Calculate equilibrium constant for the reaction :
`Mg(s) |Mg^(2+)(0.001 M) || Cu^(2+)(0.0001 M)|Cu(s)`
Given `E_((Mg^(2+)//Mg))^(@)=-2.37" V " , E_((Cu^(2+)//Cu))^(@)=0.34" V "`

Explanation

To calculate the equilibrium constant (K) for the given electrochemical cell reaction, we will follow these steps:

### Step 1: Identify the Anode and Cathode
The given cell representation is:
`Mg(s) | Mg^(2+)(0.001 M) || Cu^(2+)(0.0001 M) | Cu(s)`

From the standard reduction potentials provided:
- \( E^\circ (Mg^{2+}/Mg) = -2.37 \, V \)
- \( E^\circ (Cu^{2+}/Cu) = 0.34 \, V \)

The cathode is the electrode where reduction occurs, and the anode is where oxidation occurs. Since copper has a higher reduction potential, it will act as the cathode, and magnesium will act as the anode.

### Step 2: Calculate the Standard Cell Potential (E°cell)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Substituting the values:
\[ E^\circ_{cell} = 0.34 \, V - (-2.37 \, V) \]
\[ E^\circ_{cell} = 0.34 \, V + 2.37 \, V = 2.71 \, V \]

### Step 3: Relate Gibbs Free Energy to Cell Potential
The relationship between Gibbs free energy (ΔG), the number of moles of electrons transferred (n), Faraday's constant (F), and cell potential (E°cell) is given by:
\[ \Delta G = -nFE^\circ_{cell} \]

### Step 4: Calculate the Equilibrium Constant (K)
The relationship between Gibbs free energy and the equilibrium constant is given by:
\[ \Delta G = -RT \ln K \]

Setting the two expressions for ΔG equal gives:
\[ -nFE^\circ_{cell} = -RT \ln K \]

Rearranging this, we find:
\[ \ln K = \frac{nFE^\circ_{cell}}{RT} \]

### Step 5: Substitute Values
- n = 2 (for the transfer of 2 electrons)
- F = 96485 C/mol (Faraday's constant)
- \( E^\circ_{cell} = 2.71 \, V \)
- R = 8.314 J/(mol·K) (universal gas constant)
- T = 300 K (27°C + 273)

Substituting these values into the equation:
\[ \ln K = \frac{(2)(96485)(2.71)}{(8.314)(300)} \]

Calculating the right-hand side:
\[ \ln K = \frac{(2)(96485)(2.71)}{(8.314)(300)} \]
\[ \ln K = \frac{522,370.7}{2494.2} \approx 209.3 \]

### Step 6: Calculate K
To find K, we exponentiate both sides:
\[ K = e^{209.3} \]

Using a calculator:
\[ K \approx 10^{91.054} \]

Thus, the equilibrium constant K is:
\[ K \approx 1.12 \times 10^{91} \]

### Final Answer
The equilibrium constant for the reaction is:
\[ K \approx 1.12 \times 10^{91} \]

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