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 teh 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`
Net reaction taking place when amalgam is in contact with aluminium foil:

Options

A. `AI + O_(2) +OH^(-) rarr AI(OH^(-)) +H_(2)O`
B. `4AI +3O_(2)+12H^(+) rarr 4AI^(3+) +6H_(2)O`
C. `4AI +3O_(2) rarr 4AI_(2)O_(3)`
D. `2H_(2)+O_(2) rarr 2H_(2)O`

Answer

To solve the problem of the net reaction taking place when amalgam is in contact with aluminum foil, we will break down the steps involved in the electrochemical reactions occurring at the anode and cathode.

### Step 1: Identify the half-reactions
In this scenario, we have two half-reactions: one occurring at the anode (oxidation of aluminum) and one at the cathode (reduction of oxygen).

**Anode Reaction (Oxidation of Aluminum):**
\[ \text{Al} \rightarrow \text{Al}^{3+} + 3e^- \]

**Cathode Reaction (Reduction of Oxygen):**
\[ \text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O} \]

### Step 2: Balance the half-reactions
Next, we need to balance the half-reactions in terms of electrons transferred. The anode reaction produces 3 electrons, while the cathode reaction consumes 4 electrons. To balance the number of electrons, we can multiply the anode reaction by 4 and the cathode reaction by 3.

**Balanced Anode Reaction:**
\[ 4\text{Al} \rightarrow 4\text{Al}^{3+} + 12e^- \]

**Balanced Cathode Reaction:**
\[ 3\text{O}_2 + 12\text{H}^+ + 12e^- \rightarrow 6\text{H}_2\text{O} \]

### Step 3: Combine the half-reactions
Now that both half-reactions have the same number of electrons, we can combine them to get the net reaction.

\[ 4\text{Al} + 3\text{O}_2 + 12\text{H}^+ \rightarrow 4\text{Al}^{3+} + 6\text{H}_2\text{O} \]

### Step 4: Write the final net reaction
The final net reaction, which represents the overall process occurring when aluminum foil comes in contact with the amalgam filling, is:

\[ 4\text{Al} + 3\text{O}_2 + 12\text{H}^+ \rightarrow 4\text{Al}^{3+} + 6\text{H}_2\text{O} \]