Standard electrode potential `Au^(+3)//Au` couple is 1.428 V and that
for `Na^(+)//Na` couple `-2.714 V` These two couples in their
standard state are connected to make a cell. The cell potential will
be :-

Options

A. 4.142
B. 1.286V
C. `-1.286 V`
D. `3.376 V`

Explanation

To find the cell potential (E°cell) for the electrochemical cell formed by the `Au^(+3)//Au` and `Na^(+)//Na` couples, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Standard Electrode Potentials**:
- For the `Au^(+3)//Au` couple, the standard electrode potential (E°) is given as:
\[ E^\circ_{\text{Au}^{+3}/\text{Au}} = 1.428 \, \text{V} \]
- For the `Na^(+)//Na` couple, the standard electrode potential (E°) is given as:
\[ E^\circ_{\text{Na}^+/\text{Na}} = -2.714 \, \text{V} \]

2. **Determine the Cathode and Anode**:
- The cathode is where reduction occurs, and the anode is where oxidation occurs.
- Since gold (Au) has a higher standard electrode potential, it will act as the cathode:
- Cathode reaction:
\[ \text{Au}^{+3} + 3e^- \rightarrow \text{Au} \]
- Sodium (Na), with a lower standard electrode potential, will act as the anode:
- Anode reaction:
\[ \text{Na} \rightarrow \text{Na}^+ + e^- \]

3. **Calculate the Cell Potential (E°cell)**:
- The formula for the cell potential is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substituting the values:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Au}^{+3}/\text{Au}} - E^\circ_{\text{Na}^+/\text{Na}} \]
\[ E^\circ_{\text{cell}} = 1.428 \, \text{V} - (-2.714 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 1.428 \, \text{V} + 2.714 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 4.142 \, \text{V} \]

4. **Final Answer**:
- The cell potential (E°cell) for the electrochemical cell is:
\[ E^\circ_{\text{cell}} = 4.142 \, \text{V} \]