Given that `E^(@)(Zn^(2+)//Zn)=0.76V, E^(@)(H^(+)//H_(2))=0.00" V "`. What is the value of electrode potential of `Mg^(+)//Mg` electrode when it is dipped in a solution in which concentration of `Mg^(+)` is 0.01 M ? (Given `E_((Mg^(2+)//Mg))^(@)=-2.36" V "`

Explanation

To find the electrode potential of the Mg^(2+)//Mg electrode when it is dipped in a solution with a concentration of 0.01 M Mg^(2+), we can use the Nernst equation. The Nernst equation relates the standard electrode potential to the concentration of the ions involved in the half-reaction.

### Step-by-Step Solution:

1. **Identify the Standard Electrode Potential**:
The standard electrode potential for the Mg^(2+)//Mg half-reaction is given as:
\[ E^0(Mg^{2+}/Mg) = -2.36 \, V \]

2. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^0 - \frac{RT}{nF} \ln Q \]
where:
- \(E\) = electrode potential
- \(E^0\) = standard electrode potential
- \(R\) = universal gas constant = 8.314 J/(mol·K)
- \(T\) = temperature in Kelvin (assume 298 K for standard conditions)
- \(n\) = number of moles of electrons transferred in the half-reaction (for Mg, \(n = 2\))
- \(F\) = Faraday's constant = 96485 C/mol
- \(Q\) = reaction quotient, which for the reduction of Mg is given by \(\frac{1}{[Mg^{2+}]}\)

3. **Calculate the Reaction Quotient (Q)**:
Given that the concentration of \(Mg^{2+}\) is 0.01 M, we can write:
\[ Q = \frac{1}{[Mg^{2+}]} = \frac{1}{0.01} = 100 \]

4. **Substitute Values into the Nernst Equation**:
Now we can substitute the known values into the Nernst equation. First, we need to calculate \(\frac{RT}{nF}\):
\[ \frac{RT}{nF} = \frac{(8.314 \, J/(mol \cdot K))(298 \, K)}{(2)(96485 \, C/mol)} \approx 0.01285 \, V \]

Now substituting into the Nernst equation:
\[ E = -2.36 \, V - 0.01285 \, V \cdot \ln(100) \]

5. **Calculate \(\ln(100)\)**:
\[ \ln(100) = 4.605 \]

6. **Final Calculation**:
Now substitute \(\ln(100)\) back into the equation:
\[ E = -2.36 \, V - 0.01285 \, V \cdot 4.605 \]
\[ E = -2.36 \, V - 0.0592 \, V \]
\[ E \approx -2.419 \, V \]

### Final Answer:
The electrode potential of the Mg^(2+)//Mg electrode when dipped in a 0.01 M solution of Mg^(2+) is approximately:
\[ E \approx -2.419 \, V \]