If the standed electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction
`Zn(s)+Cu^(2+)(aq) hArrZn^(2+)(aq)+Cu(s)`
at 300 K is approximately
`(R=8JK^(-1)mol^(-1),F=96000Cmol^(-1))`

Explanation

To find the equilibrium constant (K) for the reaction:

\[ \text{Zn(s)} + \text{Cu}^{2+}(aq) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{Cu(s)} \]

given that the standard electrode potential (E°) is 2 V at 300 K, we will use the following relationships:

1. **Gibbs Free Energy and Equilibrium Constant**:
\[ \Delta G = -RT \ln K \]

2. **Gibbs Free Energy and Electrode Potential**:
\[ \Delta G = -nFE^\circ \]

Where:
- \( R \) = 8 J/K·mol (gas constant)
- \( T \) = 300 K (temperature)
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = 96000 C/mol (Faraday's constant)
- \( E^\circ \) = 2 V (standard electrode potential)

### Step-by-Step Solution:

**Step 1: Determine the number of electrons transferred (n)**

In the given reaction:
- Zinc (Zn) loses 2 electrons to form Zn²⁺.
- Copper (Cu²⁺) gains 2 electrons to form Cu.

Thus, \( n = 2 \).

**Step 2: Use the second equation to find ΔG**

Substituting the values into the equation:
\[ \Delta G = -nFE^\circ \]
\[ \Delta G = -2 \times 96000 \, \text{C/mol} \times 2 \, \text{V} \]
\[ \Delta G = -384000 \, \text{J/mol} \]

**Step 3: Use the first equation to relate ΔG and K**

Now, we can substitute ΔG into the first equation:
\[ -384000 = -RT \ln K \]
\[ 384000 = (8 \, \text{J/K·mol}) \times (300 \, \text{K}) \ln K \]

**Step 4: Calculate RT**

Calculating \( RT \):
\[ RT = 8 \times 300 = 2400 \, \text{J/mol} \]

**Step 5: Solve for ln K**

Now substituting back:
\[ 384000 = 2400 \ln K \]
\[ \ln K = \frac{384000}{2400} \]
\[ \ln K = 160 \]

**Step 6: Calculate K**

To find K, we exponentiate both sides:
\[ K = e^{160} \]

### Final Answer:

The equilibrium constant \( K \) for the reaction at 300 K is approximately:
\[ K \approx e^{160} \]