What is `E^(@)` for electrode represented by `Pt,O_(2)(1atm)//2H^(+)(1m)`

Options

A. Unpredictable
B. Zero
C. 0.018V
D. 0.118V

Explanation

To find the standard electrode potential \( E^\circ \) for the electrode represented by \( \text{Pt, O}_2(1 \text{ atm}) // 2\text{H}^+(1 \text{ M}) \), we can follow these steps:

### Step 1: Identify the half-reaction
The half-reaction for the reduction of oxygen in acidic medium can be represented as:
\[ \text{O}_2 + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \]

### Step 2: Standard electrode potential for the half-reaction
The standard electrode potential \( E^\circ \) for this half-reaction is known to be:
\[ E^\circ = +1.23 \text{ V} \]
This value is typically found in standard electrochemical tables.

### Step 3: Consider the hydrogen electrode
The standard hydrogen electrode (SHE) is defined to have a potential of:
\[ E^\circ_{\text{H}_2} = 0 \text{ V} \]
The reaction for the hydrogen half-cell is:
\[ 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 \]

### Step 4: Combine the half-reactions
In the cell notation \( \text{Pt, O}_2(1 \text{ atm}) // 2\text{H}^+(1 \text{ M}) \), the oxygen half-reaction occurs at the cathode and the hydrogen half-reaction occurs at the anode.

### Step 5: Calculate the overall cell potential
The overall cell potential \( E^\circ_{\text{cell}} \) can be calculated using:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the known values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{O}_2} - E^\circ_{\text{H}_2} = 1.23 \text{ V} - 0 \text{ V} = 1.23 \text{ V} \]

### Conclusion
Thus, the standard electrode potential \( E^\circ \) for the electrode represented by \( \text{Pt, O}_2(1 \text{ atm}) // 2\text{H}^+(1 \text{ M}) \) is:
\[ E^\circ = 1.23 \text{ V} \]