Find the equilibrium constant of the following reaction
`Cu^(2+) (aq) +Sn^(2+) (aq) hArr Cu(s) +Sn^(4+) (aq)`
at `25^(@)C, E_(Cu^(2+)//Cu)^(Theta)=0.34V, E_(Sn^(4+)//Sn^(2+))^(Theta)=0.155V`

Explanation

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

\[ \text{Cu}^{2+} (aq) + \text{Sn}^{2+} (aq) \rightleftharpoons \text{Cu} (s) + \text{Sn}^{4+} (aq) \]

at \( 25^\circ C \), we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials

1. The half-reaction for copper is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^\circ = 0.34 \, V) \]

2. The half-reaction for tin is:
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \quad (E^\circ = 0.155 \, V) \]

### Step 2: Determine the oxidation and reduction reactions

- In the given reaction, tin is oxidized (loses electrons) and copper is reduced (gains electrons).
- Therefore, the oxidation half-reaction is:
\[ \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2e^- \]
- The reduction half-reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

### Step 3: Calculate the standard cell potential (E°cell)

Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Here, the cathode is copper and the anode is tin:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Sn}} \]
\[ E^\circ_{\text{cell}} = 0.34 \, V - 0.155 \, V = 0.185 \, V \]

### Step 4: Use the Nernst equation to find Kc

At equilibrium, the cell potential (Ecell) is 0. Thus, we can use the Nernst equation in its logarithmic form:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_c \]

Where \( n \) is the number of moles of electrons transferred in the balanced equation. Here, \( n = 2 \).

Rearranging gives:
\[ \log K_c = \frac{n \cdot E^\circ_{\text{cell}}}{0.0591} \]
\[ \log K_c = \frac{2 \cdot 0.185}{0.0591} \]

### Step 5: Calculate log Kc and then Kc

Calculating:
\[ \log K_c = \frac{0.37}{0.0591} \approx 6.26 \]

Now, to find \( K_c \):
\[ K_c = 10^{6.26} \approx 1.819 \times 10^6 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction at \( 25^\circ C \) is approximately:
\[ K_c \approx 1.819 \times 10^6 \]

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