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 \( E^0_{\text{cell}} \) for the given reaction:

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

with an equilibrium constant \( K = 10^6 \), we can use the Nernst equation in its simplified form for standard conditions:

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

### Step 1: Identify the number of electrons transferred (n)
In the given reaction, copper (\( \text{Cu}^{2+} \)) is reduced to copper metal (\( \text{Cu} \)), which involves the gain of 2 electrons:

\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

Tin (\( \text{Sn}^{2+} \)) is oxidized to \( \text{Sn}^{4+} \), which involves the loss of 2 electrons:

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

Thus, the total number of electrons transferred in the reaction is \( n = 2 \).

### Step 2: Substitute the values into the equation
Now we can substitute \( n = 2 \) and \( K = 10^6 \) into the equation:

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

### Step 3: Calculate the logarithm
Using the property of logarithms:

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

### Step 4: Substitute and calculate \( E^0_{\text{cell}} \)
Now substituting the value of \( \log(10^6) \):

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

Calculating this:

\[ E^0_{\text{cell}} = 0.0591 \times 3 = 0.1773 \, \text{V} \]

### Final Answer
Thus, the standard cell potential \( E^0_{\text{cell}} \) for the reaction is:

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