A solution containing `H^(+)` and `D^(+)` ions is in equilibrium with a mixture of `H_(2)` and `D_(2)` gases at `25^(@)`C. If the partial pressures of both gases are 1.0 atm, find the ratio of [`D^(+)`]`/[H^(+)`] :
(Given : `E_(D^(+)//D_(2))` =-0.003V]

Options

A. 1.23
B. 1.12
C. 0.11
D. 1

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the Reaction
The equilibrium reaction involving the ions and gases can be written as:
\[ \text{H}^+ + \text{D}_2 \rightleftharpoons \text{D}^+ + \text{H}_2 \]

### Step 2: Identify Standard Electrode Potentials
We are given the standard electrode potential for the reaction:
\[ E^\circ_{\text{D}^+/\text{D}_2} = -0.003 \, \text{V} \]
Since we are interested in the reverse reaction (from D2 to D+), we can write:
\[ E^\circ_{\text{D}_2/\text{D}^+} = +0.003 \, \text{V} \]

For hydrogen, the standard electrode potential is:
\[ E^\circ_{\text{H}^+/\text{H}_2} = 0 \, \text{V} \]

### Step 3: Calculate the Standard Cell Potential
The standard cell potential \( E^\circ_{\text{cell}} \) can be calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}} \]
Here, the reduction potential for H+ to H2 is 0 V, and the oxidation potential for D2 to D+ is 0.003 V:
\[ E^\circ_{\text{cell}} = 0 - (-0.003) = 0.003 \, \text{V} \]

### Step 4: Use the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\text{D}^+]^2}{[\text{H}^+]^2} \cdot \frac{P_{\text{H}_2}}{P_{\text{D}_2}} \right) \]
Since the partial pressures of both gases are 1 atm, we can simplify:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \left( \frac{[\text{D}^+]^2}{[\text{H}^+]^2} \right) \]

### Step 5: Determine the Number of Electrons Transferred
In the balanced reaction, 2 electrons are transferred:
\[ n = 2 \]

### Step 6: Set Up the Equation
At equilibrium, \( E_{\text{cell}} = 0 \):
\[ 0 = 0.003 - \frac{0.059}{2} \log \left( \frac{[\text{D}^+]^2}{[\text{H}^+]^2} \right) \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ \frac{0.059}{2} \log \left( \frac{[\text{D}^+]^2}{[\text{H}^+]^2} \right) = 0.003 \]

### Step 8: Solve for the Logarithm
Multiplying both sides by \( \frac{2}{0.059} \):
\[ \log \left( \frac{[\text{D}^+]^2}{[\text{H}^+]^2} \right) = \frac{0.003 \cdot 2}{0.059} \]

### Step 9: Calculate the Logarithm Value
Calculating the right side:
\[ \log \left( \frac{[\text{D}^+]^2}{[\text{H}^+]^2} \right) = \frac{0.006}{0.059} \approx 0.1017 \]

### Step 10: Solve for the Concentration Ratio
Taking antilog:
\[ \frac{[\text{D}^+]}{[\text{H}^+]} = 10^{0.1017} \]

### Step 11: Final Calculation
Calculating the antilog gives:
\[ \frac{[\text{D}^+]}{[\text{H}^+]} \approx 1.12 \]

### Final Answer
The ratio of \([\text{D}^+]/[\text{H}^+]\) is approximately **1.12**.

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