For the half-cell given Pt(`H_(2)`, 1 atm ) |`H^+` pH = 2, the cell potential is :

Options

A. `0.0591 V`
B. `0.0295 V`
C. `0.1182 V`
D. `0.00V`

Answer

To calculate the cell potential for the half-cell given by Pt(`H2`, 1 atm) | `H^+` at pH = 2, we can follow these steps:

### Step 1: Determine the concentration of H⁺ ions
Given that pH = 2, we can find the concentration of H⁺ ions using the formula:

\[ \text{pH} = -\log[H^+] \]

Rearranging this gives:

\[ [H^+] = 10^{-\text{pH}} = 10^{-2} \, \text{M} \]

### Step 2: Identify the standard electrode potential (E°)
For the hydrogen electrode, the standard electrode potential (E°) is defined as:

\[ E^\circ = 0 \, \text{V} \]

### Step 3: Use the Nernst equation
The Nernst equation is given by:

\[ E = E^\circ - \frac{0.0591}{n} \log Q \]

Where:
- \(E\) = cell potential
- \(E^\circ\) = standard cell potential
- \(n\) = number of moles of electrons transferred in the half-reaction
- \(Q\) = reaction quotient

For the half-reaction of hydrogen:

\[ H_2 \rightarrow 2H^+ + 2e^- \]

Here, \(n = 2\).

### Step 4: Calculate the reaction quotient (Q)
The reaction quotient \(Q\) for the half-reaction can be expressed as:

\[ Q = \frac{[H^+]^2}{P_{H_2}} = \frac{(10^{-2})^2}{1} = 10^{-4} \]

### Step 5: Substitute values into the Nernst equation
Now we can substitute the values into the Nernst equation:

\[ E = 0 - \frac{0.0591}{2} \log(10^{-4}) \]

### Step 6: Simplify the equation
Calculating the logarithm:

\[ \log(10^{-4}) = -4 \]

Substituting this back into the equation gives:

\[ E = -\frac{0.0591}{2} \times (-4) \]

### Step 7: Calculate the final cell potential
Now we can calculate:

\[ E = \frac{0.0591 \times 4}{2} = 0.1182 \, \text{V} \]

Thus, the cell potential for the half-cell is:

\[ \boxed{0.1182 \, \text{V}} \]

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