Calculate the equilibrium constant for the reaction at 298K:
`NiO_(2)+2Cl^(-)+4H^(+)toCl^(2)+Ni^(2+)+2H_(2)O` if `E_(cell)^(@)=0.320V`.

Explanation

To calculate the equilibrium constant (Kc) for the given reaction at 298 K, we can use the relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant, as well as the relationship between ΔG° and the standard cell potential (E°cell).

### Step-by-Step Solution:

1. **Write the relationship between ΔG° and Kc**:
\[ \Delta G° = -RT \ln K_c \]
where:
- R = universal gas constant = 8.314 J/(mol·K)
- T = temperature in Kelvin (298 K)

2. **Write the relationship between ΔG° and E°cell**:
\[ \Delta G° = -nFE°_{cell} \]
where:
- n = number of moles of electrons transferred in the balanced equation
- F = Faraday's constant = 96500 C/mol
- E°cell = standard cell potential (given as 0.320 V)

3. **Equate the two expressions for ΔG°**:
\[ -nFE°_{cell} = -RT \ln K_c \]
This simplifies to:
\[ nFE°_{cell} = RT \ln K_c \]

4. **Solve for ln Kc**:
\[ \ln K_c = \frac{nFE°_{cell}}{RT} \]

5. **Determine the number of electrons (n)**:
In the given reaction:
\[ \text{NiO}_2 + 2\text{Cl}^- + 4\text{H}^+ \rightarrow \text{Cl}_2 + \text{Ni}^{2+} + 2\text{H}_2\text{O} \]
The half-reaction for the reduction of Cl⁻ to Cl₂ involves the transfer of 2 electrons. Thus, n = 2.

6. **Substitute the values**:
\[ \ln K_c = \frac{(2)(96500 \, \text{C/mol})(0.320 \, \text{V})}{(8.314 \, \text{J/(mol·K)})(298 \, \text{K})} \]

7. **Calculate the right-hand side**:
\[ \ln K_c = \frac{(2)(96500)(0.320)}{(8.314)(298)} \]
\[ \ln K_c = \frac{61760}{2477.572} \approx 24.93 \]

8. **Exponentiate to find Kc**:
\[ K_c = e^{\ln K_c} = e^{24.93} \approx 6.747 \times 10^{10} \]

### Final Answer:
The equilibrium constant \( K_c \) for the reaction at 298 K is approximately \( 6.747 \times 10^{10} \).