The electrode potential of electrode
`M(s) rightarrow M^(n+) (aq) (2M) + ne^(-)` at 298 K is E_(1). When temperature is doubled and concentration is made half, then the electrode potential becomes `E_(2)`. Which of the following represents the correct relationship between `E_(1)` and `E_(2)`?

Options

A. `E_(1)gtE_(2)`
B. `E_(1)ltE_(2)`
C. `E_(1)=E_(2)`
D. Can't be predicted.

Explanation

To solve the problem, we need to analyze the relationship between the electrode potentials \( E_1 \) and \( E_2 \) using the Nernst equation.

### Step-by-Step Solution:

1. **Understanding the Nernst Equation**:
The Nernst equation for the electrode potential is given by:
\[ E = E^0 - \frac{RT}{nF} \ln Q \]
where \( E \) is the electrode potential, \( E^0 \) is the standard electrode potential, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant, and \( Q \) is the reaction quotient.

2. **Calculating \( E_1 \)**:
For the initial conditions at 298 K and a concentration of 2 M, the Nernst equation becomes:
\[ E_1 = E^0 - \frac{RT}{nF} \ln(2) \]
Here, \( Q = [M^{n+}] = 2 \) M.

3. **Calculating \( E_2 \)**:
When the temperature is doubled (to 596 K) and the concentration is halved (to 1 M), the Nernst equation becomes:
\[ E_2 = E^0 - \frac{R(2T)}{nF} \ln(1) \]
Since \( \ln(1) = 0 \), this simplifies to:
\[ E_2 = E^0 \]

4. **Comparing \( E_1 \) and \( E_2 \)**:
Now we can compare the two equations:
- From \( E_1 = E^0 - \frac{RT}{nF} \ln(2) \)
- From \( E_2 = E^0 \)

Since \( \frac{RT}{nF} \ln(2) \) is a positive value (as \( \ln(2) > 0 \)), it follows that:
\[ E_1 < E_2 \]

5. **Final Relationship**:
Therefore, the correct relationship between \( E_1 \) and \( E_2 \) is:
\[ E_2 > E_1 \]