Calculate `K_c` for the reaction :
`NiO_2 + 2 Cl^- + 4H^+ hArr Cl_2 + Ni^(2) + 2 H_2O` at 298 K if `E_(cell)^(Theta)` is 0.320 V .

Explanation

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

\[ \text{NiO}_2 + 2 \text{Cl}^- + 4 \text{H}^+ \rightleftharpoons \text{Cl}_2 + \text{Ni}^{2+} + 2 \text{H}_2\text{O} \]

at 298 K, given that the standard cell potential \( E^\circ_{\text{cell}} \) is 0.320 V, we can follow these steps:

### Step 1: Identify the number of electrons transferred (N)
In the given reaction, we need to determine how many electrons are involved in the oxidation and reduction processes.

- **Reduction half-reaction**: Nickel goes from \( \text{Ni}^{4+} \) to \( \text{Ni}^{2+} \). This involves the gain of 2 electrons:
\[ \text{Ni}^{4+} + 2e^- \rightarrow \text{Ni}^{2+} \]

- **Oxidation half-reaction**: Chloride ions \( \text{Cl}^- \) are oxidized to chlorine gas \( \text{Cl}_2 \). Each \( \text{Cl}^- \) loses 1 electron, and since there are 2 chloride ions, this involves the loss of 2 electrons:
\[ 2 \text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \]

Thus, the total number of electrons transferred, \( N \), is 2.

### Step 2: Use the Nernst equation to relate \( E^\circ_{\text{cell}} \) and \( K_c \)
The relationship between the standard cell potential and the equilibrium constant is given by the equation:
\[ E^\circ_{\text{cell}} = \frac{0.0591}{N} \log K_c \]

### Step 3: Substitute the known values into the equation
We know:
- \( E^\circ_{\text{cell}} = 0.320 \, \text{V} \)
- \( N = 2 \)

Substituting these values into the equation:
\[ 0.320 = \frac{0.0591}{2} \log K_c \]

### Step 4: Solve for \( \log K_c \)
Rearranging the equation to isolate \( \log K_c \):
\[ \log K_c = \frac{0.320 \times 2}{0.0591} \]
\[ \log K_c = \frac{0.640}{0.0591} \approx 10.81 \]

### Step 5: Calculate \( K_c \)
To find \( K_c \), we take the antilogarithm:
\[ K_c = 10^{10.81} \approx 6.46 \times 10^{10} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction at 298 K is approximately:
\[ K_c \approx 6.46 \times 10^{10} \]

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