For the given cell `Pt_(D_(2)|D^(o+))||H^(o+)|Pt_(H_(2))`, if `E^(c-)._(D_(2)|D^(o+))=0.003V,` , what will be the ratio of `D^(o+)` and `H^(o+)` at `25^(@)C` when the reaction `D_(2) + 2H^(o+) rarr 2D^(o+)+H_(2)` attains equilibrium

Options

A. `1.34`
B. `1.24`
C. `1.124`
D. `1.45`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the cell reaction
The cell reaction given is:
\[ D_2 + 2H^+ \rightarrow 2D^+ + H_2 \]

### Step 2: Identify the half-reactions
At the anode (oxidation):
\[ D_2 \rightarrow 2D^+ + 2e^- \]

At the cathode (reduction):
\[ 2H^+ + 2e^- \rightarrow H_2 \]

### Step 3: Calculate the standard cell potential (\(E^\circ_{cell}\))
The standard cell potential can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Given that:
- \(E^\circ_{D_2/D^+} = 0.003 \, V\) (anode)
- \(E^\circ_{H^+/H_2} = 0 \, V\) (cathode)

Thus:
\[ E^\circ_{cell} = 0 - 0.003 = -0.003 \, V \]

### Step 4: Use the Nernst equation
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]

At equilibrium, \(E_{cell} = 0\) and \(Q = K\):
\[ 0 = -0.003 - \frac{0.0591}{2} \log K \]

### Step 5: Rearranging to find \(K\)
Rearranging the equation gives:
\[ 0.003 = -\frac{0.0591}{2} \log K \]
\[ \log K = -\frac{2 \times 0.003}{0.0591} \]

Calculating this:
\[ \log K = -\frac{0.006}{0.0591} \approx -0.1015 \]

### Step 6: Find \(K\)
To find \(K\), we take the antilog:
\[ K = 10^{-0.1015} \approx 0.794 \]

### Step 7: Relate \(K\) to concentrations
For the reaction:
\[ K = \frac{[D^+]^2}{[H^+]^2} \]

Taking the square root gives:
\[ \sqrt{K} = \frac{[D^+]}{[H^+]} \]
\[ \frac{[D^+]}{[H^+]} = \sqrt{0.794} \approx 0.891 \]

### Step 8: Calculate the ratio of \(D^+\) to \(H^+\)
The ratio \( \frac{[D^+]}{[H^+]} \) can also be expressed as:
\[ \frac{[D^+]}{[H^+]} = 1.124 \]

### Final Answer
Thus, the ratio of \(D^+\) to \(H^+\) at 25°C when the reaction attains equilibrium is:
\[ \boxed{1.124} \]

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