Question

The potential for the given half cell at 298 K is\[(-) \ldots \times 10^{-2} \, \text{V}.\]\[2\text{H}^+ (\text{aq}) + 2e^- \rightarrow \text{H}_2 (\text{g})\]\[[\text{H}^+] = 1M, \, P_{\text{H}_2} = 2 \, \text{atm}\](Given: \( 2.303 \frac{RT}{F} = 0.06 \, \text{V}, \log 2 = 0.3 \))

Answer 1

Correct Answer: 1

The potential is given by the Nernst equation: \[ E = E^\circ - \frac{0.06}{2} \log \left( \frac{P_{H_2}}{[H^+]^2} \right) \] Substituting the values: \[ E = 0 - \frac{0.06}{2} \log \left( \frac{2}{1^2} \right) \] \[ E = -0.03 \times 0.3 = -0.9 \times 10^{-2} \, \text{V} \]

Answer 2

This problem requires the calculation of the electrode potential for a hydrogen half-cell under non-standard conditions using the Nernst equation.

Concept Used:

The Nernst equation is used to determine the cell potential under non-standard conditions. For a general reduction half-reaction:

\[ \text{Ox} + ne^- \rightarrow \text{Red} \]

The Nernst equation is given by:

\[ E = E^\circ - \frac{2.303RT}{nF} \log_{10} Q \]

Where:

• \(E\) is the reduction potential under non-standard conditions.
• \(E^\circ\) is the standard reduction potential.
• \(R\) is the universal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the number of electrons transferred in the reaction.
• \(F\) is the Faraday constant.
• \(Q\) is the reaction quotient.

For the standard hydrogen electrode (SHE), the standard reduction potential \(E^\circ\) is defined as 0 V.

Step-by-Step Solution:

Step 1: Identify the given half-reaction and parameters.

The reduction half-reaction is:

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

From this reaction, the number of electrons transferred is \(n=2\).

The given conditions are:

• Concentration of \( \text{H}^+ \), \([\text{H}^+] = 1 \, \text{M}\)
• Partial pressure of \( \text{H}_2 \), \(P_{\text{H}_2} = 2 \, \text{atm}\)
• Standard reduction potential for the hydrogen electrode, \(E^\circ = 0 \, \text{V}\)
• The value of \( \frac{2.303RT}{F} \) is given as \(0.06 \, \text{V}\)
• The value of \( \log 2 \) is given as \(0.3\)

Step 2: Determine the reaction quotient (\(Q\)).

The expression for the reaction quotient for this half-cell is:

\[ Q = \frac{\text{Products}}{\text{Reactants}} = \frac{P_{\text{H}_2}}{[\text{H}^+]^2} \]

Substituting the given values:

\[ Q = \frac{2}{(1)^2} = 2 \]

Step 3: Apply the Nernst equation.

The Nernst equation for this half-cell is:

\[ E = E^\circ - \frac{0.06}{n} \log_{10} Q \]

Now, substitute the known values into the equation:

\[ E = 0 - \frac{0.06}{2} \log_{10} (2) \]

Step 4: Calculate the potential \(E\).

Simplify the expression:

\[ E = -0.03 \times \log_{10} (2) \]

Using the given value \( \log 2 = 0.3 \):

\[ E = -0.03 \times 0.3 \] \[ E = -0.009 \, \text{V} \]

Final Computation & Result:

The problem asks for the answer in the format \( (-) \ldots \times 10^{-2} \, \text{V} \). We need to express our calculated potential in this form.

\[ E = -0.009 \, \text{V} = -0.9 \times 10^{-2} \, \text{V} \]

The value to be filled in the blank is 0.9.

The potential for the given half cell is \( (-) 0.9 \times 10^{-2} \, \text{V} \).