`E^@` of `In^(3+), In^(+) and Cu^(2+)m Cu^(+) are -0.4 V and -0.42 V` respectively, Calculate the equilibrium constant for the reaction. `In^(2+) + Cu^(2+) to In^(3+) + Cu^(+) ` at `25^@C`.

Answer

To calculate the equilibrium constant \( K_c \) for the reaction:

\[ \text{In}^{2+} + \text{Cu}^{2+} \rightarrow \text{In}^{3+} + \text{Cu}^{+} \]

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

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

The half-reactions and their standard electrode potentials (\( E^\circ \)) are given as follows:

1. For In:
\[ \text{In}^{3+} + 3e^- \rightarrow \text{In} \quad E^\circ = -0.4 \, \text{V} \]
(This means that the reduction potential for In is -0.4 V)

2. For Cu:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E^\circ = -0.42 \, \text{V} \]
(This means that the reduction potential for Cu is -0.42 V)

### Step 2: Write the oxidation and reduction half-reactions

From the overall reaction, we can see that:
- In is being oxidized from \(\text{In}^{2+}\) to \(\text{In}^{3+}\).
- Cu is being reduced from \(\text{Cu}^{2+}\) to \(\text{Cu}^{+}\).

The half-reactions can be written as:
1. Oxidation (In):
\[ \text{In}^{2+} \rightarrow \text{In}^{3+} + e^- \]
(This half-reaction involves the loss of one electron)

2. Reduction (Cu):
\[ \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^{+} \]
(This half-reaction involves the gain of one electron)

### Step 3: Calculate the standard cell potential \( E^\circ_{\text{cell}} \)

The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Here, the cathode is the reduction half-reaction (Cu) and the anode is the oxidation half-reaction (In):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}^{+}} - E^\circ_{\text{In}^{3+}/\text{In}^{2+}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (-0.42 \, \text{V}) - (-0.4 \, \text{V}) = -0.42 + 0.4 = -0.02 \, \text{V} \]

### Step 4: Relate \( E^\circ_{\text{cell}} \) to the Gibbs free energy change \( \Delta G^\circ \)

The relationship between the standard cell potential and the Gibbs free energy change is given by:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where:
- \( n \) = number of moles of electrons transferred (which is 1 in this case)
- \( F \) = Faraday's constant \( \approx 96485 \, \text{C/mol} \)

Substituting the values:
\[ \Delta G^\circ = -1 \times 96485 \times (-0.02) = 1929.7 \, \text{J/mol} \]

### Step 5: Calculate the equilibrium constant \( K_c \)

Using the relationship:
\[ \Delta G^\circ = -RT \ln K_c \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \) (25°C)

Rearranging gives:
\[ \ln K_c = -\frac{\Delta G^\circ}{RT} \]
Substituting the values:
\[ \ln K_c = -\frac{1929.7}{8.314 \times 298} \approx -0.778 \]
Now, exponentiating both sides to find \( K_c \):
\[ K_c = e^{-0.778} \approx 0.458 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction at \( 25^\circ C \) is approximately:
\[ K_c \approx 0.458 \]