Determine the value of `E^0` cell for the following reaction
`Cu^(+2)+Sn^(+2)toCu+Sn^(+4)`
Equilibrium constant is `10^6`
`Cu^(++)+Sn^(++)toCu+Sn^(+4)`

Options

A. 0.1773
B. 0.01773
C. 0.2153
D. 1.773

Answer

To determine the standard cell potential \( E^0 \) for the given reaction, we can use the relationship between the standard cell potential and the equilibrium constant. The equation we will use is:

\[ E^0 = \frac{0.0591}{n} \log K \]

where:
- \( E^0 \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred in the reaction,
- \( K \) is the equilibrium constant.

### Step 1: Identify the number of electrons transferred (n)
In the reaction:

\[ \text{Cu}^{2+} + \text{Sn}^{2+} \rightarrow \text{Cu} + \text{Sn}^{4+} \]

Copper is reduced from \( \text{Cu}^{2+} \) to \( \text{Cu} \), gaining 2 electrons, and tin is oxidized from \( \text{Sn}^{2+} \) to \( \text{Sn}^{4+} \), losing 2 electrons. Therefore, the total number of electrons transferred \( n \) is:

\[ n = 2 \]

### Step 2: Substitute the values into the equation
We are given that the equilibrium constant \( K = 10^6 \). Now we can substitute \( n \) and \( K \) into the equation:

\[ E^0 = \frac{0.0591}{2} \log(10^6) \]

### Step 3: Calculate the logarithm
The logarithm of \( 10^6 \) is:

\[ \log(10^6) = 6 \]

### Step 4: Substitute the logarithm back into the equation
Now substituting this value back into the equation gives:

\[ E^0 = \frac{0.0591}{2} \times 6 \]

### Step 5: Perform the multiplication
Calculating the right side:

\[ E^0 = \frac{0.0591 \times 6}{2} = \frac{0.3546}{2} = 0.1773 \]

### Conclusion
Thus, the standard cell potential \( E^0 \) for the reaction is:

\[ E^0 = 0.1773 \, \text{V} \]

### Final Answer
The value of \( E^0 \) cell is \( 0.1773 \, \text{V} \).

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