The equillibrium constant for thefollowing general reaction is `10^(30)`. Calculate `E^(@)` for the cell at 298 K.
`2X_(2)(s)+3Y^(2+)(aq)to2X_(2)^(3+)(aq)+3Y(s)`

Options

A. `+0.105V`
B. `+0.2955 V`
C. 0.0985 V
D. `-0.2955` V

Answer

To calculate the standard cell potential \( E^\circ \) for the given reaction at 298 K, we will follow these steps:

### Step 1: Write down the Nernst equation
The Nernst equation relates the standard cell potential to the equilibrium constant:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K \]
where \( K \) is the equilibrium constant and \( n \) is the number of moles of electrons transferred in the reaction.

### Step 2: Identify the equilibrium constant
From the problem, we know that the equilibrium constant \( K \) is given as:
\[ K = 10^{30} \]

### Step 3: Determine the number of electrons transferred (n)
In the reaction:
\[ 2X_2(s) + 3Y^{2+}(aq) \rightarrow 2X^{3+}(aq) + 3Y(s) \]
- For \( X_2 \), the oxidation state changes from 0 to +3. Each \( X \) atom undergoes a change of +3, and since there are 2 atoms, the total change is:
\[ \Delta \text{oxidation state for } X = 2 \times 3 = 6 \]
- For \( Y^{2+} \), the oxidation state changes from +2 to 0. Each \( Y \) atom undergoes a change of -2, and since there are 3 atoms, the total change is:
\[ \Delta \text{oxidation state for } Y = 3 \times 2 = 6 \]
Thus, the total number of electrons transferred \( n \) is:
\[ n = 6 \]

### Step 4: Substitute values into the Nernst equation
Now we can substitute \( K \) and \( n \) into the Nernst equation:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{6} \log(10^{30}) \]

### Step 5: Calculate \( \log(10^{30}) \)
Using the property of logarithms:
\[ \log(10^{30}) = 30 \]

### Step 6: Substitute and calculate \( E^\circ_{\text{cell}} \)
Now substituting this value back into the equation:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{6} \times 30 \]
Calculating this gives:
\[ E^\circ_{\text{cell}} = 0.0591 \times 5 = 0.2955 \, \text{V} \]

### Final Answer
Thus, the standard cell potential \( E^\circ \) for the cell at 298 K is:
\[ E^\circ_{\text{cell}} = +0.2955 \, \text{V} \]

---