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 solve the problem of calculating the equilibrium constant for the reaction:

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

we will follow these steps:

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

The given half-reaction and its standard potential is:

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

For the overall reaction, we need to reverse this half-reaction to represent oxidation:

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

The reduction half-reaction for hydrogen is:

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

### Step 2: Calculate the standard cell potential (E°cell)

The standard cell potential (E°cell) can be calculated by adding the oxidation potential of cobalt and the reduction potential of hydrogen:

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

\[ E^\circ_{\text{cell}} = 0.277 \, \text{V} + 0.00 \, \text{V} = 0.277 \, \text{V} \]

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

The Nernst equation at equilibrium is given by:

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

Where:
- \(E^\circ_{\text{cell}} = 0.277 \, \text{V}\)
- \(n\) is the number of moles of electrons transferred in the balanced equation. Here, \(n = 2\) (from Co to Co²⁺ and from 2H⁺ to H₂).

Substituting the values into the Nernst equation:

\[ 0.277 = \frac{0.0591}{2} \log K \]

### Step 4: Solve for log K

Rearranging the equation to solve for \(\log K\):

\[ \log K = \frac{2 \times 0.277}{0.0591} \]

Calculating the right side:

\[ \log K = \frac{0.554}{0.0591} \approx 9.37 \]

### Step 5: Calculate K

To find \(K\), we take the antilog:

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

### Final Answer

Thus, the equilibrium constant \(K\) for the reaction is:

\[ \boxed{2.3 \times 10^9} \]

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