Accidentally chewing on a stray fragmet of aluminium foil can causes a sharp tooth path if the aluminium comes in contact with an amalgam filing. The filing, an alloy of silver, tin and mercury, acts as the cathode of a tiny galvanic cell, the aluminium behaves as the anode, and salivas serves as the electrolyte. when the aluminium and the filling come in contact, and electric current passage from the aluminium to the filling which is sensed by a nerve in the tooth. Aluminium is oxidized at the anode, and `O_(2)` gas is reduced to water at the cathode.
`E_(AI^(3+)//AI)^(@) =- 1.66 E_(O_(2)H^(+)//H_(2)O)6^(@) = 1.23 V`
The standard reduction potential of the reaction,
`H_(2)O +e^(-) rarr (1)/(2)H_(2)+OH^(-)` at `298K` is:

Options

A. `E^(@) =(RT)/(2F)In K_(w)`
B. `E^(@) =(RT)/(F)In ]P_(H_(2))]^(1//2) [OH^(-)]`
C. `E^(@) =(RT)/(F)In ([P_(H_(2))]^(1//2))/([H^(+)])`
D. `E^(@) =(RT)/(F)In K_(w)`

Answer

To solve the problem, we need to determine the standard reduction potential of the reaction given the provided standard reduction potentials of aluminum and oxygen. Here's a step-by-step solution:

### Step 1: Identify the half-reactions
The problem states that aluminum is oxidized at the anode and oxygen is reduced at the cathode. We can write the half-reactions as follows:

1. **Oxidation of Aluminum:**
\[ \text{Al} \rightarrow \text{Al}^{3+} + 3e^{-} \quad (E^\circ = -1.66 \, \text{V}) \]

2. **Reduction of Oxygen:**
\[ \text{O}_2 + 4e^{-} + 2\text{H}_2\text{O} \rightarrow 4\text{OH}^- \quad (E^\circ = 1.23 \, \text{V}) \]

### Step 2: Determine the overall cell reaction
To find the overall cell reaction, we need to balance the number of electrons transferred in both half-reactions. The oxidation of aluminum involves the transfer of 3 electrons, while the reduction of oxygen involves 4 electrons. The least common multiple of 3 and 4 is 12. Therefore, we will multiply the aluminum half-reaction by 4 and the oxygen half-reaction by 3:

1. **Oxidation of Aluminum (multiplied by 4):**
\[ 4\text{Al} \rightarrow 4\text{Al}^{3+} + 12e^{-} \]

2. **Reduction of Oxygen (multiplied by 3):**
\[ 3\text{O}_2 + 12e^{-} + 6\text{H}_2\text{O} \rightarrow 12\text{OH}^- \]

### Step 3: Write the overall cell reaction
Combining the two half-reactions gives us the overall reaction:
\[ 4\text{Al} + 3\text{O}_2 + 6\text{H}_2\text{O} \rightarrow 4\text{Al}^{3+} + 12\text{OH}^- \]

### Step 4: Calculate the standard cell potential
The standard cell potential \(E^\circ_{\text{cell}}\) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = 1.23 \, \text{V} - (-1.66 \, \text{V}) = 1.23 \, \text{V} + 1.66 \, \text{V} = 2.89 \, \text{V} \]

### Step 5: Determine the standard reduction potential for the reaction
The standard reduction potential for the reaction \( \text{H}_2\text{O} + e^- \rightarrow \frac{1}{2}\text{H}_2 + \text{OH}^- \) at 298 K can be derived from the Nernst equation. Given that the standard reduction potential for hydrogen is 0 V, we can express it as:
\[ E^\circ = E^\circ_{\text{cell}} - \left( \frac{RT}{nF} \ln Q \right) \]
Where \(Q\) is the reaction quotient. For this reaction, \(n = 1\).

### Final Answer
The standard reduction potential for the reaction at 298 K is:
\[ E^\circ = 2.89 \, \text{V} \]