \), and the reduction potential increases by approximately 0.6512 V from its standard state.