Using the standerd half-cell potential listed, calculate the equilibrium constant for the reaction :
`Co(s)+2H^(+)(aq)toCo^(2+)(aq)+H_(2)(g)" at 298 K"`
`Co^(2+)(aq)+2e^(-)toCo(s) E^(@)=-0.277 V`

Options

A. `2.3xx10^(9)`
B. `4.8xx10^(4)`
C. `4.8xx10^(7)`
D. `4.8xx10^(11)`

Answer

To calculate the equilibrium constant (K) for the reaction:

\[ \text{Co(s)} + 2\text{H}^+(aq) \rightarrow \text{Co}^{2+}(aq) + \text{H}_2(g) \]

we will use the standard half-cell potentials provided and the Nernst equation.

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

From the problem, we have the half-reaction for cobalt:

\[ \text{Co}^{2+}(aq) + 2e^- \rightarrow \text{Co(s)} \quad E^\circ = -0.277 \, \text{V} \]

The reduction half-reaction for hydrogen ions is:

\[ 2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) \quad E^\circ = 0.00 \, \text{V} \]

### Step 2: Determine the overall cell reaction and the standard cell potential

The overall cell reaction can be written as:

\[ \text{Co(s)} + 2\text{H}^+(aq) \rightarrow \text{Co}^{2+}(aq) + \text{H}_2(g) \]

To find the standard cell potential (\(E^\circ_{cell}\)), we need to consider the oxidation and reduction potentials:

- Oxidation: \(\text{Co(s)} \rightarrow \text{Co}^{2+}(aq) + 2e^-\) (reverse of the given half-reaction)
- Reduction: \(2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\)

The standard cell potential is calculated as follows:

\[ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} \]

Substituting the values:

\[ E^\circ_{cell} = 0.00 \, \text{V} - (-0.277 \, \text{V}) = 0.277 \, \text{V} \]

### Step 3: Use the Nernst equation to find the equilibrium constant

The Nernst equation at equilibrium states that:

\[ E^\circ_{cell} = \frac{0.0591}{n} \log K \]

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

Rearranging the equation to find \(K\):

\[ K = 10^{\frac{n \cdot E^\circ_{cell}}{0.0591}} \]

Substituting the values:

\[ K = 10^{\frac{2 \cdot 0.277}{0.0591}} \]

Calculating the exponent:

\[ \frac{2 \cdot 0.277}{0.0591} \approx 9.37 \]

Thus,

\[ K = 10^{9.37} \approx 2.3 \times 10^9 \]

### Final Answer

The equilibrium constant \(K\) for the reaction at 298 K is approximately:

\[ K \approx 2.3 \times 10^9 \]